\(\int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx\) [1501]

Optimal result

Integrand size = 21, antiderivative size = 202 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {(a+b) \left (8 a^2+37 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}-\frac {(a-b) \left (8 a^2-37 a b+35 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {b \left (24 a^2+35 b^2\right ) \sin (c+d x)}{8 d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d} \]

[Out]

-1/16*(a+b)*(8*a^2+37*a*b+35*b^2)*ln(1-sin(d*x+c))/d-1/16*(a-b)*(8*a^2-37*a*b+35*b^2)*ln(1+sin(d*x+c))/d-1/8*b
*(24*a^2+35*b^2)*sin(d*x+c)/d-3/2*a*b^2*sin(d*x+c)^2/d-1/3*b^3*sin(d*x+c)^3/d+1/4*sec(d*x+c)^4*(a+b*sin(d*x+c)
)^3/d-1/8*sec(d*x+c)^2*(a+b*sin(d*x+c))^2*(8*a+11*b*sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2800, 1659, 1643, 647, 31} \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {b \left (24 a^2+35 b^2\right ) \sin (c+d x)}{8 d}-\frac {(a+b) \left (8 a^2+37 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}-\frac {(a-b) \left (8 a^2-37 a b+35 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}-\frac {b^3 \sin ^3(c+d x)}{3 d} \]

[In]

Int[(a + b*Sin[c + d*x])^3*Tan[c + d*x]^5,x]

[Out]

-1/16*((a + b)*(8*a^2 + 37*a*b + 35*b^2)*Log[1 - Sin[c + d*x]])/d - ((a - b)*(8*a^2 - 37*a*b + 35*b^2)*Log[1 +
Sin[c + d*x]])/(16*d) - (b*(24*a^2 + 35*b^2)*Sin[c + d*x])/(8*d) - (3*a*b^2*Sin[c + d*x]^2)/(2*d) - (b^3*Sin[
c + d*x]^3)/(3*d) + (Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3)/(4*d) - (Sec[c + d*x]^2*(a + b*Sin[c + d*x])^2*(8*
a + 11*b*Sin[c + d*x]))/(8*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 1643

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1659

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, Simp[(d + e*x)^m*(a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c*(p + 1))), x] + Dist[1/(2*a*c*(p
+ 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*c*(p + 1)*(d + e*x)*Q - a*e*g*m + c*d*f*(2*p
+ 3) + c*e*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^5 (a+x)^3}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}+\frac {\text {Subst}\left (\int \frac {(a+x)^2 \left (-3 b^6-4 a b^4 x-4 b^4 x^2-4 a b^2 x^3-4 b^2 x^4\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b^2 d} \\ & = \frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}+\frac {\text {Subst}\left (\int \frac {(a+x) \left (21 a b^6+b^4 \left (8 a^2+27 b^2\right ) x+16 a b^4 x^2+8 b^4 x^3\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^4 d} \\ & = \frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}+\frac {\text {Subst}\left (\int \left (-24 a^2 b^4-35 b^6-24 a b^4 x-8 b^4 x^2+\frac {5 b^6 \left (9 a^2+7 b^2\right )+8 a b^4 \left (a^2+9 b^2\right ) x}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^4 d} \\ & = -\frac {b \left (24 a^2+35 b^2\right ) \sin (c+d x)}{8 d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}+\frac {\text {Subst}\left (\int \frac {5 b^6 \left (9 a^2+7 b^2\right )+8 a b^4 \left (a^2+9 b^2\right ) x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^4 d} \\ & = -\frac {b \left (24 a^2+35 b^2\right ) \sin (c+d x)}{8 d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}+\frac {\left ((a-b) \left (8 a^2-37 a b+35 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {\left ((a+b) \left (8 a^2+37 a b+35 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d} \\ & = -\frac {(a+b) \left (8 a^2+37 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}-\frac {(a-b) \left (8 a^2-37 a b+35 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {b \left (24 a^2+35 b^2\right ) \sin (c+d x)}{8 d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.99 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {3 (a+b) \left (8 a^2+37 a b+35 b^2\right ) \log (1-\sin (c+d x))+3 (a-b) \left (8 a^2-37 a b+35 b^2\right ) \log (1+\sin (c+d x))-\frac {3 (a+b)^3}{(-1+\sin (c+d x))^2}-\frac {3 (a+b)^2 (7 a+13 b)}{-1+\sin (c+d x)}+144 b \left (a^2+b^2\right ) \sin (c+d x)+72 a b^2 \sin ^2(c+d x)+16 b^3 \sin ^3(c+d x)-\frac {3 (a-b)^3}{(1+\sin (c+d x))^2}+\frac {3 (7 a-13 b) (a-b)^2}{1+\sin (c+d x)}}{48 d} \]

