\(\int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [71]

Optimal result

Integrand size = 21, antiderivative size = 178 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\left (a^2+6 b^2\right ) \cot (c+d x)}{a^5 d}+\frac {3 b \cot ^2(c+d x)}{2 a^4 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}-\frac {b \left (3 a^2+10 b^2\right ) \log (\tan (c+d x))}{a^6 d}+\frac {b \left (3 a^2+10 b^2\right ) \log (a+b \tan (c+d x))}{a^6 d}-\frac {b \left (a^2+b^2\right )}{2 a^4 d (a+b \tan (c+d x))^2}-\frac {2 b \left (a^2+2 b^2\right )}{a^5 d (a+b \tan (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3597, 908} \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {3 b \cot ^2(c+d x)}{2 a^4 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}-\frac {b \left (3 a^2+10 b^2\right ) \log (\tan (c+d x))}{a^6 d}+\frac {b \left (3 a^2+10 b^2\right ) \log (a+b \tan (c+d x))}{a^6 d}-\frac {2 b \left (a^2+2 b^2\right )}{a^5 d (a+b \tan (c+d x))}-\frac {\left (a^2+6 b^2\right ) \cot (c+d x)}{a^5 d}-\frac {b \left (a^2+b^2\right )}{2 a^4 d (a+b \tan (c+d x))^2} \]

[In]

Int[Csc[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]

[Out]

-(((a^2 + 6*b^2)*Cot[c + d*x])/(a^5*d)) + (3*b*Cot[c + d*x]^2)/(2*a^4*d) - Cot[c + d*x]^3/(3*a^3*d) - (b*(3*a^
2 + 10*b^2)*Log[Tan[c + d*x]])/(a^6*d) + (b*(3*a^2 + 10*b^2)*Log[a + b*Tan[c + d*x]])/(a^6*d) - (b*(a^2 + b^2)
)/(2*a^4*d*(a + b*Tan[c + d*x])^2) - (2*b*(a^2 + 2*b^2))/(a^5*d*(a + b*Tan[c + d*x]))

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 {b^2+x^2}{x^4 (a+x)^3} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = \frac {b \text {Subst}\left (\int \left (\frac {b^2}{a^3 x^4}-\frac {3 b^2}{a^4 x^3}+\frac {a^2+6 b^2}{a^5 x^2}+\frac {-3 a^2-10 b^2}{a^6 x}+\frac {a^2+b^2}{a^4 (a+x)^3}+\frac {2 \left (a^2+2 b^2\right )}{a^5 (a+x)^2}+\frac {3 a^2+10 b^2}{a^6 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{d} \\ & = -\frac {\left (a^2+6 b^2\right ) \cot (c+d x)}{a^5 d}+\frac {3 b \cot ^2(c+d x)}{2 a^4 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}-\frac {b \left (3 a^2+10 b^2\right ) \log (\tan (c+d x))}{a^6 d}+\frac {b \left (3 a^2+10 b^2\right ) \log (a+b \tan (c+d x))}{a^6 d}-\frac {b \left (a^2+b^2\right )}{2 a^4 d (a+b \tan (c+d x))^2}-\frac {2 b \left (a^2+2 b^2\right )}{a^5 d (a+b \tan (c+d x))} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(456\) vs. \(2(178)=356\).

Time = 6.67 (sec) , antiderivative size = 456, normalized size of antiderivative = 2.56 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {b^3 \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))}{2 a^4 d (a+b \tan (c+d x))^3}-\frac {\csc ^3(c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{3 a^3 d (a+b \tan (c+d x))^3}-\frac {2 \left (a^2 \cos (c+d x)+9 b^2 \cos (c+d x)\right ) \csc (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{3 a^5 d (a+b \tan (c+d x))^3}+\frac {3 b \csc ^2(c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{2 a^4 d (a+b \tan (c+d x))^3}+\frac {\left (-3 a^2 b-10 b^3\right ) \log (\sin (c+d x)) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{a^6 d (a+b \tan (c+d x))^3}+\frac {\left (3 a^2 b+10 b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x)) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{a^6 d (a+b \tan (c+d x))^3}+\frac {\sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \left (3 a^2 b^2 \sin (c+d x)+4 b^4 \sin (c+d x)\right )}{a^6 d (a+b \tan (c+d x))^3} \]

[In]

Integrate[Csc[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]

[Out]

