\(\int \frac {\sin ^5(c+d x)}{a+b \tan (c+d x)} \, dx\) [51]

Optimal result

Integrand size = 21, antiderivative size = 274 \[ \int \frac {\sin ^5(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {a^5 b \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2} d}+\frac {a^3 b^2 \cos (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac {a b^2 \cos (c+d x)}{\left (a^2+b^2\right )^2 d}-\frac {a \cos (c+d x)}{\left (a^2+b^2\right ) d}-\frac {a b^2 \cos ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac {2 a \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}-\frac {a \cos ^5(c+d x)}{5 \left (a^2+b^2\right ) d}+\frac {a^4 b \sin (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac {a^2 b \sin ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac {b \sin ^5(c+d x)}{5 \left (a^2+b^2\right ) d} \]

[Out]

a^5*b*arctanh((b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^(1/2))/(a^2+b^2)^(7/2)/d+a^3*b^2*cos(d*x+c)/(a^2+b^2)^3/d+
a*b^2*cos(d*x+c)/(a^2+b^2)^2/d-a*cos(d*x+c)/(a^2+b^2)/d-1/3*a*b^2*cos(d*x+c)^3/(a^2+b^2)^2/d+2/3*a*cos(d*x+c)^
3/(a^2+b^2)/d-1/5*a*cos(d*x+c)^5/(a^2+b^2)/d+a^4*b*sin(d*x+c)/(a^2+b^2)^3/d+1/3*a^2*b*sin(d*x+c)^3/(a^2+b^2)^2
/d+1/5*b*sin(d*x+c)^5/(a^2+b^2)/d

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3599, 3188, 2644, 30, 2713, 3178, 3153, 212, 2718} \[ \int \frac {\sin ^5(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {b \sin ^5(c+d x)}{5 d \left (a^2+b^2\right )}+\frac {a^2 b \sin ^3(c+d x)}{3 d \left (a^2+b^2\right )^2}-\frac {a \cos ^5(c+d x)}{5 d \left (a^2+b^2\right )}+\frac {2 a \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )}-\frac {a b^2 \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )^2}-\frac {a \cos (c+d x)}{d \left (a^2+b^2\right )}+\frac {a b^2 \cos (c+d x)}{d \left (a^2+b^2\right )^2}+\frac {a^5 b \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{7/2}}+\frac {a^4 b \sin (c+d x)}{d \left (a^2+b^2\right )^3}+\frac {a^3 b^2 \cos (c+d x)}{d \left (a^2+b^2\right )^3} \]

[In]

Int[Sin[c + d*x]^5/(a + b*Tan[c + d*x]),x]

[Out]

(a^5*b*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(7/2)*d) + (a^3*b^2*Cos[c + d*
x])/((a^2 + b^2)^3*d) + (a*b^2*Cos[c + d*x])/((a^2 + b^2)^2*d) - (a*Cos[c + d*x])/((a^2 + b^2)*d) - (a*b^2*Cos
[c + d*x]^3)/(3*(a^2 + b^2)^2*d) + (2*a*Cos[c + d*x]^3)/(3*(a^2 + b^2)*d) - (a*Cos[c + d*x]^5)/(5*(a^2 + b^2)*
d) + (a^4*b*Sin[c + d*x])/((a^2 + b^2)^3*d) + (a^2*b*Sin[c + d*x]^3)/(3*(a^2 + b^2)^2*d) + (b*Sin[c + d*x]^5)/
(5*(a^2 + b^2)*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3153

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Dist[-d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 3178

Int[sin[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[(-a)*(Sin[c + d*x]^(m - 1)/(d*(a^2 + b^2)*(m - 1))), x] + (Dist[a^2/(a^2 + b^2), Int[Sin[c + d*x]^(m - 2
)/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x] + Dist[b/(a^2 + b^2), Int[Sin[c + d*x]^(m - 1), x], x]) /; FreeQ[{
a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[m, 1]

Rule 3188

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Dist[
a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Dist[a*(b/(a^2 + b^2)), Int[Cos[c + d*x]^(m -
1)*(Sin[c + d*x]^(n - 1)/(a*Cos[c + d*x] + b*Sin[c + d*x])), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3599

