3.65 \(\int \frac{\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=140 \[ -\frac{b \left (a^2+b^2\right )}{a^4 d (a+b \tan (c+d x))}-\frac{\left (a^2+3 b^2\right ) \cot (c+d x)}{a^4 d}-\frac{2 b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{a^5 d}+\frac{2 b \left (a^2+2 b^2\right ) \log (a+b \tan (c+d x))}{a^5 d}+\frac{b \cot ^2(c+d x)}{a^3 d}-\frac{\cot ^3(c+d x)}{3 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.122497, antiderivative size = 140, 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, 894} \[ -\frac{b \left (a^2+b^2\right )}{a^4 d (a+b \tan (c+d x))}-\frac{\left (a^2+3 b^2\right ) \cot (c+d x)}{a^4 d}-\frac{2 b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{a^5 d}+\frac{2 b \left (a^2+2 b^2\right ) \log (a+b \tan (c+d x))}{a^5 d}+\frac{b \cot ^2(c+d x)}{a^3 d}-\frac{\cot ^3(c+d x)}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 + 3*b^2)*Cot[c + d*x])/(a^4*d)) + (b*Cot[c + d*x]^2)/(a^3*d) - Cot[c + d*x]^3/(3*a^2*d) - (2*b*(a^2 +
2*b^2)*Log[Tan[c + d*x]])/(a^5*d) + (2*b*(a^2 + 2*b^2)*Log[a + b*Tan[c + d*x]])/(a^5*d) - (b*(a^2 + b^2))/(a^4
*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 894

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]))

Rubi steps

\begin{align*} \int \frac{\csc ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{b^2+x^2}{x^4 (a+x)^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac{b \operatorname{Subst}\left (\int \left (\frac{b^2}{a^2 x^4}-\frac{2 b^2}{a^3 x^3}+\frac{a^2+3 b^2}{a^4 x^2}-\frac{2 \left (a^2+2 b^2\right )}{a^5 x}+\frac{a^2+b^2}{a^4 (a+x)^2}+\frac{2 \left (a^2+2 b^2\right )}{a^5 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac{\left (a^2+3 b^2\right ) \cot (c+d x)}{a^4 d}+\frac{b \cot ^2(c+d x)}{a^3 d}-\frac{\cot ^3(c+d x)}{3 a^2 d}-\frac{2 b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{a^5 d}+\frac{2 b \left (a^2+2 b^2\right ) \log (a+b \tan (c+d x))}{a^5 d}-\frac{b \left (a^2+b^2\right )}{a^4 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.50639, size = 244, normalized size = 1.74 \[ \frac{-\cot ^2(c+d x) \left (9 a^2 b^2+a^4 \csc ^2(c+d x)+2 a^4\right )+3 b^2 \left (-2 \left (a^2+2 b^2\right ) \log (\sin (c+d x))+2 a^2 \log (a \cos (c+d x)+b \sin (c+d x))+a^2 \csc ^2(c+d x)+a^2+4 b^2 \log (a \cos (c+d x)+b \sin (c+d x))+b^2\right )+a b \cot (c+d x) \left (-6 \left (a^2+2 b^2\right ) \log (\sin (c+d x))+6 a^2 \log (a \cos (c+d x)+b \sin (c+d x))+2 a^2 \csc ^2(c+d x)-2 a^2+12 b^2 \log (a \cos (c+d x)+b \sin (c+d x))-9 b^2\right )}{3 a^5 d (a \cot (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Cot[c + d*x]^2*(2*a^4 + 9*a^2*b^2 + a^4*Csc[c + d*x]^2)) + 3*b^2*(a^2 + b^2 + a^2*Csc[c + d*x]^2 - 2*(a^2 +
2*b^2)*Log[Sin[c + d*x]] + 2*a^2*Log[a*Cos[c + d*x] + b*Sin[c + d*x]] + 4*b^2*Log[a*Cos[c + d*x] + b*Sin[c +
d*x]]) + a*b*Cot[c + d*x]*(-2*a^2 - 9*b^2 + 2*a^2*Csc[c + d*x]^2 - 6*(a^2 + 2*b^2)*Log[Sin[c + d*x]] + 6*a^2*L
og[a*Cos[c + d*x] + b*Sin[c + d*x]] + 12*b^2*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]))/(3*a^5*d*(b + a*Cot[c + d*
x]))