-1/2*(b^3*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(a^4*d*(a + b*Tan[c + d*x])^3) - (Csc[c + d*x]^3*S
ec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(3*a^3*d*(a + b*Tan[c + d*x])^3) - (2*(a^2*Cos[c + d*x] + 9
*b^2*Cos[c + d*x])*Csc[c + d*x]*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(3*a^5*d*(a + b*Tan[c + d*
x])^3) + (3*b*Csc[c + d*x]^2*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(2*a^4*d*(a + b*Tan[c + d*x])
^3) + ((-3*a^2*b - 10*b^3)*Log[Sin[c + d*x]]*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(a^6*d*(a + b
*Tan[c + d*x])^3) + ((3*a^2*b + 10*b^3)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]*Sec[c + d*x]^3*(a*Cos[c + d*x] +
b*Sin[c + d*x])^3)/(a^6*d*(a + b*Tan[c + d*x])^3) + (Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^2*(3*a^2
*b^2*Sin[c + d*x] + 4*b^4*Sin[c + d*x]))/(a^6*d*(a + b*Tan[c + d*x])^3)

Maple [A] (verified)

Time = 2.47 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {-\frac {1}{3 a^{3} \tan \left (d x +c \right )^{3}}-\frac {a^{2}+6 b^{2}}{a^{5} \tan \left (d x +c \right )}+\frac {3 b}{2 a^{4} \tan \left (d x +c \right )^{2}}-\frac {b \left (3 a^{2}+10 b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{6}}+\frac {b \left (3 a^{2}+10 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{6}}-\frac {\left (a^{2}+b^{2}\right ) b}{2 a^{4} \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {2 b \left (a^{2}+2 b^{2}\right )}{a^{5} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(158\)
default	\(\frac {-\frac {1}{3 a^{3} \tan \left (d x +c \right )^{3}}-\frac {a^{2}+6 b^{2}}{a^{5} \tan \left (d x +c \right )}+\frac {3 b}{2 a^{4} \tan \left (d x +c \right )^{2}}-\frac {b \left (3 a^{2}+10 b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{6}}+\frac {b \left (3 a^{2}+10 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{6}}-\frac {\left (a^{2}+b^{2}\right ) b}{2 a^{4} \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {2 b \left (a^{2}+2 b^{2}\right )}{a^{5} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(158\)
risch	\(-\frac {2 i \left (29 a^{2} b^{3}-2 i a^{5}+30 b^{5}+90 i a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+18 i a^{3} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-150 i a \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}-9 a^{4} b \,{\mathrm e}^{8 i \left (d x +c \right )}-a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}+15 a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 i a^{5} {\mathrm e}^{6 i \left (d x +c \right )}+2 a^{4} b +60 i a \,b^{4} {\mathrm e}^{8 i \left (d x +c \right )}-45 i a^{3} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+29 i a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+30 i a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+27 i a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 a^{4} b \,{\mathrm e}^{6 i \left (d x +c \right )}-21 a^{2} b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-106 a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-29 i a^{3} b^{2}-30 i a \,b^{4}+10 i a^{5} {\mathrm e}^{4 i \left (d x +c \right )}+2 i a^{5} {\mathrm e}^{2 i \left (d x +c \right )}-6 a^{2} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+104 a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+30 b^{5} {\mathrm e}^{8 i \left (d x +c \right )}-120 b^{5} {\mathrm e}^{6 i \left (d x +c \right )}-120 b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+180 b^{5} {\mathrm e}^{4 i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} \left (i a +b \right ) a^{5} d}+\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{4} d}+\frac {10 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{6} d}-\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}-\frac {10 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{6} d}\)	\(595\)

[In]

int(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 811 vs. \(2 (172) = 344\).

