\(\int \sin ^2(c+d x) (a+b \tan (c+d x))^3 \, dx\) [33]

Optimal result

Integrand size = 21, antiderivative size = 103 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {1}{2} a \left (a^2-9 b^2\right ) x-\frac {b \left (3 a^2-2 b^2\right ) \log (\cos (c+d x))}{d}+\frac {9 a b^2 \tan (c+d x)}{2 d}+\frac {b^3 \tan ^2(c+d x)}{d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{2 d} \]

[Out]

1/2*a*(a^2-9*b^2)*x-b*(3*a^2-2*b^2)*ln(cos(d*x+c))/d+9/2*a*b^2*tan(d*x+c)/d+b^3*tan(d*x+c)^2/d-1/2*cos(d*x+c)*
sin(d*x+c)*(a+b*tan(d*x+c))^3/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3597, 1659, 815, 649, 209, 266} \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {b \left (3 a^2-2 b^2\right ) \log (\cos (c+d x))}{d}+\frac {1}{2} a x \left (a^2-9 b^2\right )+\frac {9 a b^2 \tan (c+d x)}{2 d}-\frac {\sin (c+d x) \cos (c+d x) (a+b \tan (c+d x))^3}{2 d}+\frac {b^3 \tan ^2(c+d x)}{d} \]

[In]

Int[Sin[c + d*x]^2*(a + b*Tan[c + d*x])^3,x]

[Out]

(a*(a^2 - 9*b^2)*x)/2 - (b*(3*a^2 - 2*b^2)*Log[Cos[c + d*x]])/d + (9*a*b^2*Tan[c + d*x])/(2*d) + (b^3*Tan[c +
d*x]^2)/d - (Cos[c + d*x]*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(2*d)

Rule 209

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

Rule 266

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

Rule 649

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

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 3597

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

Rubi steps \begin{align*} \text {integral}& = \frac {b \text {Subst}\left (\int \frac {x^2 (a+x)^3}{\left (b^2+x^2\right )^2} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = -\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{2 d}-\frac {\text {Subst}\left (\int \frac {(a+x)^2 \left (-a b^2-4 b^2 x\right )}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 b d} \\ & = -\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{2 d}-\frac {\text {Subst}\left (\int \left (-9 a b^2-4 b^2 x-\frac {a b^2 \left (a^2-9 b^2\right )+2 b^2 \left (3 a^2-2 b^2\right ) x}{b^2+x^2}\right ) \, dx,x,b \tan (c+d x)\right )}{2 b d} \\ & = \frac {9 a b^2 \tan (c+d x)}{2 d}+\frac {b^3 \tan ^2(c+d x)}{d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{2 d}+\frac {\text {Subst}\left (\int \frac {a b^2 \left (a^2-9 b^2\right )+2 b^2 \left (3 a^2-2 b^2\right ) x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 b d} \\ & = \frac {9 a b^2 \tan (c+d x)}{2 d}+\frac {b^3 \tan ^2(c+d x)}{d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{2 d}+\frac {\left (a b \left (a^2-9 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 d}+\frac {\left (b \left (3 a^2-2 b^2\right )\right ) \text {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = \frac {1}{2} a \left (a^2-9 b^2\right ) x-\frac {b \left (3 a^2-2 b^2\right ) \log (\cos (c+d x))}{d}+\frac {9 a b^2 \tan (c+d x)}{2 d}+\frac {b^3 \tan ^2(c+d x)}{d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.79 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.97 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {b \left (-\frac {a \left (a^2-3 b^2\right ) \arctan (\tan (c+d x))}{b}+\left (3 a^2-b^2\right ) \cos ^2(c+d x)+\left (3 a^2-2 b^2+\frac {a^3-6 a b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+\left (3 a^2-2 b^2+\frac {-a^3+6 a b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )-\frac {a \left (a^2-3 b^2\right ) \sin (2 (c+d x))}{2 b}+6 a b \tan (c+d x)+b^2 \tan ^2(c+d x)\right )}{2 d} \]

[In]

Integrate[Sin[c + d*x]^2*(a + b*Tan[c + d*x])^3,x]

[Out]

(b*(-((a*(a^2 - 3*b^2)*ArcTan[Tan[c + d*x]])/b) + (3*a^2 - b^2)*Cos[c + d*x]^2 + (3*a^2 - 2*b^2 + (a^3 - 6*a*b
^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] + (3*a^2 - 2*b^2 + (-a^3 + 6*a*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2
] + b*Tan[c + d*x]] - (a*(a^2 - 3*b^2)*Sin[2*(c + d*x)])/(2*b) + 6*a*b*Tan[c + d*x] + b^2*Tan[c + d*x]^2))/(2*
d)

Maple [A] (verified)