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Maple [A]  time = 0.153, size = 189, normalized size = 1.4 \begin{align*} -{\frac{1}{3\,{a}^{2}d \left ( \tan \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{{a}^{2}d\tan \left ( dx+c \right ) }}-3\,{\frac{{b}^{2}}{d{a}^{4}\tan \left ( dx+c \right ) }}+{\frac{b}{d{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}-2\,{\frac{b\ln \left ( \tan \left ( dx+c \right ) \right ) }{d{a}^{3}}}-4\,{\frac{{b}^{3}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d{a}^{5}}}-{\frac{b}{{a}^{2}d \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{{b}^{3}}{d{a}^{4} \left ( a+b\tan \left ( dx+c \right ) \right ) }}+2\,{\frac{b\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d{a}^{3}}}+4\,{\frac{{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d{a}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.09184, size = 194, normalized size = 1.39 \begin{align*} \frac{\frac{2 \, a^{2} b \tan \left (d x + c\right ) - 6 \,{\left (a^{2} b + 2 \, b^{3}\right )} \tan \left (d x + c\right )^{3} - a^{3} - 3 \,{\left (a^{3} + 2 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{a^{4} b \tan \left (d x + c\right )^{4} + a^{5} \tan \left (d x + c\right )^{3}} + \frac{6 \,{\left (a^{2} b + 2 \, b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{5}} - \frac{6 \,{\left (a^{2} b + 2 \, b^{3}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{5}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.37406, size = 1004, normalized size = 7.17 \begin{align*} \frac{2 \,{\left (a^{4} + 6 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{4} + 6 \, a^{2} b^{2} - 3 \,{\left (a^{4} + 6 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left ({\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{2} b^{2} + 2 \, b^{4} - 2 \,{\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} -{\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )\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 ({\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{2} b^{2} + 2 \, b^{4} - 2 \,{\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} -{\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac{1}{4} \, \cos \left (d x + c\right )^{2} + \frac{1}{4}\right ) - 2 \,{\left (6 \, a b^{3} \cos \left (d x + c\right ) -{\left (a^{3} b + 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3 \,{\left (a^{5} b d \cos \left (d x + c\right )^{4} - 2 \, a^{5} b d \cos \left (d x + c\right )^{2} + a^{5} b d -{\left (a^{6} d \cos \left (d x + c\right )^{3} - a^{6} d \cos \left (d x + c\right )\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)^4/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.29818, size = 274, normalized size = 1.96 \begin{align*} -\frac{\frac{6 \,{\left (a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{5}} - \frac{6 \,{\left (a^{2} b^{2} + 2 \, b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{5} b} + \frac{3 \,{\left (2 \, a^{2} b^{2} \tan \left (d x + c\right ) + 4 \, b^{4} \tan \left (d x + c\right ) + 3 \, a^{3} b + 5 \, a b^{3}\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )} a^{5}} - \frac{11 \, a^{2} b \tan \left (d x + c\right )^{3} + 22 \, b^{3} \tan \left (d x + c\right )^{3} - 3 \, a^{3} \tan \left (d x + c\right )^{2} - 9 \, a b^{2} \tan \left (d x + c\right )^{2} + 3 \, a^{2} b \tan \left (d x + c\right ) - a^{3}}{a^{5} \tan \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*(a^2*b + 2*b^3)*log(abs(tan(d*x + c)))/a^5 - 6*(a^2*b^2 + 2*b^4)*log(abs(b*tan(d*x + c) + a))/(a^5*b)
+ 3*(2*a^2*b^2*tan(d*x + c) + 4*b^4*tan(d*x + c) + 3*a^3*b + 5*a*b^3)/((b*tan(d*x + c) + a)*a^5) - (11*a^2*b*t
an(d*x + c)^3 + 22*b^3*tan(d*x + c)^3 - 3*a^3*tan(d*x + c)^2 - 9*a*b^2*tan(d*x + c)^2 + 3*a^2*b*tan(d*x + c) -
a^3)/(a^5*tan(d*x + c)^3))/d