3.53 \(\int \frac {\sin ^3(c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=168 \[ \frac {b \sin ^3(c+d x)}{3 d \left (a^2+b^2\right )}+\frac {a^2 b \sin (c+d x)}{d \left (a^2+b^2\right )^2}+\frac {a \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )}-\frac {a \cos (c+d x)}{d \left (a^2+b^2\right )}+\frac {a b^2 \cos (c+d x)}{d \left (a^2+b^2\right )^2}+\frac {a^3 b \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3518, 3109, 2564, 30, 2633, 3099, 3074, 206, 2638} \[ \frac {b \sin ^3(c+d x)}{3 d \left (a^2+b^2\right )}+\frac {a^2 b \sin (c+d x)}{d \left (a^2+b^2\right )^2}+\frac {a \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )}-\frac {a \cos (c+d x)}{d \left (a^2+b^2\right )}+\frac {a b^2 \cos (c+d x)}{d \left (a^2+b^2\right )^2}+\frac {a^3 b \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*b*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(5/2)*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*Cos[c + d*x]^3)/(3*(a^2 + b^2)*d) + (a^2*b*Sin[c +
d*x])/((a^2 + b^2)^2*d) + (b*Sin[c + d*x]^3)/(3*(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 206

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

Rule 2564

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 2633

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 2638

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

Rule 3074

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 3099

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 3109

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 3518

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*} \int \frac {\sin ^3(c+d x)}{a+b \tan (c+d x)} \, dx &=\int \frac {\cos (c+d x) \sin ^3(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx\\ &=\frac {a \int \sin ^3(c+d x) \, dx}{a^2+b^2}+\frac {b \int \cos (c+d x) \sin ^2(c+d x) \, dx}{a^2+b^2}-\frac {(a b) \int \frac {\sin ^2(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{a^2+b^2}\\ &=\frac {a^2 b \sin (c+d x)}{\left (a^2+b^2\right )^2 d}-\frac {\left (a^3 b\right ) \int \frac {1}{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 (c+d x) \, dx}{\left (a^2+b^2\right )^2}-\frac {a \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{\left (a^2+b^2\right ) d}+\frac {b \operatorname {Subst}\left (\int x^2 \, dx,x,\sin (c+d x)\right )}{\left (a^2+b^2\right ) 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 \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}+\frac {a^2 b \sin (c+d x)}{\left (a^2+b^2\right )^2 d}+\frac {b \sin ^3(c+d x)}{3 \left (a^2+b^2\right ) d}+\frac {\left (a^3 b\right ) \operatorname {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 )^2 d}\\ &=\frac {a^3 b \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2} 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 \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}+\frac {a^2 b \sin (c+d x)}{\left (a^2+b^2\right )^2 d}+\frac {b \sin ^3(c+d x)}{3 \left (a^2+b^2\right ) d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.89, size = 139, normalized size = 0.83 \[ \frac {\sqrt {a^2+b^2} \left (\left (3 a b^2-9 a^3\right ) \cos (c+d x)+a \left (a^2+b^2\right ) \cos (3 (c+d x))-2 b \sin (c+d x) \left (\left (a^2+b^2\right ) \cos (2 (c+d x))-7 a^2-b^2\right )\right )-24 a^3 b \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{12 d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*a^3*b*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] + Sqrt[a^2 + b^2]*((-9*a^3 + 3*a*b^2)*Cos[c + d*
x] + a*(a^2 + b^2)*Cos[3*(c + d*x)] - 2*b*(-7*a^2 - b^2 + (a^2 + b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/(12*(a^
2 + b^2)^(5/2)*d)

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 261, normalized size = 1.55 \[ \frac {3 \, \sqrt {a^{2} + b^{2}} a^{3} b \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (a^{5} + a^{3} b^{2}\right )} \cos \left (d x + c\right ) + 2 \, {\left (4 \, a^{4} b + 5 \, a^{2} b^{3} + b^{5} - {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*sqrt(a^2 + b^2)*a^3*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)) + 2*(a^5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c)^3 - 6*(a^5 + a^3*b^2)*cos(d*x + c) + 2*(4*a^4*b + 5*a
^2*b^3 + b^5 - (a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d)

