\(\int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [62]

Optimal result

Integrand size = 21, antiderivative size = 217 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\left (3 a^6-33 a^4 b^2+13 a^2 b^4+b^6\right ) x}{8 \left (a^2+b^2\right )^4}+\frac {2 a^3 b \left (a^2-2 b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^4 d}-\frac {a^4 b}{\left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^4(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{4 \left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) \left (16 a^3 b+\left (5 a^4-12 a^2 b^2-b^4\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^3 d} \]

[Out]

1/8*(3*a^6-33*a^4*b^2+13*a^2*b^4+b^6)*x/(a^2+b^2)^4+2*a^3*b*(a^2-2*b^2)*ln(a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2
)^4/d-a^4*b/(a^2+b^2)^3/d/(a+b*tan(d*x+c))+1/4*cos(d*x+c)^4*(2*a*b+(a^2-b^2)*tan(d*x+c))/(a^2+b^2)^2/d-1/8*cos
(d*x+c)^2*(16*a^3*b+(5*a^4-12*a^2*b^2-b^4)*tan(d*x+c))/(a^2+b^2)^3/d

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3597, 1661, 1643, 649, 209, 266} \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\cos ^4(c+d x) \left (\left (a^2-b^2\right ) \tan (c+d x)+2 a b\right )}{4 d \left (a^2+b^2\right )^2}-\frac {a^4 b}{d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}+\frac {2 a^3 b \left (a^2-2 b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}+\frac {x \left (3 a^6-33 a^4 b^2+13 a^2 b^4+b^6\right )}{8 \left (a^2+b^2\right )^4}-\frac {\cos ^2(c+d x) \left (16 a^3 b+\left (5 a^4-12 a^2 b^2-b^4\right ) \tan (c+d x)\right )}{8 d \left (a^2+b^2\right )^3} \]

[In]

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

[Out]

((3*a^6 - 33*a^4*b^2 + 13*a^2*b^4 + b^6)*x)/(8*(a^2 + b^2)^4) + (2*a^3*b*(a^2 - 2*b^2)*Log[a*Cos[c + d*x] + b*
Sin[c + d*x]])/((a^2 + b^2)^4*d) - (a^4*b)/((a^2 + b^2)^3*d*(a + b*Tan[c + d*x])) + (Cos[c + d*x]^4*(2*a*b + (
a^2 - b^2)*Tan[c + d*x]))/(4*(a^2 + b^2)^2*d) - (Cos[c + d*x]^2*(16*a^3*b + (5*a^4 - 12*a^2*b^2 - b^4)*Tan[c +
d*x]))/(8*(a^2 + b^2)^3*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 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 1661

