3.70 \(\int \frac{\csc ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=95 \[ -\frac{2 b}{a^3 d (a+b \tan (c+d x))}-\frac{b}{2 a^2 d (a+b \tan (c+d x))^2}-\frac{3 b \log (\tan (c+d x))}{a^4 d}+\frac{3 b \log (a+b \tan (c+d x))}{a^4 d}-\frac{\cot (c+d x)}{a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.0774468, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3516, 44} \[ -\frac{2 b}{a^3 d (a+b \tan (c+d x))}-\frac{b}{2 a^2 d (a+b \tan (c+d x))^2}-\frac{3 b \log (\tan (c+d x))}{a^4 d}+\frac{3 b \log (a+b \tan (c+d x))}{a^4 d}-\frac{\cot (c+d x)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3516

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]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\csc ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{1}{x^2 (a+x)^3} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac{b \operatorname{Subst}\left (\int \left (\frac{1}{a^3 x^2}-\frac{3}{a^4 x}+\frac{1}{a^2 (a+x)^3}+\frac{2}{a^3 (a+x)^2}+\frac{3}{a^4 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac{\cot (c+d x)}{a^3 d}-\frac{3 b \log (\tan (c+d x))}{a^4 d}+\frac{3 b \log (a+b \tan (c+d x))}{a^4 d}-\frac{b}{2 a^2 d (a+b \tan (c+d x))^2}-\frac{2 b}{a^3 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [B]  time = 2.58893, size = 241, normalized size = 2.54 \[ \frac{b \left (a^2 \left (-b^2\right ) \sec ^2(c+d x)-2 a^2 \left (a^2+b^2\right ) (-3 \log (a \cos (c+d x)+b \sin (c+d x))+3 \log (\sin (c+d x))+2)-2 b^2 \tan ^2(c+d x) \left (3 \left (a^2+b^2\right ) \log (\sin (c+d x))-3 \left (a^2+b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))-3 a^2-2 b^2\right )+2 a b \tan (c+d x) \left (-6 \left (a^2+b^2\right ) \log (\sin (c+d x))+6 \left (a^2+b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))+2 a^2+b^2\right )\right )-2 a^3 \left (a^2+b^2\right ) \cot (c+d x)}{2 a^4 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.119, size = 96, normalized size = 1. \begin{align*} -{\frac{1}{d{a}^{3}\tan \left ( dx+c \right ) }}-3\,{\frac{b\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{4}d}}-{\frac{b}{2\,{a}^{2}d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{b\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{{a}^{4}d}}-2\,{\frac{b}{d{a}^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.14572, size = 146, normalized size = 1.54 \begin{align*} -\frac{\frac{6 \, b^{2} \tan \left (d x + c\right )^{2} + 9 \, a b \tan \left (d x + c\right ) + 2 \, a^{2}}{a^{3} b^{2} \tan \left (d x + c\right )^{3} + 2 \, a^{4} b \tan \left (d x + c\right )^{2} + a^{5} \tan \left (d x + c\right )} - \frac{6 \, b \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4}} + \frac{6 \, b \log \left (\tan \left (d x + c\right )\right )}{a^{4}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.4596, size = 1219, normalized size = 12.83 \begin{align*} \frac{2 \,{\left (a^{7} + 4 \, a^{5} b^{2} - 2 \, a^{3} b^{4} - 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \,{\left (2 \, a^{5} b^{2} - 3 \, a^{3} b^{4} - 3 \, a b^{6}\right )} \cos \left (d x + c\right ) + 3 \,{\left (2 \,{\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \,{\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right ) -{\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7} +{\left (a^{6} b + a^{4} b^{3} - a^{2} b^{5} - 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 (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \,{\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right ) -{\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7} +{\left (a^{6} b + a^{4} b^{3} - a^{2} b^{5} - 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 (5 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - 4 \,{\left (a^{6} b + 5 \, a^{4} b^{3} + 3 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (2 \,{\left (a^{9} b + 2 \, a^{7} b^{3} + a^{5} b^{5}\right )} d \cos \left (d x + c\right )^{3} - 2 \,{\left (a^{9} b + 2 \, a^{7} b^{3} + a^{5} b^{5}\right )} d \cos \left (d x + c\right ) -{\left ({\left (a^{10} + a^{8} b^{2} - a^{6} b^{4} - a^{4} b^{6}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{8} b^{2} + 2 \, a^{6} b^{4} + a^{4} b^{6}\right )} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a^7 + 4*a^5*b^2 - 2*a^3*b^4 - 3*a*b^6)*cos(d*x + c)^3 - 2*(2*a^5*b^2 - 3*a^3*b^4 - 3*a*b^6)*cos(d*x +
c) + 3*(2*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c)^3 - 2*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c) - (a^4*b
^3 + 2*a^2*b^5 + b^7 + (a^6*b + a^4*b^3 - a^2*b^5 - 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*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c)^3 - 2*(a^5*
b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c) - (a^4*b^3 + 2*a^2*b^5 + b^7 + (a^6*b + a^4*b^3 - a^2*b^5 - b^7)*cos(d*x
+ c)^2)*sin(d*x + c))*log(-1/4*cos(d*x + c)^2 + 1/4) - (5*a^4*b^3 + 3*a^2*b^5 - 4*(a^6*b + 5*a^4*b^3 + 3*a^2*
b^5)*cos(d*x + c)^2)*sin(d*x + c))/(2*(a^9*b + 2*a^7*b^3 + a^5*b^5)*d*cos(d*x + c)^3 - 2*(a^9*b + 2*a^7*b^3 +
a^5*b^5)*d*cos(d*x + c) - ((a^10 + a^8*b^2 - a^6*b^4 - a^4*b^6)*d*cos(d*x + c)^2 + (a^8*b^2 + 2*a^6*b^4 + a^4*
b^6)*d)*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.26987, size = 153, normalized size = 1.61 \begin{align*} \frac{\frac{6 \, b \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4}} - \frac{6 \, b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{4}} + \frac{2 \,{\left (3 \, b \tan \left (d x + c\right ) - a\right )}}{a^{4} \tan \left (d x + c\right )} - \frac{9 \, b^{3} \tan \left (d x + c\right )^{2} + 22 \, a b^{2} \tan \left (d x + c\right ) + 14 \, a^{2} b}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} a^{4}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*b*log(abs(b*tan(d*x + c) + a))/a^4 - 6*b*log(abs(tan(d*x + c)))/a^4 + 2*(3*b*tan(d*x + c) - a)/(a^4*tan
(d*x + c)) - (9*b^3*tan(d*x + c)^2 + 22*a*b^2*tan(d*x + c) + 14*a^2*b)/((b*tan(d*x + c) + a)^2*a^4))/d