________________________________________________________________________________________

giac [A]  time = 0.98, size = 241, normalized size = 1.43 \[ \frac {\frac {3 \, a^{3} b \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{3} + a b^{2}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a^3*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^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) + 2*(3*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 3*a*b^2*ta
n(1/2*d*x + 1/2*c)^4 + 10*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 4*b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a^3*tan(1/2*d*x + 1/
2*c)^2 + 3*a^2*b*tan(1/2*d*x + 1/2*c) - 2*a^3 + a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*(tan(1/2*d*x + 1/2*c)^2 + 1)^3
))/d

________________________________________________________________________________________

maple [A]  time = 0.36, size = 205, normalized size = 1.22 \[ \frac {-\frac {16 a^{3} b \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (8 a^{4}+16 a^{2} b^{2}+8 b^{4}\right ) \sqrt {a^{2}+b^{2}}}-\frac {2 \left (-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b -b^{2} a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {10}{3} a^{2} b -\frac {4}{3} b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 a^{3}}{3}-\frac {b^{2} a}{3}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-16*a^3*b/(8*a^4+16*a^2*b^2+8*b^4)/(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)*(-tan(1/2*d*x+1/2*c)^5*a^2*b-b^2*a*tan(1/2*d*x+1/2*c)^4+(-10/3*a^2*b-4/3*b^3)*tan(1/
2*d*x+1/2*c)^3+2*a^3*tan(1/2*d*x+1/2*c)^2-a^2*b*tan(1/2*d*x+1/2*c)+2/3*a^3-1/3*b^2*a)/(1+tan(1/2*d*x+1/2*c)^2)
^3)

________________________________________________________________________________________

maxima [B]  time = 0.64, size = 364, normalized size = 2.17 \[ \frac {\frac {3 \, a^{3} b \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (2 \, a^{3} - a b^{2} - \frac {3 \, a^{2} b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {3 \, a b^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3 \, a^{2} b \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {2 \, {\left (5 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a^3*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^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(2*a^3 - a*b^2 - 3*a^2*b*sin(d*x + c)/(c
os(d*x + c) + 1) + 6*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 3*a*b^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 3
*a^2*b*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 2*(5*a^2*b + 2*b^3)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 + 2
*a^2*b^2 + b^4 + 3*(a^4 + 2*a^2*b^2 + b^4)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*(a^4 + 2*a^2*b^2 + b^4)*sin
(d*x + c)^4/(cos(d*x + c) + 1)^4 + (a^4 + 2*a^2*b^2 + b^4)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6))/d

________________________________________________________________________________________

mupad [B]  time = 6.50, size = 324, normalized size = 1.93 \[ \frac {\frac {\frac {2\,a\,b^2}{3}-\frac {4\,a^3}{3}}{a^4+2\,a^2\,b^2+b^4}-\frac {4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4+2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {20\,a^2\,b}{3}+\frac {8\,b^3}{3}\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^4+2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a^3\,b\,\mathrm {atanh}\left (\frac {a^4\,b+b^5+2\,a^2\,b^3-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{{\left (a^2+b^2\right )}^{5/2}}\right )}{d\,{\left (a^2+b^2\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((2*a*b^2)/3 - (4*a^3)/3)/(a^4 + b^4 + 2*a^2*b^2) - (4*a^3*tan(c/2 + (d*x)/2)^2)/(a^4 + b^4 + 2*a^2*b^2) + (t
an(c/2 + (d*x)/2)^3*((20*a^2*b)/3 + (8*b^3)/3))/(a^4 + b^4 + 2*a^2*b^2) + (2*a^2*b*tan(c/2 + (d*x)/2))/(a^4 +
b^4 + 2*a^2*b^2) + (2*a*b^2*tan(c/2 + (d*x)/2)^4)/(a^4 + b^4 + 2*a^2*b^2) + (2*a^2*b*tan(c/2 + (d*x)/2)^5)/(a^
4 + b^4 + 2*a^2*b^2))/(d*(3*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (2*a^
3*b*atanh((a^4*b + b^5 + 2*a^2*b^3 - a*tan(c/2 + (d*x)/2)*(a^4 + b^4 + 2*a^2*b^2))/(a^2 + b^2)^(5/2)))/(d*(a^2
+ b^2)^(5/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{3}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**3/(a+b*tan(d*x+c)),x)

[Out]

Integral(sin(c + d*x)**3/(a + b*tan(c + d*x)), x)

________________________________________________________________________________________