[In]

Integrate[(a + b*Sin[c + d*x])^3*Tan[c + d*x]^5,x]

[Out]

-1/48*(3*(a + b)*(8*a^2 + 37*a*b + 35*b^2)*Log[1 - Sin[c + d*x]] + 3*(a - b)*(8*a^2 - 37*a*b + 35*b^2)*Log[1 +
Sin[c + d*x]] - (3*(a + b)^3)/(-1 + Sin[c + d*x])^2 - (3*(a + b)^2*(7*a + 13*b))/(-1 + Sin[c + d*x]) + 144*b*
(a^2 + b^2)*Sin[c + d*x] + 72*a*b^2*Sin[c + d*x]^2 + 16*b^3*Sin[c + d*x]^3 - (3*(a - b)^3)/(1 + Sin[c + d*x])^
2 + (3*(7*a - 13*b)*(a - b)^2)/(1 + Sin[c + d*x]))/d

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.50

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{2} b \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(304\)
default	\(\frac {a^{3} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{2} b \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(304\)
parallelrisch	\(\frac {96 \left (a^{2}+9 b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-96 \left (a^{2}+\frac {37}{8} a b +\frac {35}{8} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-96 \left (a^{2}-\frac {37}{8} a b +\frac {35}{8} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-24 a^{3}-81 a \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (18 a^{3}+108 a \,b^{2}\right ) \cos \left (4 d x +4 c \right )+\left (-270 a^{2} b -189 b^{3}\right ) \sin \left (3 d x +3 c \right )+\left (-36 a^{2} b -35 b^{3}\right ) \sin \left (5 d x +5 c \right )+9 a \,b^{2} \cos \left (6 d x +6 c \right )+b^{3} \sin \left (7 d x +7 c \right )+\left (-90 a^{2} b -105 b^{3}\right ) \sin \left (d x +c \right )+6 a^{3}-36 a \,b^{2}}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(329\)
norman	\(\frac {\frac {\left (2 a^{3}+18 a \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 a^{3}+18 a \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (16 a^{3}+48 a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (16 a^{3}+48 a \,b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \left (33 a^{2}-17 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 a^{3}+18 a \,b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 a^{3}+18 a \,b^{2}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 b \left (9 a^{2}+7 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {5 b \left (9 a^{2}+7 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {47 b \left (9 a^{2}+7 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {47 b \left (9 a^{2}+7 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {5 b \left (9 a^{2}+7 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {5 b \left (9 a^{2}+7 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a \left (a^{2}+9 b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (8 a^{3}-45 a^{2} b +72 a \,b^{2}-35 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (8 a^{3}+45 a^{2} b +72 a \,b^{2}+35 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\)	\(499\)
risch	\(9 i a \,b^{2} x +\frac {3 i b \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (16 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+72 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+27 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+13 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+16 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+96 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+3 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+5 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+16 i a^{3} {\mathrm e}^{i \left (d x +c \right )}+72 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-5 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-27 a^{2} b -13 b^{3}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+i a^{3} x -\frac {13 i b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {18 i a \,b^{2} c}{d}+\frac {13 i b^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {3 a \,b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{3} c}{d}-\frac {3 i b \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 d}-\frac {i b^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {i b^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{3}}{d}+\frac {45 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{8 d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{d}+\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{8 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {45 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{8 d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{d}-\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{8 d}\)	\(610\)