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Sin[e + f*x]^
m*((a*Cos[e + f*x] + b*Sin[e + f*x])^n/Cos[e + f*x]^n), x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos (c+d x) \sin ^5(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx \\ & = \frac {a \int \sin ^5(c+d x) \, dx}{a^2+b^2}+\frac {b \int \cos (c+d x) \sin ^4(c+d x) \, dx}{a^2+b^2}-\frac {(a b) \int \frac {\sin ^4(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{a^2+b^2} \\ & = \frac {a^2 b \sin ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}-\frac {\left (a^3 b\right ) \int \frac {\sin ^2(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{\left (a^2+b^2\right )^2}-\frac {\left (a b^2\right ) \int \sin ^3(c+d x) \, dx}{\left (a^2+b^2\right )^2}-\frac {a \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{\left (a^2+b^2\right ) d}+\frac {b \text {Subst}\left (\int x^4 \, dx,x,\sin (c+d x)\right )}{\left (a^2+b^2\right ) d} \\ & = -\frac {a \cos (c+d x)}{\left (a^2+b^2\right ) d}+\frac {2 a \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}-\frac {a \cos ^5(c+d x)}{5 \left (a^2+b^2\right ) d}+\frac {a^4 b \sin (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac {a^2 b \sin ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac {b \sin ^5(c+d x)}{5 \left (a^2+b^2\right ) d}-\frac {\left (a^5 b\right ) \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{\left (a^2+b^2\right )^3}-\frac {\left (a^3 b^2\right ) \int \sin (c+d x) \, dx}{\left (a^2+b^2\right )^3}+\frac {\left (a b^2\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{\left (a^2+b^2\right )^2 d} \\ & = \frac {a^3 b^2 \cos (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac {a b^2 \cos (c+d x)}{\left (a^2+b^2\right )^2 d}-\frac {a \cos (c+d x)}{\left (a^2+b^2\right ) d}-\frac {a b^2 \cos ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac {2 a \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}-\frac {a \cos ^5(c+d x)}{5 \left (a^2+b^2\right ) d}+\frac {a^4 b \sin (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac {a^2 b \sin ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac {b \sin ^5(c+d x)}{5 \left (a^2+b^2\right ) d}+\frac {\left (a^5 b\right ) \text {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{\left (a^2+b^2\right )^3 d} \\ & = \frac {a^5 b \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2} d}+\frac {a^3 b^2 \cos (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac {a b^2 \cos (c+d x)}{\left (a^2+b^2\right )^2 d}-\frac {a \cos (c+d x)}{\left (a^2+b^2\right ) d}-\frac {a b^2 \cos ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac {2 a \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}-\frac {a \cos ^5(c+d x)}{5 \left (a^2+b^2\right ) d}+\frac {a^4 b \sin (c+d x)}{\left (a^2+b^2\right )^3 d}+\frac {a^2 b \sin ^3(c+d x)}{3 \left (a^2+b^2\right )^2 d}+\frac {b \sin ^5(c+d x)}{5 \left (a^2+b^2\right ) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.80 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.05 \[ \int \frac {\sin ^5(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-480 a^5 b \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )+\sqrt {a^2+b^2} \left (-30 a \left (5 a^4-4 a^2 b^2-b^4\right ) \cos (c+d x)+5 a \left (5 a^4+6 a^2 b^2+b^4\right ) \cos (3 (c+d x))-3 a^5 \cos (5 (c+d x))-6 a^3 b^2 \cos (5 (c+d x))-3 a b^4 \cos (5 (c+d x))+330 a^4 b \sin (c+d x)+120 a^2 b^3 \sin (c+d x)+30 b^5 \sin (c+d x)-35 a^4 b \sin (3 (c+d x))-50 a^2 b^3 \sin (3 (c+d x))-15 b^5 \sin (3 (c+d x))+3 a^4 b \sin (5 (c+d x))+6 a^2 b^3 \sin (5 (c+d x))+3 b^5 \sin (5 (c+d x))\right )}{240 \left (a^2+b^2\right )^{7/2} d} \]

[In]

Integrate[Sin[c + d*x]^5/(a + b*Tan[c + d*x]),x]

[Out]

(-480*a^5*b*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] + Sqrt[a^2 + b^2]*(-30*a*(5*a^4 - 4*a^2*b^2 - b
^4)*Cos[c + d*x] + 5*a*(5*a^4 + 6*a^2*b^2 + b^4)*Cos[3*(c + d*x)] - 3*a^5*Cos[5*(c + d*x)] - 6*a^3*b^2*Cos[5*(
c + d*x)] - 3*a*b^4*Cos[5*(c + d*x)] + 330*a^4*b*Sin[c + d*x] + 120*a^2*b^3*Sin[c + d*x] + 30*b^5*Sin[c + d*x]
- 35*a^4*b*Sin[3*(c + d*x)] - 50*a^2*b^3*Sin[3*(c + d*x)] - 15*b^5*Sin[3*(c + d*x)] + 3*a^4*b*Sin[5*(c + d*x)
] + 6*a^2*b^3*Sin[5*(c + d*x)] + 3*b^5*Sin[5*(c + d*x)]))/(240*(a^2 + b^2)^(7/2)*d)