Time = 1.76 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.58

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+b^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(163\)
default	\(\frac {a^{3} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+b^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(163\)
risch	\(\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} a \,b^{2}}{8 d}-\frac {4 i b^{3} c}{d}+\frac {a^{3} x}{2}-\frac {9 x a \,b^{2}}{2}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} b \,a^{2}}{8 d}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} b^{3}}{8 d}+3 i x b \,a^{2}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{3}}{8 d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} b \,a^{2}}{8 d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} b^{3}}{8 d}-2 i x \,b^{3}+\frac {6 i b \,a^{2} c}{d}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{3}}{8 d}-\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )} a \,b^{2}}{8 d}+\frac {2 b^{2} \left (3 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}+3 i a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{d}+\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(286\)

[In]

int(sin(d*x+c)^2*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.45 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) + 2 \, b^{3} - {\left (3 \, a^{2} b - b^{3} - 2 \, {\left (a^{3} - 9 \, a b^{2}\right )} d x\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, a b^{2} \cos \left (d x + c\right ) - {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(2*(3*a^2*b - b^3)*cos(d*x + c)^4 - 4*(3*a^2*b - 2*b^3)*cos(d*x + c)^2*log(-cos(d*x + c)) + 2*b^3 - (3*a^2
*b - b^3 - 2*(a^3 - 9*a*b^2)*d*x)*cos(d*x + c)^2 + 2*(6*a*b^2*cos(d*x + c) - (a^3 - 3*a*b^2)*cos(d*x + c)^3)*s
in(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F]

\[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sin ^{2}{\left (c + d x \right )}\, dx \]

[In]

integrate(sin(d*x+c)**2*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*sin(c + d*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.10 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {b^{3} \tan \left (d x + c\right )^{2} + 6 \, a b^{2} \tan \left (d x + c\right ) + {\left (a^{3} - 9 \, a b^{2}\right )} {\left (d x + c\right )} + {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + \frac {3 \, a^{2} b - b^{3} - {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

[In]

integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*(b^3*tan(d*x + c)^2 + 6*a*b^2*tan(d*x + c) + (a^3 - 9*a*b^2)*(d*x + c) + (3*a^2*b - 2*b^3)*log(tan(d*x + c
)^2 + 1) + (3*a^2*b - b^3 - (a^3 - 3*a*b^2)*tan(d*x + c))/(tan(d*x + c)^2 + 1))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2370 vs. \(2 (97) = 194\).

Time = 1.18 (sec) , antiderivative size = 2370, normalized size of antiderivative = 23.01 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\text {Too large to display} \]

[In]

integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/4*(2*a^3*d*x*tan(d*x)^4*tan(c)^4 - 18*a*b^2*d*x*tan(d*x)^4*tan(c)^4 - 6*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2
*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4*tan(c)^4 + 4*b^3*log(4*(ta
n(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4*tan(c
)^4 + 2*a^3*d*x*tan(d*x)^4*tan(c)^2 - 18*a*b^2*d*x*tan(d*x)^4*tan(c)^2 - 4*a^3*d*x*tan(d*x)^3*tan(c)^3 + 36*a*
b^2*d*x*tan(d*x)^3*tan(c)^3 + 2*a^3*d*x*tan(d*x)^2*tan(c)^4 - 18*a*b^2*d*x*tan(d*x)^2*tan(c)^4 + 3*a^2*b*tan(d
*x)^4*tan(c)^4 + b^3*tan(d*x)^4*tan(c)^4 - 6*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*
x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4*tan(c)^2 + 4*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x
)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4*tan(c)^2 + 12*a^2*b*log(4*(tan(d*x
)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 -
8*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*
tan(d*x)^3*tan(c)^3 + 2*a^3*tan(d*x)^4*tan(c)^3 - 18*a*b^2*tan(d*x)^4*tan(c)^3 - 6*a^2*b*log(4*(tan(d*x)^2*tan
(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^4 + 4*b^3*
log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x
)^2*tan(c)^4 + 2*a^3*tan(d*x)^3*tan(c)^4 - 18*a*b^2*tan(d*x)^3*tan(c)^4 - 4*a^3*d*x*tan(d*x)^3*tan(c) + 36*a*b
^2*d*x*tan(d*x)^3*tan(c) + 4*a^3*d*x*tan(d*x)^2*tan(c)^2 - 36*a*b^2*d*x*tan(d*x)^2*tan(c)^2 - 3*a^2*b*tan(d*x)
^4*tan(c)^2 + 5*b^3*tan(d*x)^4*tan(c)^2 - 4*a^3*d*x*tan(d*x)*tan(c)^3 + 36*a*b^2*d*x*tan(d*x)*tan(c)^3 - 18*a^
2*b*tan(d*x)^3*tan(c)^3 + 6*b^3*tan(d*x)^3*tan(c)^3 - 3*a^2*b*tan(d*x)^2*tan(c)^4 + 5*b^3*tan(d*x)^2*tan(c)^4
+ 12*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 +
1))*tan(d*x)^3*tan(c) - 8*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d
*x)^2 + tan(c)^2 + 1))*tan(d*x)^3*tan(c) - 12*a*b^2*tan(d*x)^4*tan(c) - 12*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 -
2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + 8*b^3*log(4*(t
an(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(
c)^2 - 6*a^3*tan(d*x)^3*tan(c)^2 + 18*a*b^2*tan(d*x)^3*tan(c)^2 + 12*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(
d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c)^3 - 8*b^3*log(4*(tan(d*x)^
2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c)^3 - 6*a
^3*tan(d*x)^2*tan(c)^3 + 18*a*b^2*tan(d*x)^2*tan(c)^3 - 12*a*b^2*tan(d*x)*tan(c)^4 + 2*a^3*d*x*tan(d*x)^2 - 18
*a*b^2*d*x*tan(d*x)^2 + 2*b^3*tan(d*x)^4 - 4*a^3*d*x*tan(d*x)*tan(c) + 36*a*b^2*d*x*tan(d*x)*tan(c) + 6*a^2*b*
tan(d*x)^3*tan(c) - 2*b^3*tan(d*x)^3*tan(c) + 2*a^3*d*x*tan(c)^2 - 18*a*b^2*d*x*tan(c)^2 + 30*a^2*b*tan(d*x)^2
*tan(c)^2 - 2*b^3*tan(d*x)^2*tan(c)^2 + 6*a^2*b*tan(d*x)*tan(c)^3 - 2*b^3*tan(d*x)*tan(c)^3 + 2*b^3*tan(c)^4 -
6*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1)
)*tan(d*x)^2 + 4*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + t
an(c)^2 + 1))*tan(d*x)^2 + 12*a*b^2*tan(d*x)^3 + 12*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/
(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c) - 8*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(
d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c) + 6*a^3*tan(d*x)^2*tan(c)
- 18*a*b^2*tan(d*x)^2*tan(c) - 6*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^
2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(c)^2 + 4*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)
^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(c)^2 + 6*a^3*tan(d*x)*tan(c)^2 - 18*a*b^2*tan(d*x)*tan(c)^2 + 12
*a*b^2*tan(c)^3 + 2*a^3*d*x - 18*a*b^2*d*x - 3*a^2*b*tan(d*x)^2 + 5*b^3*tan(d*x)^2 - 18*a^2*b*tan(d*x)*tan(c)
+ 6*b^3*tan(d*x)*tan(c) - 3*a^2*b*tan(c)^2 + 5*b^3*tan(c)^2 - 6*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*
tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1)) + 4*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)
*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1)) - 2*a^3*tan(d*x) + 18*a*b^2*tan(d*x) - 2*a^3*t
an(c) + 18*a*b^2*tan(c) + 3*a^2*b + b^3)/(d*tan(d*x)^4*tan(c)^4 + d*tan(d*x)^4*tan(c)^2 - 2*d*tan(d*x)^3*tan(c
)^3 + d*tan(d*x)^2*tan(c)^4 - 2*d*tan(d*x)^3*tan(c) + 2*d*tan(d*x)^2*tan(c)^2 - 2*d*tan(d*x)*tan(c)^3 + d*tan(
d*x)^2 - 2*d*tan(d*x)*tan(c) + d*tan(c)^2 + d)

Mupad [B] (verification not implemented)

Time = 4.68 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.47 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d}+\frac {{\cos \left (c+d\,x\right )}^2\,\left (\frac {3\,a^2\,b}{2}-\frac {b^3}{2}+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {3\,a\,b^2}{2}-\frac {a^3}{2}\right )\right )}{d}+\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {3\,a^2\,b}{2}-b^3\right )}{d}+\frac {3\,a\,b^2\,\mathrm {tan}\left (c+d\,x\right )}{d}-\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (a-3\,b\right )\,\left (a+3\,b\right )}{2\,\left (\frac {9\,a\,b^2}{2}-\frac {a^3}{2}\right )}\right )\,\left (a-3\,b\right )\,\left (a+3\,b\right )}{2\,d} \]

[In]

int(sin(c + d*x)^2*(a + b*tan(c + d*x))^3,x)

[Out]

(b^3*tan(c + d*x)^2)/(2*d) + (cos(c + d*x)^2*((3*a^2*b)/2 - b^3/2 + tan(c + d*x)*((3*a*b^2)/2 - a^3/2)))/d + (
log(tan(c + d*x)^2 + 1)*((3*a^2*b)/2 - b^3))/d + (3*a*b^2*tan(c + d*x))/d - (a*atan((a*tan(c + d*x)*(a - 3*b)*
(a + 3*b))/(2*((9*a*b^2)/2 - a^3/2)))*(a - 3*b)*(a + 3*b))/(2*d)