3.32 \(\int \sin ^3(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=205 \[ \frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {a^3 \cos (c+d x)}{d}-\frac {a^2 b \sin ^3(c+d x)}{d}-\frac {3 a^2 b \sin (c+d x)}{d}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {5 b^3 \sin ^3(c+d x)}{6 d}+\frac {5 b^3 \sin (c+d x)}{2 d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac {5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d} \]

[Out]

3*a^2*b*arctanh(sin(d*x+c))/d-5/2*b^3*arctanh(sin(d*x+c))/d-a^3*cos(d*x+c)/d+6*a*b^2*cos(d*x+c)/d+1/3*a^3*cos(
d*x+c)^3/d-a*b^2*cos(d*x+c)^3/d+3*a*b^2*sec(d*x+c)/d-3*a^2*b*sin(d*x+c)/d+5/2*b^3*sin(d*x+c)/d-a^2*b*sin(d*x+c
)^3/d+5/6*b^3*sin(d*x+c)^3/d+1/2*b^3*sin(d*x+c)^3*tan(d*x+c)^2/d

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Rubi [A]  time = 0.19, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3517, 2633, 2592, 302, 206, 2590, 270, 288} \[ -\frac {a^2 b \sin ^3(c+d x)}{d}-\frac {3 a^2 b \sin (c+d x)}{d}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {a^3 \cos (c+d x)}{d}-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {5 b^3 \sin ^3(c+d x)}{6 d}+\frac {5 b^3 \sin (c+d x)}{2 d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac {5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*b*ArcTanh[Sin[c + d*x]])/d - (5*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (a^3*Cos[c + d*x])/d + (6*a*b^2*Cos[
c + d*x])/d + (a^3*Cos[c + d*x]^3)/(3*d) - (a*b^2*Cos[c + d*x]^3)/d + (3*a*b^2*Sec[c + d*x])/d - (3*a^2*b*Sin[
c + d*x])/d + (5*b^3*Sin[c + d*x])/(2*d) - (a^2*b*Sin[c + d*x]^3)/d + (5*b^3*Sin[c + d*x]^3)/(6*d) + (b^3*Sin[
c + d*x]^3*Tan[c + d*x]^2)/(2*d)

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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

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 3517

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Expand[Sin[e
+ f*x]^m*(a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \sin ^3(c+d x) (a+b \tan (c+d x))^3 \, dx &=\int \left (a^3 \sin ^3(c+d x)+3 a^2 b \sin ^3(c+d x) \tan (c+d x)+3 a b^2 \sin ^3(c+d x) \tan ^2(c+d x)+b^3 \sin ^3(c+d x) \tan ^3(c+d x)\right ) \, dx\\ &=a^3 \int \sin ^3(c+d x) \, dx+\left (3 a^2 b\right ) \int \sin ^3(c+d x) \tan (c+d x) \, dx+\left (3 a b^2\right ) \int \sin ^3(c+d x) \tan ^2(c+d x) \, dx+b^3 \int \sin ^3(c+d x) \tan ^3(c+d x) \, dx\\ &=-\frac {a^3 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (3 a b^2\right ) \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac {b^3 \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {a^3 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}+\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (3 a b^2\right ) \operatorname {Subst}\left (\int \left (-2+\frac {1}{x^2}+x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=-\frac {a^3 \cos (c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}-\frac {3 a^2 b \sin (c+d x)}{d}-\frac {a^2 b \sin ^3(c+d x)}{d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}+\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^3 \cos (c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}-\frac {3 a^2 b \sin (c+d x)}{d}+\frac {5 b^3 \sin (c+d x)}{2 d}-\frac {a^2 b \sin ^3(c+d x)}{d}+\frac {5 b^3 \sin ^3(c+d x)}{6 d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {a^3 \cos (c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}-\frac {3 a^2 b \sin (c+d x)}{d}+\frac {5 b^3 \sin (c+d x)}{2 d}-\frac {a^2 b \sin ^3(c+d x)}{d}+\frac {5 b^3 \sin ^3(c+d x)}{6 d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}\\ \end {align*}