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[(a*g - c*f*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 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^4}{(a+x)^2 \left (b^2+x^2\right )^3} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = \frac {\cos ^4(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{4 \left (a^2+b^2\right )^2 d}-\frac {\text {Subst}\left (\int \frac {\frac {a^2 b^4 \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2}-\frac {2 a b^4 \left (3 a^2+b^2\right ) x}{\left (a^2+b^2\right )^2}-\frac {b^2 \left (4 a^4+11 a^2 b^2+b^4\right ) x^2}{\left (a^2+b^2\right )^2}}{(a+x)^2 \left (b^2+x^2\right )^2} \, dx,x,b \tan (c+d x)\right )}{4 b d} \\ & = \frac {\cos ^4(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{4 \left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) \left (16 a^3 b+\left (5 a^4-12 a^2 b^2-b^4\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^3 d}+\frac {\text {Subst}\left (\int \frac {\frac {a^2 b^4 \left (3 a^4-12 a^2 b^2+b^4\right )}{\left (a^2+b^2\right )^3}-\frac {2 a b^4 \left (5 a^2-b^2\right ) x}{\left (a^2+b^2\right )^2}-\frac {b^4 \left (5 a^4-12 a^2 b^2-b^4\right ) x^2}{\left (a^2+b^2\right )^3}}{(a+x)^2 \left (b^2+x^2\right )} \, dx,x,b \tan (c+d x)\right )}{8 b^3 d} \\ & = \frac {\cos ^4(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{4 \left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) \left (16 a^3 b+\left (5 a^4-12 a^2 b^2-b^4\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^3 d}+\frac {\text {Subst}\left (\int \left (\frac {8 a^4 b^4}{\left (a^2+b^2\right )^3 (a+x)^2}+\frac {16 a^3 b^4 \left (a^2-2 b^2\right )}{\left (a^2+b^2\right )^4 (a+x)}+\frac {b^4 \left (3 a^6-33 a^4 b^2+13 a^2 b^4+b^6-16 a^3 \left (a^2-2 b^2\right ) x\right )}{\left (a^2+b^2\right )^4 \left (b^2+x^2\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{8 b^3 d} \\ & = \frac {2 a^3 b \left (a^2-2 b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}-\frac {a^4 b}{\left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^4(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{4 \left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) \left (16 a^3 b+\left (5 a^4-12 a^2 b^2-b^4\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^3 d}+\frac {b \text {Subst}\left (\int \frac {3 a^6-33 a^4 b^2+13 a^2 b^4+b^6-16 a^3 \left (a^2-2 b^2\right ) x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^4 d} \\ & = \frac {2 a^3 b \left (a^2-2 b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}-\frac {a^4 b}{\left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^4(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{4 \left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) \left (16 a^3 b+\left (5 a^4-12 a^2 b^2-b^4\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^3 d}-\frac {\left (2 a^3 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 )}{\left (a^2+b^2\right )^4 d}+\frac {\left (b \left (3 a^6-33 a^4 b^2+13 a^2 b^4+b^6\right )\right ) \text {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^4 d} \\ & = \frac {\left (3 a^6-33 a^4 b^2+13 a^2 b^4+b^6\right ) x}{8 \left (a^2+b^2\right )^4}+\frac {2 a^3 b \left (a^2-2 b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac {2 a^3 b \left (a^2-2 b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}-\frac {a^4 b}{\left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^4(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{4 \left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) \left (16 a^3 b+\left (5 a^4-12 a^2 b^2-b^4\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.47 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.81 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {b \left (\frac {3 \left (a^2-b^2\right ) \left (a^2+b^2\right )^2 \arctan (\tan (c+d x))}{b}+\frac {4 \left (a^2+b^2\right ) \left (-2 a^4+3 a^2 b^2+b^4\right ) \arctan (\tan (c+d x))}{b}-16 a^3 \left (a^2+b^2\right ) \cos ^2(c+d x)+4 a \left (a^2+b^2\right )^2 \cos ^4(c+d x)-4 a^3 \left (2 a^2-4 b^2+\frac {-a^3+5 a b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+16 a^3 \left (a^2-2 b^2\right ) \log (a+b \tan (c+d x))-4 a^3 \left (2 a^2-4 b^2+\frac {a^3-5 a b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+\frac {2 \left (a^2-b^2\right ) \left (a^2+b^2\right )^2 \cos ^3(c+d x) \sin (c+d x)}{b}+\frac {3 (a-b) (a+b) \left (a^2+b^2\right )^2 \sin (2 (c+d x))}{2 b}+\frac {2 \left (a^2+b^2\right ) \left (-2 a^4+3 a^2 b^2+b^4\right ) \sin (2 (c+d x))}{b}-\frac {8 a^4 \left (a^2+b^2\right )}{a+b \tan (c+d x)}\right )}{8 \left (a^2+b^2\right )^4 d} \]

[In]

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

[Out]