Time = 0.30 (sec) , antiderivative size = 811, normalized size of antiderivative = 4.56 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {2 \, {\left (2 \, a^{7} + 27 \, a^{5} b^{2} + a^{3} b^{4} - 30 \, a b^{6}\right )} \cos \left (d x + c\right )^{5} - 2 \, {\left (3 \, a^{7} + 43 \, a^{5} b^{2} - 8 \, a^{3} b^{4} - 60 \, a b^{6}\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (5 \, a^{5} b^{2} - 3 \, a^{3} b^{4} - 10 \, a b^{6}\right )} \cos \left (d x + c\right ) + 3 \, {\left (2 \, {\left (3 \, a^{5} b^{2} + 13 \, a^{3} b^{4} + 10 \, a b^{6}\right )} \cos \left (d x + c\right )^{5} - 4 \, {\left (3 \, a^{5} b^{2} + 13 \, a^{3} b^{4} + 10 \, a b^{6}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, a^{5} b^{2} + 13 \, a^{3} b^{4} + 10 \, a b^{6}\right )} \cos \left (d x + c\right ) + {\left (3 \, a^{4} b^{3} + 13 \, a^{2} b^{5} + 10 \, b^{7} - {\left (3 \, a^{6} b + 10 \, a^{4} b^{3} - 3 \, a^{2} b^{5} - 10 \, b^{7}\right )} \cos \left (d x + c\right )^{4} + {\left (3 \, a^{6} b + 7 \, a^{4} b^{3} - 16 \, a^{2} b^{5} - 20 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\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 ) - 3 \, {\left (2 \, {\left (3 \, a^{5} b^{2} + 13 \, a^{3} b^{4} + 10 \, a b^{6}\right )} \cos \left (d x + c\right )^{5} - 4 \, {\left (3 \, a^{5} b^{2} + 13 \, a^{3} b^{4} + 10 \, a b^{6}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, a^{5} b^{2} + 13 \, a^{3} b^{4} + 10 \, a b^{6}\right )} \cos \left (d x + c\right ) + {\left (3 \, a^{4} b^{3} + 13 \, a^{2} b^{5} + 10 \, b^{7} - {\left (3 \, a^{6} b + 10 \, a^{4} b^{3} - 3 \, a^{2} b^{5} - 10 \, b^{7}\right )} \cos \left (d x + c\right )^{4} + {\left (3 \, a^{6} b + 7 \, a^{4} b^{3} - 16 \, a^{2} b^{5} - 20 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) + {\left (24 \, a^{4} b^{3} + 30 \, a^{2} b^{5} + 4 \, {\left (2 \, a^{6} b + 29 \, a^{4} b^{3} + 30 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4} - 3 \, {\left (a^{6} b + 45 \, a^{4} b^{3} + 50 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (2 \, {\left (a^{9} b + a^{7} b^{3}\right )} d \cos \left (d x + c\right )^{5} - 4 \, {\left (a^{9} b + a^{7} b^{3}\right )} d \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{9} b + a^{7} b^{3}\right )} d \cos \left (d x + c\right ) - {\left ({\left (a^{10} - a^{6} b^{4}\right )} d \cos \left (d x + c\right )^{4} - {\left (a^{10} - a^{8} b^{2} - 2 \, a^{6} b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{8} b^{2} + a^{6} b^{4}\right )} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/6*(2*(2*a^7 + 27*a^5*b^2 + a^3*b^4 - 30*a*b^6)*cos(d*x + c)^5 - 2*(3*a^7 + 43*a^5*b^2 - 8*a^3*b^4 - 60*a*b^6
)*cos(d*x + c)^3 + 6*(5*a^5*b^2 - 3*a^3*b^4 - 10*a*b^6)*cos(d*x + c) + 3*(2*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6
)*cos(d*x + c)^5 - 4*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6)*cos(d*x + c)^3 + 2*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6
)*cos(d*x + c) + (3*a^4*b^3 + 13*a^2*b^5 + 10*b^7 - (3*a^6*b + 10*a^4*b^3 - 3*a^2*b^5 - 10*b^7)*cos(d*x + c)^4
+ (3*a^6*b + 7*a^4*b^3 - 16*a^2*b^5 - 20*b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x +
c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 3*(2*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6)*cos(d*x + c)^5 - 4*(3*a^5*b^
2 + 13*a^3*b^4 + 10*a*b^6)*cos(d*x + c)^3 + 2*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6)*cos(d*x + c) + (3*a^4*b^3 +
13*a^2*b^5 + 10*b^7 - (3*a^6*b + 10*a^4*b^3 - 3*a^2*b^5 - 10*b^7)*cos(d*x + c)^4 + (3*a^6*b + 7*a^4*b^3 - 16*a
^2*b^5 - 20*b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/4*cos(d*x + c)^2 + 1/4) + (24*a^4*b^3 + 30*a^2*b^5 + 4*(
2*a^6*b + 29*a^4*b^3 + 30*a^2*b^5)*cos(d*x + c)^4 - 3*(a^6*b + 45*a^4*b^3 + 50*a^2*b^5)*cos(d*x + c)^2)*sin(d*
x + c))/(2*(a^9*b + a^7*b^3)*d*cos(d*x + c)^5 - 4*(a^9*b + a^7*b^3)*d*cos(d*x + c)^3 + 2*(a^9*b + a^7*b^3)*d*c
os(d*x + c) - ((a^10 - a^6*b^4)*d*cos(d*x + c)^4 - (a^10 - a^8*b^2 - 2*a^6*b^4)*d*cos(d*x + c)^2 - (a^8*b^2 +
a^6*b^4)*d)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\csc ^{4}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

integrate(csc(d*x+c)**4/(a+b*tan(d*x+c))**3,x)

