3.138 \(\int \frac{\csc ^4(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=114 \[ -\frac{2 b (3 a-4 b) \tan (e+f x)}{3 a^3 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{(3 a-4 b) \cot (e+f x)}{3 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^3(e+f x)}{3 a f \sqrt{a+b \tan ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.121558, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3663, 453, 271, 191} \[ -\frac{2 b (3 a-4 b) \tan (e+f x)}{3 a^3 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{(3 a-4 b) \cot (e+f x)}{3 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^3(e+f x)}{3 a f \sqrt{a+b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

-((3*a - 4*b)*Cot[e + f*x])/(3*a^2*f*Sqrt[a + b*Tan[e + f*x]^2]) - Cot[e + f*x]^3/(3*a*f*Sqrt[a + b*Tan[e + f*
x]^2]) - (2*(3*a - 4*b)*b*Tan[e + f*x])/(3*a^3*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 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 )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^3(e+f x)}{3 a f \sqrt{a+b \tan ^2(e+f x)}}+\frac{(3 a-4 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a f}\\ &=-\frac{(3 a-4 b) \cot (e+f x)}{3 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^3(e+f x)}{3 a f \sqrt{a+b \tan ^2(e+f x)}}-\frac{(2 (3 a-4 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^2 f}\\ &=-\frac{(3 a-4 b) \cot (e+f x)}{3 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^3(e+f x)}{3 a f \sqrt{a+b \tan ^2(e+f x)}}-\frac{2 (3 a-4 b) b \tan (e+f x)}{3 a^3 f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.868828, size = 119, normalized size = 1.04 \[ \frac{\csc ^3(e+f x) \sec (e+f x) \left (-2 \left (a^2-6 a b+8 b^2\right ) \cos (2 (e+f x))+\left (a^2-5 a b+4 b^2\right ) \cos (4 (e+f x))-3 a^2-7 a b+12 b^2\right )}{6 \sqrt{2} a^3 f \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

((-3*a^2 - 7*a*b + 12*b^2 - 2*(a^2 - 6*a*b + 8*b^2)*Cos[2*(e + f*x)] + (a^2 - 5*a*b + 4*b^2)*Cos[4*(e + f*x)])
*Csc[e + f*x]^3*Sec[e + f*x])/(6*Sqrt[2]*a^3*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])

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Maple [A]  time = 0.194, size = 170, normalized size = 1.5 \begin{align*}{\frac{ \left ( 2\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}-10\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}ab+8\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{2}-3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}+16\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}ab-16\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{2}-6\,ab+8\,{b}^{2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{3\,f{a}^{3} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2} \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{3}{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)^(3/2),x)

[Out]

1/3/f/a^3/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*(2*cos(f*x+e)^4*a^2-10*cos(f*x+e)^4*a*b+8*cos(f*x+e)^4*b^2-3*cos
(f*x+e)^2*a^2+16*cos(f*x+e)^2*a*b-16*cos(f*x+e)^2*b^2-6*a*b+8*b^2)*cos(f*x+e)^3*((a*cos(f*x+e)^2-cos(f*x+e)^2*
b+b)/cos(f*x+e)^2)^(3/2)/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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 27.5993, size = 359, normalized size = 3.15 \begin{align*} -\frac{{\left (2 \,{\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{5} -{\left (3 \, a^{2} - 16 \, a b + 16 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 2 \,{\left (3 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \,{\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f -{\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )} \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)^(3/2),x, algorithm="fricas")

[Out]

-1/3*(2*(a^2 - 5*a*b + 4*b^2)*cos(f*x + e)^5 - (3*a^2 - 16*a*b + 16*b^2)*cos(f*x + e)^3 - 2*(3*a*b - 4*b^2)*co
s(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(((a^4 - a^3*b)*f*cos(f*x + e)^4 - a^3*b*f - (a^
4 - 2*a^3*b)*f*cos(f*x + e)^2)*sin(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{4}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \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)**(3/2),x)

[Out]

Integral(csc(e + f*x)**4/(a + b*tan(e + f*x)**2)**(3/2), x)

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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{3}{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)^(3/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^4/(b*tan(f*x + e)^2 + a)^(3/2), x)