(b*((3*(a^2 - b^2)*(a^2 + b^2)^2*ArcTan[Tan[c + d*x]])/b + (4*(a^2 + b^2)*(-2*a^4 + 3*a^2*b^2 + b^4)*ArcTan[Ta
n[c + d*x]])/b - 16*a^3*(a^2 + b^2)*Cos[c + d*x]^2 + 4*a*(a^2 + b^2)^2*Cos[c + d*x]^4 - 4*a^3*(2*a^2 - 4*b^2 +
(-a^3 + 5*a*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] + 16*a^3*(a^2 - 2*b^2)*Log[a + b*Tan[c + d*x]]
- 4*a^3*(2*a^2 - 4*b^2 + (a^3 - 5*a*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + (2*(a^2 - b^2)*(a^2 +
b^2)^2*Cos[c + d*x]^3*Sin[c + d*x])/b + (3*(a - b)*(a + b)*(a^2 + b^2)^2*Sin[2*(c + d*x)])/(2*b) + (2*(a^2 + b
^2)*(-2*a^4 + 3*a^2*b^2 + b^4)*Sin[2*(c + d*x)])/b - (8*a^4*(a^2 + b^2))/(a + b*Tan[c + d*x])))/(8*(a^2 + b^2)
^4*d)

Maple [A] (verified)

Time = 9.21 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(\frac {-\frac {a^{4} b}{\left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 a^{3} b \left (a^{2}-2 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{4}}+\frac {\frac {\left (-\frac {5}{8} a^{6}+\frac {7}{8} a^{4} b^{2}+\frac {13}{8} a^{2} b^{4}+\frac {1}{8} b^{6}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (-2 a^{5} b -2 a^{3} b^{3}\right ) \left (\tan ^{2}\left (d x +c \right )\right )+\left (-\frac {3}{8} a^{6}+\frac {9}{8} a^{4} b^{2}+\frac {11}{8} a^{2} b^{4}-\frac {1}{8} b^{6}\right ) \tan \left (d x +c \right )-\frac {3 a^{5} b}{2}-a^{3} b^{3}+\frac {a \,b^{5}}{2}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {\left (-16 a^{5} b +32 a^{3} b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{16}+\frac {\left (3 a^{6}-33 a^{4} b^{2}+13 a^{2} b^{4}+b^{6}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{\left (a^{2}+b^{2}\right )^{4}}}{d}\)	\(269\)
default	\(\frac {-\frac {a^{4} b}{\left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 a^{3} b \left (a^{2}-2 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{4}}+\frac {\frac {\left (-\frac {5}{8} a^{6}+\frac {7}{8} a^{4} b^{2}+\frac {13}{8} a^{2} b^{4}+\frac {1}{8} b^{6}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (-2 a^{5} b -2 a^{3} b^{3}\right ) \left (\tan ^{2}\left (d x +c \right )\right )+\left (-\frac {3}{8} a^{6}+\frac {9}{8} a^{4} b^{2}+\frac {11}{8} a^{2} b^{4}-\frac {1}{8} b^{6}\right ) \tan \left (d x +c \right )-\frac {3 a^{5} b}{2}-a^{3} b^{3}+\frac {a \,b^{5}}{2}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {\left (-16 a^{5} b +32 a^{3} b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{16}+\frac {\left (3 a^{6}-33 a^{4} b^{2}+13 a^{2} b^{4}+b^{6}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{\left (a^{2}+b^{2}\right )^{4}}}{d}\)	\(269\)
risch	\(-\frac {i x a b}{2 \left (4 i a^{3} b -4 i a \,b^{3}-a^{4}+6 a^{2} b^{2}-b^{4}\right )}-\frac {3 x \,a^{2}}{8 \left (4 i a^{3} b -4 i a \,b^{3}-a^{4}+6 a^{2} b^{2}-b^{4}\right )}-\frac {x \,b^{2}}{8 \left (4 i a^{3} b -4 i a \,b^{3}-a^{4}+6 a^{2} b^{2}-b^{4}\right )}-\frac {i {\mathrm e}^{4 i \left (d x +c \right )}}{64 \left (-2 i a b +a^{2}-b^{2}\right ) d}+\frac {i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 \left (-3 i b \,a^{2}+i b^{3}+a^{3}-3 a \,b^{2}\right ) d}-\frac {i a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 \left (2 i a b +a^{2}-b^{2}\right ) \left (i b +a \right ) d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{64 \left (2 i a b +a^{2}-b^{2}\right ) d}-\frac {4 i a^{5} b x}{a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}}+\frac {8 i a^{3} b^{3} x}{a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}}-\frac {4 i a^{5} b c}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}\right )}+\frac {8 i a^{3} b^{3} c}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}\right )}+\frac {2 i a^{4} b^{2}}{\left (i b +a \right )^{3} d \left (-i b +a \right )^{4} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +a \right )}+\frac {2 a^{5} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}\right )}-\frac {4 a^{3} b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}\right )}\)	\(646\)

