\(\int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx\) [42]

Optimal result

Integrand size = 21, antiderivative size = 139 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {1}{2} \left (a^4-18 a^2 b^2+5 b^4\right ) x-\frac {4 a b \left (a^2-2 b^2\right ) \log (\cos (c+d x))}{d}+\frac {b^2 \left (18 a^2-5 b^2\right ) \tan (c+d x)}{2 d}+\frac {4 a b^3 \tan ^2(c+d x)}{d}+\frac {5 b^4 \tan ^3(c+d x)}{6 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^4}{2 d} \]

[Out]

1/2*(a^4-18*a^2*b^2+5*b^4)*x-4*a*b*(a^2-2*b^2)*ln(cos(d*x+c))/d+1/2*b^2*(18*a^2-5*b^2)*tan(d*x+c)/d+4*a*b^3*ta
n(d*x+c)^2/d+5/6*b^4*tan(d*x+c)^3/d-1/2*cos(d*x+c)*sin(d*x+c)*(a+b*tan(d*x+c))^4/d

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 139, 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} \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {b^2 \left (18 a^2-5 b^2\right ) \tan (c+d x)}{2 d}-\frac {4 a b \left (a^2-2 b^2\right ) \log (\cos (c+d x))}{d}+\frac {1}{2} x \left (a^4-18 a^2 b^2+5 b^4\right )+\frac {4 a b^3 \tan ^2(c+d x)}{d}-\frac {\sin (c+d x) \cos (c+d x) (a+b \tan (c+d x))^4}{2 d}+\frac {5 b^4 \tan ^3(c+d x)}{6 d} \]

[In]

Int[Sin[c + d*x]^2*(a + b*Tan[c + d*x])^4,x]

[Out]

((a^4 - 18*a^2*b^2 + 5*b^4)*x)/2 - (4*a*b*(a^2 - 2*b^2)*Log[Cos[c + d*x]])/d + (b^2*(18*a^2 - 5*b^2)*Tan[c + d
*x])/(2*d) + (4*a*b^3*Tan[c + d*x]^2)/d + (5*b^4*Tan[c + d*x]^3)/(6*d) - (Cos[c + d*x]*Sin[c + d*x]*(a + b*Tan
[c + d*x])^4)/(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)^4}{\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))^4}{2 d}-\frac {\text {Subst}\left (\int \frac {(a+x)^3 \left (-a b^2-5 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))^4}{2 d}-\frac {\text {Subst}\left (\int \left (-18 a^2 b^2+5 b^4-16 a b^2 x-5 b^2 x^2-\frac {b^2 \left (a^4-18 a^2 b^2+5 b^4\right )+8 a b^2 \left (a^2-2 b^2\right ) x}{b^2+x^2}\right ) \, dx,x,b \tan (c+d x)\right )}{2 b d} \\ & = \frac {b^2 \left (18 a^2-5 b^2\right ) \tan (c+d x)}{2 d}+\frac {4 a b^3 \tan ^2(c+d x)}{d}+\frac {5 b^4 \tan ^3(c+d x)}{6 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^4}{2 d}+\frac {\text {Subst}\left (\int \frac {b^2 \left (a^4-18 a^2 b^2+5 b^4\right )+8 a b^2 \left (a^2-2 b^2\right ) x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 b d} \\ & = \frac {b^2 \left (18 a^2-5 b^2\right ) \tan (c+d x)}{2 d}+\frac {4 a b^3 \tan ^2(c+d x)}{d}+\frac {5 b^4 \tan ^3(c+d x)}{6 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^4}{2 d}+\frac {\left (4 a b \left (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 {\left (b \left (a^4-18 a^2 b^2+5 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 d} \\ & = \frac {1}{2} \left (a^4-18 a^2 b^2+5 b^4\right ) x-\frac {4 a b \left (a^2-2 b^2\right ) \log (\cos (c+d x))}{d}+\frac {b^2 \left (18 a^2-5 b^2\right ) \tan (c+d x)}{2 d}+\frac {4 a b^3 \tan ^2(c+d x)}{d}+\frac {5 b^4 \tan ^3(c+d x)}{6 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^4}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.31 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.89 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {b \left (-\frac {\left (a^4-6 a^2 b^2+b^4\right ) \arctan (\tan (c+d x))}{2 b}+2 a (a-b) (a+b) \cos ^2(c+d x)+\frac {1}{2} \left (4 a^3-8 a b^2+\frac {a^4-12 a^2 b^2+3 b^4}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+\frac {1}{2} \left (4 a^3-8 a b^2-\frac {a^4-12 a^2 b^2+3 b^4}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )-\frac {\left (a^4-6 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{2 b}+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+2 a b^2 \tan ^2(c+d x)+\frac {1}{3} b^3 \tan ^3(c+d x)\right )}{d} \]

