\(\int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx\) [59]

Optimal result

Integrand size = 21, antiderivative size = 108 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\left (a^2+b^2\right ) \cot (c+d x)}{a^3 d}+\frac {b \cot ^2(c+d x)}{2 a^2 d}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {b \left (a^2+b^2\right ) \log (\tan (c+d x))}{a^4 d}+\frac {b \left (a^2+b^2\right ) \log (a+b \tan (c+d x))}{a^4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 108, 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} \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {b \cot ^2(c+d x)}{2 a^2 d}-\frac {b \left (a^2+b^2\right ) \log (\tan (c+d x))}{a^4 d}+\frac {b \left (a^2+b^2\right ) \log (a+b \tan (c+d x))}{a^4 d}-\frac {\left (a^2+b^2\right ) \cot (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a d} \]

[In]

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

[Out]

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

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)} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = \frac {b \text {Subst}\left (\int \left (\frac {b^2}{a x^4}-\frac {b^2}{a^2 x^3}+\frac {a^2+b^2}{a^3 x^2}+\frac {-a^2-b^2}{a^4 x}+\frac {a^2+b^2}{a^4 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{d} \\ & = -\frac {\left (a^2+b^2\right ) \cot (c+d x)}{a^3 d}+\frac {b \cot ^2(c+d x)}{2 a^2 d}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {b \left (a^2+b^2\right ) \log (\tan (c+d x))}{a^4 d}+\frac {b \left (a^2+b^2\right ) \log (a+b \tan (c+d x))}{a^4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.94 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.88 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {3 a^2 b \csc ^2(c+d x)-2 \cot (c+d x) \left (2 a^3+3 a b^2+a^3 \csc ^2(c+d x)\right )-6 b \left (a^2+b^2\right ) (\log (\sin (c+d x))-\log (a \cos (c+d x)+b \sin (c+d x)))}{6 a^4 d} \]

[In]

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

[Out]

(3*a^2*b*Csc[c + d*x]^2 - 2*Cot[c + d*x]*(2*a^3 + 3*a*b^2 + a^3*Csc[c + d*x]^2) - 6*b*(a^2 + b^2)*(Log[Sin[c +
d*x]] - Log[a*Cos[c + d*x] + b*Sin[c + d*x]]))/(6*a^4*d)

Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {\frac {\left (a^{2}+b^{2}\right ) b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{4}}-\frac {1}{3 a \tan \left (d x +c \right )^{3}}-\frac {a^{2}+b^{2}}{a^{3} \tan \left (d x +c \right )}+\frac {b}{2 a^{2} \tan \left (d x +c \right )^{2}}-\frac {\left (a^{2}+b^{2}\right ) b \ln \left (\tan \left (d x +c \right )\right )}{a^{4}}}{d}\)	\(96\)
default	\(\frac {\frac {\left (a^{2}+b^{2}\right ) b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{4}}-\frac {1}{3 a \tan \left (d x +c \right )^{3}}-\frac {a^{2}+b^{2}}{a^{3} \tan \left (d x +c \right )}+\frac {b}{2 a^{2} \tan \left (d x +c \right )^{2}}-\frac {\left (a^{2}+b^{2}\right ) b \ln \left (\tan \left (d x +c \right )\right )}{a^{4}}}{d}\)	\(96\)
risch	\(-\frac {2 \left (3 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 a b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i a^{2}+3 i b^{2}\right )}{3 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{4} d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}\)	\(227\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.93 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {2 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, a^{2} b \sin \left (d x + c\right ) + 3 \, {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) - 6 \, {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )}{6 \, {\left (a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.90 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {6 \, {\left (a^{2} b + b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4}} - \frac {6 \, {\left (a^{2} b + b^{3}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{4}} + \frac {3 \, a b \tan \left (d x + c\right ) - 6 \, {\left (a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{2} - 2 \, a^{2}}{a^{3} \tan \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.33 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\frac {6 \, {\left (a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac {6 \, {\left (a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b} - \frac {11 \, a^{2} b \tan \left (d x + c\right )^{3} + 11 \, b^{3} \tan \left (d x + c\right )^{3} - 6 \, a^{3} \tan \left (d x + c\right )^{2} - 6 \, a b^{2} \tan \left (d x + c\right )^{2} + 3 \, a^{2} b \tan \left (d x + c\right ) - 2 \, a^{3}}{a^{4} \tan \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

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

[Out]

-1/6*(6*(a^2*b + b^3)*log(abs(tan(d*x + c)))/a^4 - 6*(a^2*b^2 + b^4)*log(abs(b*tan(d*x + c) + a))/(a^4*b) - (1
1*a^2*b*tan(d*x + c)^3 + 11*b^3*tan(d*x + c)^3 - 6*a^3*tan(d*x + c)^2 - 6*a*b^2*tan(d*x + c)^2 + 3*a^2*b*tan(d
*x + c) - 2*a^3)/(a^4*tan(d*x + c)^3))/d

Mupad [B] (verification not implemented)

Time = 4.36 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94 \[ \int \frac {\csc ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {2\,b\,\mathrm {atanh}\left (\frac {b\,\left (a^2+b^2\right )\,\left (a+2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a\,\left (a^2\,b+b^3\right )}\right )\,\left (a^2+b^2\right )}{a^4\,d}-\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^2+b^2\right )}{a^3}-\frac {b\,\mathrm {tan}\left (c+d\,x\right )}{2\,a^2}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^3} \]

[In]

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

[Out]

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