[In]

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

[Out]

1/d*(-a^4*b/(a^2+b^2)^3/(a+b*tan(d*x+c))+2*a^3*b*(a^2-2*b^2)/(a^2+b^2)^4*ln(a+b*tan(d*x+c))+1/(a^2+b^2)^4*(((-
5/8*a^6+7/8*a^4*b^2+13/8*a^2*b^4+1/8*b^6)*tan(d*x+c)^3+(-2*a^5*b-2*a^3*b^3)*tan(d*x+c)^2+(-3/8*a^6+9/8*a^4*b^2
+11/8*a^2*b^4-1/8*b^6)*tan(d*x+c)-3/2*a^5*b-a^3*b^3+1/2*a*b^5)/(1+tan(d*x+c)^2)^2+1/16*(-16*a^5*b+32*a^3*b^3)*
ln(1+tan(d*x+c)^2)+1/8*(3*a^6-33*a^4*b^2+13*a^2*b^4+b^6)*arctan(tan(d*x+c))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 444 vs. \(2 (211) = 422\).

Time = 0.31 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.05 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {4 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{5} - 6 \, {\left (3 \, a^{6} b + 7 \, a^{4} b^{3} + 5 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{6} b + 8 \, a^{4} b^{3} + 23 \, a^{2} b^{5} + 2 \, b^{7} + 2 \, {\left (3 \, a^{7} - 33 \, a^{5} b^{2} + 13 \, a^{3} b^{4} + a b^{6}\right )} d x\right )} \cos \left (d x + c\right ) + 16 \, {\left ({\left (a^{6} b - 2 \, a^{4} b^{3}\right )} \cos \left (d x + c\right ) + {\left (a^{5} b^{2} - 2 \, a^{3} b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) + {\left (29 \, a^{5} b^{2} + 10 \, a^{3} b^{4} - 3 \, a b^{6} + 4 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (3 \, a^{6} b - 33 \, a^{4} b^{3} + 13 \, a^{2} b^{5} + b^{7}\right )} d x - 2 \, {\left (5 \, a^{7} + 9 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left ({\left (a^{9} + 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} + 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}\right )} d \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/16*(4*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(d*x + c)^5 - 6*(3*a^6*b + 7*a^4*b^3 + 5*a^2*b^5 + b^7)*cos(d
*x + c)^3 + (3*a^6*b + 8*a^4*b^3 + 23*a^2*b^5 + 2*b^7 + 2*(3*a^7 - 33*a^5*b^2 + 13*a^3*b^4 + a*b^6)*d*x)*cos(d
*x + c) + 16*((a^6*b - 2*a^4*b^3)*cos(d*x + c) + (a^5*b^2 - 2*a^3*b^4)*sin(d*x + c))*log(2*a*b*cos(d*x + c)*si
n(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) + (29*a^5*b^2 + 10*a^3*b^4 - 3*a*b^6 + 4*(a^7 + 3*a^5*b^2 + 3*a
^3*b^4 + a*b^6)*cos(d*x + c)^4 + 2*(3*a^6*b - 33*a^4*b^3 + 13*a^2*b^5 + b^7)*d*x - 2*(5*a^7 + 9*a^5*b^2 + 3*a^
3*b^4 - a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^9 + 4*a^7*b^2 + 6*a^5*b^4 + 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)
+ (a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9)*d*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 507 vs. \(2 (211) = 422\).