[In]

Integrate[Sin[c + d*x]^2*(a + b*Tan[c + d*x])^4,x]

[Out]

(b*(-1/2*((a^4 - 6*a^2*b^2 + b^4)*ArcTan[Tan[c + d*x]])/b + 2*a*(a - b)*(a + b)*Cos[c + d*x]^2 + ((4*a^3 - 8*a
*b^2 + (a^4 - 12*a^2*b^2 + 3*b^4)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]])/2 + ((4*a^3 - 8*a*b^2 - (a^4 -
12*a^2*b^2 + 3*b^4)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]])/2 - ((a^4 - 6*a^2*b^2 + b^4)*Cos[c + d*x]*S
in[c + d*x])/(2*b) + 2*b*(3*a^2 - b^2)*Tan[c + d*x] + 2*a*b^2*Tan[c + d*x]^2 + (b^3*Tan[c + d*x]^3)/3))/d

Maple [A] (verified)

Time = 3.53 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.80

method	result	size
derivativedivides	\(\frac {a^{4} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+6 a^{2} b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+4 a \,b^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\)	\(250\)
default	\(\frac {a^{4} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+6 a^{2} b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+4 a \,b^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\)	\(250\)
risch	\(\frac {i {\mathrm e}^{2 i \left (d x +c \right )} b^{4}}{8 d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} b^{4}}{8 d}+\frac {x \,a^{4}}{2}-9 x \,a^{2} b^{2}+\frac {5 x \,b^{4}}{2}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} a^{3} b}{2 d}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} a \,b^{3}}{2 d}-8 i x a \,b^{3}+4 i x \,a^{3} b +\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{4}}{8 d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a^{3} b}{2 d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a \,b^{3}}{2 d}-\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )} a^{2} b^{2}}{4 d}-\frac {16 i b^{3} a c}{d}+\frac {8 i b \,a^{3} c}{d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{4}}{8 d}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} a^{2} b^{2}}{4 d}-\frac {2 i b^{2} \left (-18 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-36 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-18 a^{2}+7 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {4 b \,a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {8 b^{3} a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(406\)

[In]

int(sin(d*x+c)^2*(a+b*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.34 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {12 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{5} + 12 \, a b^{3} \cos \left (d x + c\right ) - 24 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) - 3 \, {\left (2 \, a^{3} b - 2 \, a b^{3} - {\left (a^{4} - 18 \, a^{2} b^{2} + 5 \, b^{4}\right )} d x\right )} \cos \left (d x + c\right )^{3} - {\left (3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, b^{4} - 2 \, {\left (18 \, a^{2} b^{2} - 7 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/6*(12*(a^3*b - a*b^3)*cos(d*x + c)^5 + 12*a*b^3*cos(d*x + c) - 24*(a^3*b - 2*a*b^3)*cos(d*x + c)^3*log(-cos(
d*x + c)) - 3*(2*a^3*b - 2*a*b^3 - (a^4 - 18*a^2*b^2 + 5*b^4)*d*x)*cos(d*x + c)^3 - (3*(a^4 - 6*a^2*b^2 + b^4)
*cos(d*x + c)^4 - 2*b^4 - 2*(18*a^2*b^2 - 7*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F]

\[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{4} \sin ^{2}{\left (c + d x \right )}\, dx \]