Maple [A] (verified)

Time = 9.86 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(\frac {-\frac {64 a^{5} b \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (32 a^{6}+96 a^{4} b^{2}+96 a^{2} b^{4}+32 b^{6}\right ) \sqrt {a^{2}+b^{2}}}+\frac {2 a^{4} b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a^{3} b^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\frac {16}{3} a^{4} b +\frac {4}{3} a^{2} b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (6 a^{3} b^{2}+2 a \,b^{4}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\frac {178}{15} a^{4} b +\frac {136}{15} a^{2} b^{3}+\frac {16}{5} b^{5}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (-\frac {16}{3} a^{5}-\frac {2}{3} a \,b^{4}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\frac {16}{3} a^{4} b +\frac {4}{3} a^{2} b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (2 a^{3} b^{2}-\frac {8}{3} a^{5}+\frac {2}{3} a \,b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a^{4} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {16 a^{5}}{15}+\frac {6 a^{3} b^{2}}{5}+\frac {4 a \,b^{4}}{15}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}}{d}\)	\(358\)
default	\(\frac {-\frac {64 a^{5} b \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (32 a^{6}+96 a^{4} b^{2}+96 a^{2} b^{4}+32 b^{6}\right ) \sqrt {a^{2}+b^{2}}}+\frac {2 a^{4} b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a^{3} b^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\frac {16}{3} a^{4} b +\frac {4}{3} a^{2} b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (6 a^{3} b^{2}+2 a \,b^{4}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\frac {178}{15} a^{4} b +\frac {136}{15} a^{2} b^{3}+\frac {16}{5} b^{5}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (-\frac {16}{3} a^{5}-\frac {2}{3} a \,b^{4}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\frac {16}{3} a^{4} b +\frac {4}{3} a^{2} b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (2 a^{3} b^{2}-\frac {8}{3} a^{5}+\frac {2}{3} a \,b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a^{4} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {16 a^{5}}{15}+\frac {6 a^{3} b^{2}}{5}+\frac {4 a \,b^{4}}{15}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}}{d}\)	\(358\)
risch	\(-\frac {i {\mathrm e}^{3 i \left (d x +c \right )} b}{32 \left (-2 i a b +a^{2}-b^{2}\right ) d}+\frac {5 \,{\mathrm e}^{3 i \left (d x +c \right )} a}{96 \left (-2 i a b +a^{2}-b^{2}\right ) d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} a b}{4 \left (-3 i b \,a^{2}+i b^{3}+a^{3}-3 a \,b^{2}\right ) d}-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{16 \left (-3 i b \,a^{2}+i b^{3}+a^{3}-3 a \,b^{2}\right ) d}+\frac {{\mathrm e}^{i \left (d x +c \right )} b^{2}}{16 \left (-3 i b \,a^{2}+i b^{3}+a^{3}-3 a \,b^{2}\right ) d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} a b}{4 \left (i b +a \right )^{3} d}-\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{16 \left (i b +a \right )^{3} d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{16 \left (i b +a \right )^{3} d}-\frac {5 \,{\mathrm e}^{-3 i \left (d x +c \right )} a}{96 \left (-i a +b \right )^{2} d}-\frac {i {\mathrm e}^{-3 i \left (d x +c \right )} b}{32 \left (-i a +b \right )^{2} d}+\frac {i b \,a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{3} d}-\frac {i b \,a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{3} d}+\frac {a \cos \left (5 d x +5 c \right )}{80 d \left (-a^{2}-b^{2}\right )}-\frac {b \sin \left (5 d x +5 c \right )}{80 d \left (-a^{2}-b^{2}\right )}\)	\(496\)