Time = 0.66 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.34 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {{\left (3 \, a^{6} - 33 \, a^{4} b^{2} + 13 \, a^{2} b^{4} + b^{6}\right )} {\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {16 \, {\left (a^{5} b - 2 \, a^{3} b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {8 \, {\left (a^{5} b - 2 \, a^{3} b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {20 \, a^{4} b - 4 \, a^{2} b^{3} + {\left (13 \, a^{4} b - 12 \, a^{2} b^{3} - b^{5}\right )} \tan \left (d x + c\right )^{4} + {\left (5 \, a^{5} + 4 \, a^{3} b^{2} - a b^{4}\right )} \tan \left (d x + c\right )^{3} + {\left (35 \, a^{4} b - 12 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{5} - a b^{4}\right )} \tan \left (d x + c\right )}{a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6} + {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \tan \left (d x + c\right )^{5} + {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \tan \left (d x + c\right )^{3} + 2 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )^{2} + {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \tan \left (d x + c\right )}}{8 \, d} \]

[In]

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

[Out]

1/8*((3*a^6 - 33*a^4*b^2 + 13*a^2*b^4 + b^6)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 16*(a
^5*b - 2*a^3*b^3)*log(b*tan(d*x + c) + a)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 8*(a^5*b - 2*a^3*b
^3)*log(tan(d*x + c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - (20*a^4*b - 4*a^2*b^3 + (13*a^4*
b - 12*a^2*b^3 - b^5)*tan(d*x + c)^4 + (5*a^5 + 4*a^3*b^2 - a*b^4)*tan(d*x + c)^3 + (35*a^4*b - 12*a^2*b^3 + b
^5)*tan(d*x + c)^2 + 3*(a^5 - a*b^4)*tan(d*x + c))/(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 + (a^6*b + 3*a^4*b^3 +
3*a^2*b^5 + b^7)*tan(d*x + c)^5 + (a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*tan(d*x + c)^4 + 2*(a^6*b + 3*a^4*b^3
+ 3*a^2*b^5 + b^7)*tan(d*x + c)^3 + 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*tan(d*x + c)^2 + (a^6*b + 3*a^4*b
^3 + 3*a^2*b^5 + b^7)*tan(d*x + c)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 513 vs. \(2 (211) = 422\).

Time = 0.53 (sec) , antiderivative size = 513, normalized size of antiderivative = 2.36 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {{\left (3 \, a^{6} - 33 \, a^{4} b^{2} + 13 \, a^{2} b^{4} + b^{6}\right )} {\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {8 \, {\left (a^{5} b - 2 \, a^{3} b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {16 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}} - \frac {8 \, {\left (2 \, a^{5} b^{2} \tan \left (d x + c\right ) - 4 \, a^{3} b^{4} \tan \left (d x + c\right ) + 3 \, a^{6} b - 3 \, a^{4} b^{3}\right )}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}} + \frac {12 \, a^{5} b \tan \left (d x + c\right )^{4} - 24 \, a^{3} b^{3} \tan \left (d x + c\right )^{4} - 5 \, a^{6} \tan \left (d x + c\right )^{3} + 7 \, a^{4} b^{2} \tan \left (d x + c\right )^{3} + 13 \, a^{2} b^{4} \tan \left (d x + c\right )^{3} + b^{6} \tan \left (d x + c\right )^{3} + 8 \, a^{5} b \tan \left (d x + c\right )^{2} - 64 \, a^{3} b^{3} \tan \left (d x + c\right )^{2} - 3 \, a^{6} \tan \left (d x + c\right ) + 9 \, a^{4} b^{2} \tan \left (d x + c\right ) + 11 \, a^{2} b^{4} \tan \left (d x + c\right ) - b^{6} \tan \left (d x + c\right ) - 32 \, a^{3} b^{3} + 4 \, a b^{5}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} {\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2}}}{8 \, d} \]

[In]

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

[Out]