[In]

integrate(sin(d*x+c)**2*(a+b*tan(d*x+c))**4,x)

[Out]

Integral((a + b*tan(c + d*x))**4*sin(c + d*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.70 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.11 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {2 \, b^{4} \tan \left (d x + c\right )^{3} + 12 \, a b^{3} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{4} - 18 \, a^{2} b^{2} + 5 \, b^{4}\right )} {\left (d x + c\right )} + 12 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (3 \, a^{2} b^{2} - b^{4}\right )} \tan \left (d x + c\right ) + \frac {3 \, {\left (4 \, a^{3} b - 4 \, a b^{3} - {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{2} + 1}}{6 \, d} \]

[In]

integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/6*(2*b^4*tan(d*x + c)^3 + 12*a*b^3*tan(d*x + c)^2 + 3*(a^4 - 18*a^2*b^2 + 5*b^4)*(d*x + c) + 12*(a^3*b - 2*a
*b^3)*log(tan(d*x + c)^2 + 1) + 12*(3*a^2*b^2 - b^4)*tan(d*x + c) + 3*(4*a^3*b - 4*a*b^3 - (a^4 - 6*a^2*b^2 +
b^4)*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. 3651 vs. \(2 (131) = 262\).

Time = 2.57 (sec) , antiderivative size = 3651, normalized size of antiderivative = 26.27 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\text {Too large to display} \]

[In]

integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/6*(3*a^4*d*x*tan(d*x)^5*tan(c)^5 - 54*a^2*b^2*d*x*tan(d*x)^5*tan(c)^5 + 15*b^4*d*x*tan(d*x)^5*tan(c)^5 - 12*
a^3*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))*t
an(d*x)^5*tan(c)^5 + 24*a*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)^5*tan(c)^5 + 3*a^4*d*x*tan(d*x)^5*tan(c)^3 - 54*a^2*b^2*d*x*tan(d*x)^5*tan(c)^
3 + 15*b^4*d*x*tan(d*x)^5*tan(c)^3 - 9*a^4*d*x*tan(d*x)^4*tan(c)^4 + 162*a^2*b^2*d*x*tan(d*x)^4*tan(c)^4 - 45*
b^4*d*x*tan(d*x)^4*tan(c)^4 + 3*a^4*d*x*tan(d*x)^3*tan(c)^5 - 54*a^2*b^2*d*x*tan(d*x)^3*tan(c)^5 + 15*b^4*d*x*
tan(d*x)^3*tan(c)^5 + 6*a^3*b*tan(d*x)^5*tan(c)^5 + 6*a*b^3*tan(d*x)^5*tan(c)^5 - 12*a^3*b*log(4*(tan(d*x)^2*t
an(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)^5*tan(c)^3 + 24*a
*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))*ta
n(d*x)^5*tan(c)^3 + 36*a^3*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 - 72*a*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)^4 + 3*a^4*tan(d*x)^5*tan(c)^4 - 54*a^2*b^2*tan
(d*x)^5*tan(c)^4 + 15*b^4*tan(d*x)^5*tan(c)^4 - 12*a^3*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)^5 + 24*a*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)^5 + 3*a^4*tan(d*x)
^4*tan(c)^5 - 54*a^2*b^2*tan(d*x)^4*tan(c)^5 + 15*b^4*tan(d*x)^4*tan(c)^5 - 9*a^4*d*x*tan(d*x)^4*tan(c)^2 + 16
2*a^2*b^2*d*x*tan(d*x)^4*tan(c)^2 - 45*b^4*d*x*tan(d*x)^4*tan(c)^2 + 12*a^4*d*x*tan(d*x)^3*tan(c)^3 - 216*a^2*
b^2*d*x*tan(d*x)^3*tan(c)^3 + 60*b^4*d*x*tan(d*x)^3*tan(c)^3 - 6*a^3*b*tan(d*x)^5*tan(c)^3 + 30*a*b^3*tan(d*x)
^5*tan(c)^3 - 9*a^4*d*x*tan(d*x)^2*tan(c)^4 + 162*a^2*b^2*d*x*tan(d*x)^2*tan(c)^4 - 45*b^4*d*x*tan(d*x)^2*tan(
c)^4 - 42*a^3*b*tan(d*x)^4*tan(c)^4 + 30*a*b^3*tan(d*x)^4*tan(c)^4 - 6*a^3*b*tan(d*x)^3*tan(c)^5 + 30*a*b^3*ta
n(d*x)^3*tan(c)^5 + 36*a^3*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 - 72*a*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 - 36*a^2*b^2*tan(d*x)^5*tan(c)^2 + 10*b^4*ta
n(d*x)^5*tan(c)^2 - 48*a^3*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 + 96*a*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 - 12*a^4*tan(d*x)^4*tan(c)^3 + 108*a^2*b^2*t
an(d*x)^4*tan(c)^3 - 30*b^4*tan(d*x)^4*tan(c)^3 + 36*a^3*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 - 72*a*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 - 12*a^4*tan(d
*x)^3*tan(c)^4 + 108*a^2*b^2*tan(d*x)^3*tan(c)^4 - 30*b^4*tan(d*x)^3*tan(c)^4 - 36*a^2*b^2*tan(d*x)^2*tan(c)^5
+ 10*b^4*tan(d*x)^2*tan(c)^5 + 9*a^4*d*x*tan(d*x)^3*tan(c) - 162*a^2*b^2*d*x*tan(d*x)^3*tan(c) + 45*b^4*d*x*t
an(d*x)^3*tan(c) + 12*a*b^3*tan(d*x)^5*tan(c) - 12*a^4*d*x*tan(d*x)^2*tan(c)^2 + 216*a^2*b^2*d*x*tan(d*x)^2*ta
n(c)^2 - 60*b^4*d*x*tan(d*x)^2*tan(c)^2 + 18*a^3*b*tan(d*x)^4*tan(c)^2 - 42*a*b^3*tan(d*x)^4*tan(c)^2 + 9*a^4*
d*x*tan(d*x)*tan(c)^3 - 162*a^2*b^2*d*x*tan(d*x)*tan(c)^3 + 45*b^4*d*x*tan(d*x)*tan(c)^3 + 96*a^3*b*tan(d*x)^3
*tan(c)^3 - 48*a*b^3*tan(d*x)^3*tan(c)^3 + 18*a^3*b*tan(d*x)^2*tan(c)^4 - 42*a*b^3*tan(d*x)^2*tan(c)^4 + 12*a*
b^3*tan(d*x)*tan(c)^5 - 2*b^4*tan(d*x)^5 - 36*a^3*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) + 72*a*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) + 72*a^2*b^2*tan(d*x)^4*t
an(c) - 30*b^4*tan(d*x)^4*tan(c) + 48*a^3*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*ta