[In]

int(sec(d*x+c)^5*sin(d*x+c)^5*(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^3*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c)))+3*a^2*b*(1/4*sin(d*x+c)^7/cos(d*x+c)^4-3/8*sin(d*x
+c)^7/cos(d*x+c)^2-3/8*sin(d*x+c)^5-5/8*sin(d*x+c)^3-15/8*sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c)))+3*a*b^2*(
1/4*sin(d*x+c)^8/cos(d*x+c)^4-1/2*sin(d*x+c)^8/cos(d*x+c)^2-1/2*sin(d*x+c)^6-3/4*sin(d*x+c)^4-3/2*sin(d*x+c)^2
-3*ln(cos(d*x+c)))+b^3*(1/4*sin(d*x+c)^9/cos(d*x+c)^4-5/8*sin(d*x+c)^9/cos(d*x+c)^2-5/8*sin(d*x+c)^7-7/8*sin(d
*x+c)^5-35/24*sin(d*x+c)^3-35/8*sin(d*x+c)+35/8*ln(sec(d*x+c)+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.18 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=\frac {72 \, a b^{2} \cos \left (d x + c\right )^{6} - 36 \, a b^{2} \cos \left (d x + c\right )^{4} - 3 \, {\left (8 \, a^{3} - 45 \, a^{2} b + 72 \, a b^{2} - 35 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, a^{3} + 45 \, a^{2} b + 72 \, a b^{2} + 35 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 12 \, a^{3} + 36 \, a b^{2} - 24 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, b^{3} \cos \left (d x + c\right )^{6} - 8 \, {\left (9 \, a^{2} b + 10 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 18 \, a^{2} b + 6 \, b^{3} - 3 \, {\left (27 \, a^{2} b + 13 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/48*(72*a*b^2*cos(d*x + c)^6 - 36*a*b^2*cos(d*x + c)^4 - 3*(8*a^3 - 45*a^2*b + 72*a*b^2 - 35*b^3)*cos(d*x + c
)^4*log(sin(d*x + c) + 1) - 3*(8*a^3 + 45*a^2*b + 72*a*b^2 + 35*b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 1
2*a^3 + 36*a*b^2 - 24*(2*a^3 + 9*a*b^2)*cos(d*x + c)^2 + 2*(8*b^3*cos(d*x + c)^6 - 8*(9*a^2*b + 10*b^3)*cos(d*
x + c)^4 + 18*a^2*b + 6*b^3 - 3*(27*a^2*b + 13*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**5*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.07 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {16 \, b^{3} \sin \left (d x + c\right )^{3} + 72 \, a b^{2} \sin \left (d x + c\right )^{2} + 3 \, {\left (8 \, a^{3} - 45 \, a^{2} b + 72 \, a b^{2} - 35 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (8 \, a^{3} + 45 \, a^{2} b + 72 \, a b^{2} + 35 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, {\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right ) - \frac {6 \, {\left ({\left (27 \, a^{2} b + 13 \, b^{3}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{3} - 30 \, a b^{2} + 4 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (21 \, a^{2} b + 11 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{48 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/48*(16*b^3*sin(d*x + c)^3 + 72*a*b^2*sin(d*x + c)^2 + 3*(8*a^3 - 45*a^2*b + 72*a*b^2 - 35*b^3)*log(sin(d*x
+ c) + 1) + 3*(8*a^3 + 45*a^2*b + 72*a*b^2 + 35*b^3)*log(sin(d*x + c) - 1) + 144*(a^2*b + b^3)*sin(d*x + c) -
6*((27*a^2*b + 13*b^3)*sin(d*x + c)^3 - 6*a^3 - 30*a*b^2 + 4*(2*a^3 + 9*a*b^2)*sin(d*x + c)^2 - (21*a^2*b + 11
*b^3)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.24 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {16 \, b^{3} \sin \left (d x + c\right )^{3} + 72 \, a b^{2} \sin \left (d x + c\right )^{2} + 144 \, a^{2} b \sin \left (d x + c\right ) + 144 \, b^{3} \sin \left (d x + c\right ) + 3 \, {\left (8 \, a^{3} - 45 \, a^{2} b + 72 \, a b^{2} - 35 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (8 \, a^{3} + 45 \, a^{2} b + 72 \, a b^{2} + 35 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (6 \, a^{3} \sin \left (d x + c\right )^{4} + 54 \, a b^{2} \sin \left (d x + c\right )^{4} + 27 \, a^{2} b \sin \left (d x + c\right )^{3} + 13 \, b^{3} \sin \left (d x + c\right )^{3} - 4 \, a^{3} \sin \left (d x + c\right )^{2} - 72 \, a b^{2} \sin \left (d x + c\right )^{2} - 21 \, a^{2} b \sin \left (d x + c\right ) - 11 \, b^{3} \sin \left (d x + c\right ) + 24 \, a b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{48 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/48*(16*b^3*sin(d*x + c)^3 + 72*a*b^2*sin(d*x + c)^2 + 144*a^2*b*sin(d*x + c) + 144*b^3*sin(d*x + c) + 3*(8*
a^3 - 45*a^2*b + 72*a*b^2 - 35*b^3)*log(abs(sin(d*x + c) + 1)) + 3*(8*a^3 + 45*a^2*b + 72*a*b^2 + 35*b^3)*log(
abs(sin(d*x + c) - 1)) - 6*(6*a^3*sin(d*x + c)^4 + 54*a*b^2*sin(d*x + c)^4 + 27*a^2*b*sin(d*x + c)^3 + 13*b^3*
sin(d*x + c)^3 - 4*a^3*sin(d*x + c)^2 - 72*a*b^2*sin(d*x + c)^2 - 21*a^2*b*sin(d*x + c) - 11*b^3*sin(d*x + c)
+ 24*a*b^2)/(sin(d*x + c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 12.02 (sec) , antiderivative size = 512, normalized size of antiderivative = 2.53 \[ \int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^3+9\,a\,b^2\right )}{d}-\frac {\left (-\frac {45\,a^2\,b}{4}-\frac {35\,b^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+\left (-2\,a^3-18\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\left (\frac {15\,a^2\,b}{2}+\frac {35\,b^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (2\,a^3+18\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {141\,a^2\,b}{4}+\frac {329\,b^3}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (16\,a^3+48\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (33\,a^2\,b-17\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (16\,a^3+48\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {141\,a^2\,b}{4}+\frac {329\,b^3}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (2\,a^3+18\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {15\,a^2\,b}{2}+\frac {35\,b^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,a^3-18\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-\frac {45\,a^2\,b}{4}-\frac {35\,b^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-b\right )\,\left (8\,a^2-37\,a\,b+35\,b^2\right )}{8\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a+b\right )\,\left (8\,a^2+37\,a\,b+35\,b^2\right )}{8\,d} \]