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Mathematica [B]  time = 6.30, size = 771, normalized size = 3.76 \[ \frac {\left (5 b^3-6 a^2 b\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {\left (6 a^2 b-5 b^3\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {3 a \left (a^2-7 b^2\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^3}{4 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {b \left (3 a^2-b^2\right ) \sin (3 (c+d x)) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{12 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {a \left (a^2-3 b^2\right ) \cos (3 (c+d x)) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{12 d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {3 b \left (5 a^2-3 b^2\right ) \sin (c+d x) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {b^3 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {b^3 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {3 a b^2 \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {3 a b^2 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac {3 a b^2 \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*b^2*Cos[c + d*x]^3*(a + b*Tan[c + d*x])^3)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (3*a*(a^2 - 7*b^2)*C
os[c + d*x]^4*(a + b*Tan[c + d*x])^3)/(4*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (a*(a^2 - 3*b^2)*Cos[c + d*x
]^3*Cos[3*(c + d*x)]*(a + b*Tan[c + d*x])^3)/(12*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + ((-6*a^2*b + 5*b^3)*
Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3)/(2*d*(a*Cos[c + d*x] + b*Sin[c
+ d*x])^3) + ((6*a^2*b - 5*b^3)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^
3)/(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (b^3*Cos[c + d*x]^3*(a + b*Tan[c + d*x])^3)/(4*d*(Cos[(c + d*x)
/2] - Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (3*a*b^2*Cos[c + d*x]^3*Sin[(c + d*x)/2]*(a +
b*Tan[c + d*x])^3)/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (b^3*Cos[c
+ d*x]^3*(a + b*Tan[c + d*x])^3)/(4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x
])^3) - (3*a*b^2*Cos[c + d*x]^3*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^3)/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (3*b*(5*a^2 - 3*b^2)*Cos[c + d*x]^3*Sin[c + d*x]*(a + b*Tan[c + d*x
])^3)/(4*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (b*(3*a^2 - b^2)*Cos[c + d*x]^3*Sin[3*(c + d*x)]*(a + b*Tan[
c + d*x])^3)/(12*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)

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fricas [A]  time = 0.47, size = 188, normalized size = 0.92 \[ \frac {4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 36 \, a b^{2} \cos \left (d x + c\right ) - 12 \, {\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (6 \, a^{2} b - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (6 \, a^{2} b - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, b^{3} - 2 \, {\left (12 \, a^{2} b - 7 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(a^3 - 3*a*b^2)*cos(d*x + c)^5 + 36*a*b^2*cos(d*x + c) - 12*(a^3 - 6*a*b^2)*cos(d*x + c)^3 + 3*(6*a^2*
b - 5*b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*(6*a^2*b - 5*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) +
2*(2*(3*a^2*b - b^3)*cos(d*x + c)^4 + 3*b^3 - 2*(12*a^2*b - 7*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x +
c)^2)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.44, size = 271, normalized size = 1.32 \[ -\frac {\cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{3}}{3 d}-\frac {2 a^{3} \cos \left (d x +c \right )}{3 d}-\frac {a^{2} b \left (\sin ^{3}\left (d x +c \right )\right )}{d}-\frac {3 a^{2} b \sin \left (d x +c \right )}{d}+\frac {3 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 b^{2} a \left (\sin ^{6}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {8 a \,b^{2} \cos \left (d x +c \right )}{d}+\frac {3 b^{2} a \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{d}+\frac {4 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a \,b^{2}}{d}+\frac {b^{3} \left (\sin ^{7}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {b^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{2 d}+\frac {5 b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{6 d}+\frac {5 b^{3} \sin \left (d x +c \right )}{2 d}-\frac {5 b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.71, size = 173, normalized size = 0.84 \[ \frac {4 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{3} - 6 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{2} b - 12 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a b^{2} + {\left (4 \, \sin \left (d x + c\right )^{3} - \frac {6 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 24 \, \sin \left (d x + c\right )\right )} b^{3}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^3 - 6*(2*sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x
+ c) - 1) + 6*sin(d*x + c))*a^2*b - 12*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*a*b^2 + (4*sin(d*x +
c)^3 - 6*sin(d*x + c)/(sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1) + 24*sin(d*x
+ c))*b^3)/d

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mupad [B]  time = 6.44, size = 291, normalized size = 1.42 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (6\,a^2\,b-5\,b^3\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2\,b-5\,b^3\right )+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-16\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (16\,a\,b^2-\frac {4\,a^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (32\,a\,b^2-\frac {20\,a^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (6\,a^2\,b-5\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (8\,a^2\,b-\frac {20\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (8\,a^2\,b-\frac {20\,b^3}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (28\,a^2\,b-\frac {22\,b^3}{3}\right )+\frac {4\,a^3}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh(tan(c/2 + (d*x)/2))*(6*a^2*b - 5*b^3))/d - (tan(c/2 + (d*x)/2)*(6*a^2*b - 5*b^3) + 4*a^3*tan(c/2 + (d*x
)/2)^6 - 16*a*b^2 - tan(c/2 + (d*x)/2)^2*(16*a*b^2 - (4*a^3)/3) + tan(c/2 + (d*x)/2)^4*(32*a*b^2 - (20*a^3)/3)
+ tan(c/2 + (d*x)/2)^9*(6*a^2*b - 5*b^3) + tan(c/2 + (d*x)/2)^3*(8*a^2*b - (20*b^3)/3) + tan(c/2 + (d*x)/2)^7
*(8*a^2*b - (20*b^3)/3) - tan(c/2 + (d*x)/2)^5*(28*a^2*b - (22*b^3)/3) + (4*a^3)/3)/(d*(tan(c/2 + (d*x)/2)^2 -
2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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