n(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 - 96*a*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*ta
n(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + 18*a^4*tan(d*x)^3*tan(c)^2
- 108*a^2*b^2*tan(d*x)^3*tan(c)^2 + 10*b^4*tan(d*x)^3*tan(c)^2 - 36*a^3*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 + 72*a*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 + 1
8*a^4*tan(d*x)^2*tan(c)^3 - 108*a^2*b^2*tan(d*x)^2*tan(c)^3 + 10*b^4*tan(d*x)^2*tan(c)^3 + 72*a^2*b^2*tan(d*x)
*tan(c)^4 - 30*b^4*tan(d*x)*tan(c)^4 - 2*b^4*tan(c)^5 - 3*a^4*d*x*tan(d*x)^2 + 54*a^2*b^2*d*x*tan(d*x)^2 - 15*
b^4*d*x*tan(d*x)^2 - 12*a*b^3*tan(d*x)^4 + 9*a^4*d*x*tan(d*x)*tan(c) - 162*a^2*b^2*d*x*tan(d*x)*tan(c) + 45*b^
4*d*x*tan(d*x)*tan(c) - 18*a^3*b*tan(d*x)^3*tan(c) + 42*a*b^3*tan(d*x)^3*tan(c) - 3*a^4*d*x*tan(c)^2 + 54*a^2*
b^2*d*x*tan(c)^2 - 15*b^4*d*x*tan(c)^2 - 96*a^3*b*tan(d*x)^2*tan(c)^2 + 48*a*b^3*tan(d*x)^2*tan(c)^2 - 18*a^3*
b*tan(d*x)*tan(c)^3 + 42*a*b^3*tan(d*x)*tan(c)^3 - 12*a*b^3*tan(c)^4 + 12*a^3*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 - 24*a*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 - 36*a^2*b^
2*tan(d*x)^3 + 10*b^4*tan(d*x)^3 - 36*a^3*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*ta
n(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c) + 72*a*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) - 12*a^4*tan(d*x)^2*tan(c) + 108*a^2*
b^2*tan(d*x)^2*tan(c) - 30*b^4*tan(d*x)^2*tan(c) + 12*a^3*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 - 24*a*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 - 12*a^4*tan(d*x)*tan(c)^2 + 108*a^
2*b^2*tan(d*x)*tan(c)^2 - 30*b^4*tan(d*x)*tan(c)^2 - 36*a^2*b^2*tan(c)^3 + 10*b^4*tan(c)^3 - 3*a^4*d*x + 54*a^
2*b^2*d*x - 15*b^4*d*x + 6*a^3*b*tan(d*x)^2 - 30*a*b^3*tan(d*x)^2 + 42*a^3*b*tan(d*x)*tan(c) - 30*a*b^3*tan(d*
x)*tan(c) + 6*a^3*b*tan(c)^2 - 30*a*b^3*tan(c)^2 + 12*a^3*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)) - 24*a*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)) + 3*a^4*tan(d*x) - 54*a^2*b^2*tan(d*x) + 15*b^4*tan(d
*x) + 3*a^4*tan(c) - 54*a^2*b^2*tan(c) + 15*b^4*tan(c) - 6*a^3*b - 6*a*b^3)/(d*tan(d*x)^5*tan(c)^5 + d*tan(d*x
)^5*tan(c)^3 - 3*d*tan(d*x)^4*tan(c)^4 + d*tan(d*x)^3*tan(c)^5 - 3*d*tan(d*x)^4*tan(c)^2 + 4*d*tan(d*x)^3*tan(
c)^3 - 3*d*tan(d*x)^2*tan(c)^4 + 3*d*tan(d*x)^3*tan(c) - 4*d*tan(d*x)^2*tan(c)^2 + 3*d*tan(d*x)*tan(c)^3 - d*t
an(d*x)^2 + 3*d*tan(d*x)*tan(c) - d*tan(c)^2 - d)

Mupad [B] (verification not implemented)

Time = 4.98 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.16 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=x\,\left (\frac {a^4}{2}-9\,a^2\,b^2+\frac {5\,b^4}{2}\right )-\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (4\,a\,b^3-2\,a^3\,b\right )}{d}-\frac {{\cos \left (c+d\,x\right )}^2\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {a^4}{2}-3\,a^2\,b^2+\frac {b^4}{2}\right )+2\,a\,b^3-2\,a^3\,b\right )}{d}+\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,b^4-6\,a^2\,b^2\right )}{d}+\frac {2\,a\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d} \]

[In]

int(sin(c + d*x)^2*(a + b*tan(c + d*x))^4,x)

[Out]

x*(a^4/2 + (5*b^4)/2 - 9*a^2*b^2) - (log(tan(c + d*x)^2 + 1)*(4*a*b^3 - 2*a^3*b))/d - (cos(c + d*x)^2*(tan(c +
d*x)*(a^4/2 + b^4/2 - 3*a^2*b^2) + 2*a*b^3 - 2*a^3*b))/d + (b^4*tan(c + d*x)^3)/(3*d) - (tan(c + d*x)*(2*b^4
- 6*a^2*b^2))/d + (2*a*b^3*tan(c + d*x)^2)/d