[In]

int((sin(c + d*x)^5*(a + b*sin(c + d*x))^3)/cos(c + d*x)^5,x)

[Out]

(log(tan(c/2 + (d*x)/2)^2 + 1)*(9*a*b^2 + a^3))/d - (tan(c/2 + (d*x)/2)^4*(18*a*b^2 + 2*a^3) - tan(c/2 + (d*x)
/2)^2*(18*a*b^2 + 2*a^3) - tan(c/2 + (d*x)/2)*((45*a^2*b)/4 + (35*b^3)/4) + tan(c/2 + (d*x)/2)^10*(18*a*b^2 +
2*a^3) - tan(c/2 + (d*x)/2)^12*(18*a*b^2 + 2*a^3) + tan(c/2 + (d*x)/2)^6*(48*a*b^2 + 16*a^3) + tan(c/2 + (d*x)
/2)^8*(48*a*b^2 + 16*a^3) + tan(c/2 + (d*x)/2)^7*(33*a^2*b - 17*b^3) + tan(c/2 + (d*x)/2)^3*((15*a^2*b)/2 + (3
5*b^3)/6) + tan(c/2 + (d*x)/2)^11*((15*a^2*b)/2 + (35*b^3)/6) - tan(c/2 + (d*x)/2)^13*((45*a^2*b)/4 + (35*b^3)
/4) + tan(c/2 + (d*x)/2)^5*((141*a^2*b)/4 + (329*b^3)/12) + tan(c/2 + (d*x)/2)^9*((141*a^2*b)/4 + (329*b^3)/12
))/(d*(tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 + 3*tan
(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 - tan(c/2 + (d*x)/2)^14 - 1)) - (log(tan(c/2 + (d*x)/2) + 1)*(a - b
)*(8*a^2 - 37*a*b + 35*b^2))/(8*d) - (log(tan(c/2 + (d*x)/2) - 1)*(a + b)*(37*a*b + 8*a^2 + 35*b^2))/(8*d)