3.150 \(\int \frac{\csc ^4(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=146 \[ -\frac{8 b (a-2 b) \tan (e+f x)}{3 a^4 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{4 b (a-2 b) \tan (e+f x)}{3 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(a-2 b) \cot (e+f x)}{a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x)}{3 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

[Out]

-(((a - 2*b)*Cot[e + f*x])/(a^2*f*(a + b*Tan[e + f*x]^2)^(3/2))) - Cot[e + f*x]^3/(3*a*f*(a + b*Tan[e + f*x]^2
)^(3/2)) - (4*(a - 2*b)*b*Tan[e + f*x])/(3*a^3*f*(a + b*Tan[e + f*x]^2)^(3/2)) - (8*(a - 2*b)*b*Tan[e + f*x])/
(3*a^4*f*Sqrt[a + b*Tan[e + f*x]^2])

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Rubi [A]  time = 0.149029, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3663, 453, 271, 192, 191} \[ -\frac{8 b (a-2 b) \tan (e+f x)}{3 a^4 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{4 b (a-2 b) \tan (e+f x)}{3 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(a-2 b) \cot (e+f x)}{a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x)}{3 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-(((a - 2*b)*Cot[e + f*x])/(a^2*f*(a + b*Tan[e + f*x]^2)^(3/2))) - Cot[e + f*x]^3/(3*a*f*(a + b*Tan[e + f*x]^2
)^(3/2)) - (4*(a - 2*b)*b*Tan[e + f*x])/(3*a^3*f*(a + b*Tan[e + f*x]^2)^(3/2)) - (8*(a - 2*b)*b*Tan[e + f*x])/
(3*a^4*f*Sqrt[a + b*Tan[e + f*x]^2])

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{x^4 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^3(e+f x)}{3 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(a-2 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=-\frac{(a-2 b) \cot (e+f x)}{a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x)}{3 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(4 (a-2 b) b) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{a^2 f}\\ &=-\frac{(a-2 b) \cot (e+f x)}{a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x)}{3 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{4 (a-2 b) b \tan (e+f x)}{3 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(8 (a-2 b) b) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a^3 f}\\ &=-\frac{(a-2 b) \cot (e+f x)}{a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\cot ^3(e+f x)}{3 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{4 (a-2 b) b \tan (e+f x)}{3 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{8 (a-2 b) b \tan (e+f x)}{3 a^4 f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.02165, size = 140, normalized size = 0.96 \[ \frac{\sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\frac{2 b \sin (2 (e+f x)) \left (\left (-3 a^2+7 a b-4 b^2\right ) \cos (2 (e+f x))-3 a^2+2 a b+4 b^2\right )}{((a-b) \cos (2 (e+f x))+a+b)^2}-\cot (e+f x) \left (a \csc ^2(e+f x)+2 a-8 b\right )\right )}{3 \sqrt{2} a^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(-(Cot[e + f*x]*(2*a - 8*b + a*Csc[e + f*x]^2)) + (2*
b*(-3*a^2 + 2*a*b + 4*b^2 + (-3*a^2 + 7*a*b - 4*b^2)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[
2*(e + f*x)])^2))/(3*Sqrt[2]*a^4*f)

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Maple [A]  time = 0.297, size = 245, normalized size = 1.7 \begin{align*}{\frac{ \left ( 2\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{3}-18\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{2}b+32\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}a{b}^{2}-16\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{b}^{3}-3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{3}+30\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}b-72\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}a{b}^{2}+48\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{3}-12\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}b+48\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}a{b}^{2}-48\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{3}-8\,a{b}^{2}+16\,{b}^{3} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{5}}{3\,f{a}^{4} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{4} \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ({\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x)

[Out]

1/3/f/a^4/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^4*(2*cos(f*x+e)^6*a^3-18*cos(f*x+e)^6*a^2*b+32*cos(f*x+e)^6*a*b^2-
16*cos(f*x+e)^6*b^3-3*cos(f*x+e)^4*a^3+30*cos(f*x+e)^4*a^2*b-72*cos(f*x+e)^4*a*b^2+48*cos(f*x+e)^4*b^3-12*cos(
f*x+e)^2*a^2*b+48*cos(f*x+e)^2*a*b^2-48*cos(f*x+e)^2*b^3-8*a*b^2+16*b^3)*cos(f*x+e)^5*((a*cos(f*x+e)^2-cos(f*x
+e)^2*b+b)/cos(f*x+e)^2)^(5/2)/sin(f*x+e)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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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(f*x+e)**4/(a+b*tan(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )^{4}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^4/(b*tan(f*x + e)^2 + a)^(5/2), x)