1/8*((3*a^6 - 33*a^4*b^2 + 13*a^2*b^4 + b^6)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 8*(a^
5*b - 2*a^3*b^3)*log(tan(d*x + c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 16*(a^5*b^2 - 2*a^3
*b^4)*log(abs(b*tan(d*x + c) + a))/(a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9) - 8*(2*a^5*b^2*tan(d*x +
c) - 4*a^3*b^4*tan(d*x + c) + 3*a^6*b - 3*a^4*b^3)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(b*tan(d*x
+ c) + a)) + (12*a^5*b*tan(d*x + c)^4 - 24*a^3*b^3*tan(d*x + c)^4 - 5*a^6*tan(d*x + c)^3 + 7*a^4*b^2*tan(d*x
+ c)^3 + 13*a^2*b^4*tan(d*x + c)^3 + b^6*tan(d*x + c)^3 + 8*a^5*b*tan(d*x + c)^2 - 64*a^3*b^3*tan(d*x + c)^2 -
3*a^6*tan(d*x + c) + 9*a^4*b^2*tan(d*x + c) + 11*a^2*b^4*tan(d*x + c) - b^6*tan(d*x + c) - 32*a^3*b^3 + 4*a*b
^5)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(tan(d*x + c)^2 + 1)^2))/d

Mupad [B] (verification not implemented)

Time = 5.82 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.22 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a\,b^2-5\,a^3\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (-13\,a^4\,b+12\,a^2\,b^3+b^5\right )}{8\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {3\,\mathrm {tan}\left (c+d\,x\right )\,\left (a\,b^2-a^3\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (35\,a^4\,b-12\,a^2\,b^3+b^5\right )}{8\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {a\,\left (a\,b^3-5\,a^3\,b\right )}{2\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^5+a\,{\mathrm {tan}\left (c+d\,x\right )}^4+2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3+2\,a\,{\mathrm {tan}\left (c+d\,x\right )}^2+b\,\mathrm {tan}\left (c+d\,x\right )+a\right )}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {2\,a\,b}{{\left (a^2+b^2\right )}^2}-\frac {8\,a\,b^3}{{\left (a^2+b^2\right )}^3}+\frac {6\,a\,b^5}{{\left (a^2+b^2\right )}^4}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (3\,a^2-a\,b\,4{}\mathrm {i}+b^2\right )}{16\,d\,\left (a^4\,1{}\mathrm {i}-4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}+4\,a\,b^3+b^4\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (3\,a^2+a\,b\,4{}\mathrm {i}+b^2\right )}{16\,d\,\left (a^4\,1{}\mathrm {i}+4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}-4\,a\,b^3+b^4\,1{}\mathrm {i}\right )} \]

[In]

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

[Out]

((tan(c + d*x)^3*(a*b^2 - 5*a^3))/(8*(a^4 + b^4 + 2*a^2*b^2)) + (tan(c + d*x)^4*(b^5 - 13*a^4*b + 12*a^2*b^3))
/(8*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (3*tan(c + d*x)*(a*b^2 - a^3))/(8*(a^4 + b^4 + 2*a^2*b^2)) - (tan(c
+ d*x)^2*(35*a^4*b + b^5 - 12*a^2*b^3))/(8*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)) + (a*(a*b^3 - 5*a^3*b))/(2*(a
^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)))/(d*(a + b*tan(c + d*x) + 2*a*tan(c + d*x)^2 + a*tan(c + d*x)^4 + 2*b*tan(c
+ d*x)^3 + b*tan(c + d*x)^5)) + (log(a + b*tan(c + d*x))*((2*a*b)/(a^2 + b^2)^2 - (8*a*b^3)/(a^2 + b^2)^3 + (
6*a*b^5)/(a^2 + b^2)^4))/d + (log(tan(c + d*x) - 1i)*(3*a^2 - a*b*4i + b^2))/(16*d*(4*a*b^3 - 4*a^3*b + a^4*1i
+ b^4*1i - a^2*b^2*6i)) - (log(tan(c + d*x) + 1i)*(a*b*4i + 3*a^2 + b^2))/(16*d*(4*a^3*b - 4*a*b^3 + a^4*1i +
b^4*1i - a^2*b^2*6i))