[Out]

Integral(csc(c + d*x)**4/(a + b*tan(c + d*x))**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.08 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {5 \, a^{3} b \tan \left (d x + c\right ) - 6 \, {\left (3 \, a^{2} b^{2} + 10 \, b^{4}\right )} \tan \left (d x + c\right )^{4} - 2 \, a^{4} - 9 \, {\left (3 \, a^{3} b + 10 \, a b^{3}\right )} \tan \left (d x + c\right )^{3} - 2 \, {\left (3 \, a^{4} + 10 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{a^{5} b^{2} \tan \left (d x + c\right )^{5} + 2 \, a^{6} b \tan \left (d x + c\right )^{4} + a^{7} \tan \left (d x + c\right )^{3}} + \frac {6 \, {\left (3 \, a^{2} b + 10 \, b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6}} - \frac {6 \, {\left (3 \, a^{2} b + 10 \, b^{3}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{6}}}{6 \, d} \]

[In]

integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/6*((5*a^3*b*tan(d*x + c) - 6*(3*a^2*b^2 + 10*b^4)*tan(d*x + c)^4 - 2*a^4 - 9*(3*a^3*b + 10*a*b^3)*tan(d*x +
c)^3 - 2*(3*a^4 + 10*a^2*b^2)*tan(d*x + c)^2)/(a^5*b^2*tan(d*x + c)^5 + 2*a^6*b*tan(d*x + c)^4 + a^7*tan(d*x +
c)^3) + 6*(3*a^2*b + 10*b^3)*log(b*tan(d*x + c) + a)/a^6 - 6*(3*a^2*b + 10*b^3)*log(tan(d*x + c))/a^6)/d

Giac [A] (verification not implemented)

none

Time = 0.60 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.33 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\frac {6 \, {\left (3 \, a^{2} b + 10 \, b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{6}} - \frac {6 \, {\left (3 \, a^{2} b^{2} + 10 \, b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b} + \frac {3 \, {\left (9 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} + 30 \, b^{5} \tan \left (d x + c\right )^{2} + 22 \, a^{3} b^{2} \tan \left (d x + c\right ) + 68 \, a b^{4} \tan \left (d x + c\right ) + 14 \, a^{4} b + 39 \, a^{2} b^{3}\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} a^{6}} - \frac {33 \, a^{2} b \tan \left (d x + c\right )^{3} + 110 \, b^{3} \tan \left (d x + c\right )^{3} - 6 \, a^{3} \tan \left (d x + c\right )^{2} - 36 \, a b^{2} \tan \left (d x + c\right )^{2} + 9 \, a^{2} b \tan \left (d x + c\right ) - 2 \, a^{3}}{a^{6} \tan \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/6*(6*(3*a^2*b + 10*b^3)*log(abs(tan(d*x + c)))/a^6 - 6*(3*a^2*b^2 + 10*b^4)*log(abs(b*tan(d*x + c) + a))/(a
^6*b) + 3*(9*a^2*b^3*tan(d*x + c)^2 + 30*b^5*tan(d*x + c)^2 + 22*a^3*b^2*tan(d*x + c) + 68*a*b^4*tan(d*x + c)
+ 14*a^4*b + 39*a^2*b^3)/((b*tan(d*x + c) + a)^2*a^6) - (33*a^2*b*tan(d*x + c)^3 + 110*b^3*tan(d*x + c)^3 - 6*
a^3*tan(d*x + c)^2 - 36*a*b^2*tan(d*x + c)^2 + 9*a^2*b*tan(d*x + c) - 2*a^3)/(a^6*tan(d*x + c)^3))/d

Mupad [B] (verification not implemented)

Time = 5.05 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {2\,b\,\mathrm {atanh}\left (\frac {b\,\left (3\,a^2+10\,b^2\right )\,\left (a+2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a\,\left (3\,a^2\,b+10\,b^3\right )}\right )\,\left (3\,a^2+10\,b^2\right )}{a^6\,d}-\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (3\,a^2+10\,b^2\right )}{3\,a^3}-\frac {5\,b\,\mathrm {tan}\left (c+d\,x\right )}{6\,a^2}+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (3\,a^2+10\,b^2\right )}{a^5}+\frac {3\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,a^2+10\,b^2\right )}{2\,a^4}}{d\,\left (a^2\,{\mathrm {tan}\left (c+d\,x\right )}^3+2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^4+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^5\right )} \]

[In]

int(1/(sin(c + d*x)^4*(a + b*tan(c + d*x))^3),x)

[Out]

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