[In]

int(sin(d*x+c)^5/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-64*a^5*b/(32*a^6+96*a^4*b^2+96*a^2*b^4+32*b^6)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/
(a^2+b^2)^(1/2))+2/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)*(a^4*b*tan(1/2*d*x+1/2*c)^9+a^3*b^2*tan(1/2*d*x+1/2*c)^8+(16/
3*a^4*b+4/3*a^2*b^3)*tan(1/2*d*x+1/2*c)^7+(6*a^3*b^2+2*a*b^4)*tan(1/2*d*x+1/2*c)^6+(178/15*a^4*b+136/15*a^2*b^
3+16/5*b^5)*tan(1/2*d*x+1/2*c)^5+(-16/3*a^5-2/3*a*b^4)*tan(1/2*d*x+1/2*c)^4+(16/3*a^4*b+4/3*a^2*b^3)*tan(1/2*d
*x+1/2*c)^3+(2*a^3*b^2-8/3*a^5+2/3*a*b^4)*tan(1/2*d*x+1/2*c)^2+a^4*b*tan(1/2*d*x+1/2*c)-8/15*a^5+3/5*a^3*b^2+2
/15*a*b^4)/(1+tan(1/2*d*x+1/2*c)^2)^5)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.35 \[ \int \frac {\sin ^5(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {15 \, \sqrt {a^{2} + b^{2}} a^{5} b \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 6 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{5} + 10 \, {\left (2 \, a^{7} + 5 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 30 \, {\left (a^{7} + a^{5} b^{2}\right )} \cos \left (d x + c\right ) + 2 \, {\left (23 \, a^{6} b + 34 \, a^{4} b^{3} + 14 \, a^{2} b^{5} + 3 \, b^{7} + 3 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4} - {\left (11 \, a^{6} b + 28 \, a^{4} b^{3} + 23 \, a^{2} b^{5} + 6 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} d} \]

[In]

integrate(sin(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/30*(15*sqrt(a^2 + b^2)*a^5*b*log((2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2
- 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x
+ c)^2 + b^2)) - 6*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(d*x + c)^5 + 10*(2*a^7 + 5*a^5*b^2 + 4*a^3*b^4 +
a*b^6)*cos(d*x + c)^3 - 30*(a^7 + a^5*b^2)*cos(d*x + c) + 2*(23*a^6*b + 34*a^4*b^3 + 14*a^2*b^5 + 3*b^7 + 3*(
a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(d*x + c)^4 - (11*a^6*b + 28*a^4*b^3 + 23*a^2*b^5 + 6*b^7)*cos(d*x + c
)^2)*sin(d*x + c))/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^5(c+d x)}{a+b \tan (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(sin(d*x+c)**5/(a+b*tan(d*x+c)),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 658 vs. \(2 (260) = 520\).

Time = 0.31 (sec) , antiderivative size = 658, normalized size of antiderivative = 2.40 \[ \int \frac {\sin ^5(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {15 \, a^{5} b \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (8 \, a^{5} - 9 \, a^{3} b^{2} - 2 \, a b^{4} - \frac {15 \, a^{4} b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {15 \, a^{3} b^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {15 \, a^{4} b \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {10 \, {\left (4 \, a^{5} - 3 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {20 \, {\left (4 \, a^{4} b + a^{2} b^{3}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {10 \, {\left (8 \, a^{5} + a b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {2 \, {\left (89 \, a^{4} b + 68 \, a^{2} b^{3} + 24 \, b^{5}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {30 \, {\left (3 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {20 \, {\left (4 \, a^{4} b + a^{2} b^{3}\right )} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6} + \frac {5 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}}}{15 \, d} \]

[In]

integrate(sin(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/15*(15*a^5*b*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c)/(cos(d*x + c)
+ 1) - sqrt(a^2 + b^2)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sqrt(a^2 + b^2)) - 2*(8*a^5 - 9*a^3*b^2 - 2*a*b
^4 - 15*a^4*b*sin(d*x + c)/(cos(d*x + c) + 1) - 15*a^3*b^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 15*a^4*b*sin(
d*x + c)^9/(cos(d*x + c) + 1)^9 + 10*(4*a^5 - 3*a^3*b^2 - a*b^4)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 20*(4*a
^4*b + a^2*b^3)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 10*(8*a^5 + a*b^4)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 -
2*(89*a^4*b + 68*a^2*b^3 + 24*b^5)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 30*(3*a^3*b^2 + a*b^4)*sin(d*x + c)^
6/(cos(d*x + c) + 1)^6 - 20*(4*a^4*b + a^2*b^3)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^6 + 3*a^4*b^2 + 3*a^2*
b^4 + b^6 + 5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*(a^6 + 3*a^4*b^2 +
3*a^2*b^4 + b^6)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sin(d*x + c)^6/(
cos(d*x + c) + 1)^6 + 5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + (a^6 + 3*a^4
*b^2 + 3*a^2*b^4 + b^6)*sin(d*x + c)^10/(cos(d*x + c) + 1)^10))/d

Giac [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 464, normalized size of antiderivative = 1.69 \[ \int \frac {\sin ^5(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {15 \, a^{5} b \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (15 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 15 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 80 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 20 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 90 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 30 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 178 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 136 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 48 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 80 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 30 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a^{5} + 9 \, a^{3} b^{2} + 2 \, a b^{4}\right )}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{15 \, d} \]

[In]

integrate(sin(d*x+c)^5/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

1/15*(15*a^5*b*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b
+ 2*sqrt(a^2 + b^2)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sqrt(a^2 + b^2)) + 2*(15*a^4*b*tan(1/2*d*x + 1/2*c)
^9 + 15*a^3*b^2*tan(1/2*d*x + 1/2*c)^8 + 80*a^4*b*tan(1/2*d*x + 1/2*c)^7 + 20*a^2*b^3*tan(1/2*d*x + 1/2*c)^7 +
90*a^3*b^2*tan(1/2*d*x + 1/2*c)^6 + 30*a*b^4*tan(1/2*d*x + 1/2*c)^6 + 178*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 136*
a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 48*b^5*tan(1/2*d*x + 1/2*c)^5 - 80*a^5*tan(1/2*d*x + 1/2*c)^4 - 10*a*b^4*tan(
1/2*d*x + 1/2*c)^4 + 80*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 20*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 40*a^5*tan(1/2*d*x
+ 1/2*c)^2 + 30*a^3*b^2*tan(1/2*d*x + 1/2*c)^2 + 10*a*b^4*tan(1/2*d*x + 1/2*c)^2 + 15*a^4*b*tan(1/2*d*x + 1/2*
c) - 8*a^5 + 9*a^3*b^2 + 2*a*b^4)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(tan(1/2*d*x + 1/2*c)^2 + 1)^5))/d

Mupad [B] (verification not implemented)

Time = 8.30 (sec) , antiderivative size = 683, normalized size of antiderivative = 2.49 \[ \int \frac {\sin ^5(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {2\,\left (-8\,a^5+9\,a^3\,b^2+2\,a\,b^4\right )}{15\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a^4\,b+a^2\,b^3\right )}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (3\,a^3\,b^2+a\,b^4\right )}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-4\,a^5+3\,a^3\,b^2+a\,b^4\right )}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (89\,a^4\,b+68\,a^2\,b^3+24\,b^5\right )}{15\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,a^5+a\,b^4\right )}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {2\,a^3\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+\frac {8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (4\,a^4+a^2\,b^2\right )}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {2\,a^4\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+\frac {2\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a^5\,b\,\mathrm {atanh}\left (\frac {2\,a^6\,b+2\,b^7+6\,a^2\,b^5+6\,a^4\,b^3-2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}{2\,{\left (a^2+b^2\right )}^{7/2}}\right )}{d\,{\left (a^2+b^2\right )}^{7/2}} \]

[In]

int(sin(c + d*x)^5/(a + b*tan(c + d*x)),x)

[Out]

((2*(2*a*b^4 - 8*a^5 + 9*a^3*b^2))/(15*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (8*tan(c/2 + (d*x)/2)^3*(4*a^4*b
+ a^2*b^3))/(3*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (4*tan(c/2 + (d*x)/2)^6*(a*b^4 + 3*a^3*b^2))/(a^6 + b^6
+ 3*a^2*b^4 + 3*a^4*b^2) + (4*tan(c/2 + (d*x)/2)^2*(a*b^4 - 4*a^5 + 3*a^3*b^2))/(3*(a^6 + b^6 + 3*a^2*b^4 + 3
*a^4*b^2)) + (4*tan(c/2 + (d*x)/2)^5*(89*a^4*b + 24*b^5 + 68*a^2*b^3))/(15*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)
) - (4*tan(c/2 + (d*x)/2)^4*(a*b^4 + 8*a^5))/(3*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (2*a^3*b^2*tan(c/2 + (d
*x)/2)^8)/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (8*b*tan(c/2 + (d*x)/2)^7*(4*a^4 + a^2*b^2))/(3*(a^6 + b^6 + 3
*a^2*b^4 + 3*a^4*b^2)) + (2*a^4*b*tan(c/2 + (d*x)/2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (2*a^4*b*tan(c/2 +
(d*x)/2)^9)/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*ta
n(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (2*a^5*b*atanh((2*a^6*b + 2*b^7 +
6*a^2*b^5 + 6*a^4*b^3 - 2*a*tan(c/2 + (d*x)/2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))/(2*(a^2 + b^2)^(7/2))))/(d
*(a^2 + b^2)^(7/2))