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

Optimal. Leaf size=106 \[ -\frac {8 b \tan (e+f x)}{3 f (a+b)^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {4 b \tan (e+f x)}{3 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {\cot (e+f x)}{f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

[Out]

-8/3*b*tan(f*x+e)/(a+b)^3/f/(a+b+b*tan(f*x+e)^2)^(1/2)-cot(f*x+e)/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)-4/3*b*tan
(f*x+e)/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(3/2)

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Rubi [A]  time = 0.11, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4132, 271, 192, 191} \[ -\frac {8 b \tan (e+f x)}{3 f (a+b)^3 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {4 b \tan (e+f x)}{3 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac {\cot (e+f x)}{f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 4132

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

Rubi steps

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

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Mathematica [A]  time = 1.97, size = 108, normalized size = 1.02 \[ -\frac {\tan ^3(e+f x) \sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-6 \left (a^2-b^2\right ) \csc ^2(e+f x)+3 a^2+3 (a+b)^2 \csc ^4(e+f x)-6 a b-b^2\right )}{6 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*((a + 2*b + a*Cos[2*(e + f*x)])*(3*a^2 - 6*a*b - b^2 - 6*(a^2 - b^2)*Csc[e + f*x]^2 + 3*(a + b)^2*Csc[e +
f*x]^4)*Sec[e + f*x]^2*Tan[e + f*x]^3)/((a + b)^3*f*(a + b*Sec[e + f*x]^2)^(5/2))

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fricas [A]  time = 1.46, size = 192, normalized size = 1.81 \[ -\frac {{\left ({\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{5} + 4 \, {\left (3 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \, b^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left ({\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/3*((3*a^2 - 6*a*b - b^2)*cos(f*x + e)^5 + 4*(3*a*b - b^2)*cos(f*x + e)^3 + 8*b^2*cos(f*x + e))*sqrt((a*cos(
f*x + e)^2 + b)/cos(f*x + e)^2)/(((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*f*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^
2 + 3*a^2*b^3 + a*b^4)*f*cos(f*x + e)^2 + (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f)*sin(f*x + e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{2}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.61, size = 146, normalized size = 1.38 \[ -\frac {\left (3 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2}-6 \left (\cos ^{4}\left (f x +e \right )\right ) a b -\left (\cos ^{4}\left (f x +e \right )\right ) b^{2}+12 \left (\cos ^{2}\left (f x +e \right )\right ) a b -4 b^{2} \left (\cos ^{2}\left (f x +e \right )\right )+8 b^{2}\right ) \left (\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}\right )^{\frac {5}{2}} \left (\cos ^{5}\left (f x +e \right )\right )}{3 f \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{4} \sin \left (f x +e \right ) \left (a^{2}+2 a b +b^{2}\right ) \left (a +b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/f/(b+a*cos(f*x+e)^2)^4*(3*cos(f*x+e)^4*a^2-6*cos(f*x+e)^4*a*b-cos(f*x+e)^4*b^2+12*cos(f*x+e)^2*a*b-4*b^2*
cos(f*x+e)^2+8*b^2)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(5/2)*cos(f*x+e)^5/sin(f*x+e)/(a^2+2*a*b+b^2)/(a+b)

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maxima [A]  time = 0.36, size = 94, normalized size = 0.89 \[ -\frac {\frac {8 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{3}} + \frac {4 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}^{2}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(8*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^3) + 4*b*tan(f*x + e)/((b*tan(f*x + e)^2 + a +
b)^(3/2)*(a + b)^2) + 3/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)*tan(f*x + e)))/f

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mupad [B]  time = 16.30, size = 336, normalized size = 3.17 \[ -\frac {\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {a+\frac {b}{{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}\right )}^2}}\,\left (-a\,b\,6{}\mathrm {i}+a^2\,3{}\mathrm {i}-b^2\,1{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,12{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,18{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,12{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,3{}\mathrm {i}-b^2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,20{}\mathrm {i}+b^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,90{}\mathrm {i}-b^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,20{}\mathrm {i}-b^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,1{}\mathrm {i}+a\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,24{}\mathrm {i}+a\,b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,60{}\mathrm {i}+a\,b\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,24{}\mathrm {i}-a\,b\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,6{}\mathrm {i}\right )}{3\,f\,{\left (a+b\right )}^3\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )\,{\left (a+2\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}+4\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^2*(a + b/cos(e + f*x)^2)^(5/2)),x)

[Out]

-((exp(e*2i + f*x*2i) + 1)*(a + b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2)^2)^(1/2)*(a^2*3i - a*b*6i -
b^2*1i + a^2*exp(e*2i + f*x*2i)*12i + a^2*exp(e*4i + f*x*4i)*18i + a^2*exp(e*6i + f*x*6i)*12i + a^2*exp(e*8i +
f*x*8i)*3i - b^2*exp(e*2i + f*x*2i)*20i + b^2*exp(e*4i + f*x*4i)*90i - b^2*exp(e*6i + f*x*6i)*20i - b^2*exp(e
*8i + f*x*8i)*1i + a*b*exp(e*2i + f*x*2i)*24i + a*b*exp(e*4i + f*x*4i)*60i + a*b*exp(e*6i + f*x*6i)*24i - a*b*
exp(e*8i + f*x*8i)*6i))/(3*f*(a + b)^3*(exp(e*2i + f*x*2i) - 1)*(a + 2*a*exp(e*2i + f*x*2i) + a*exp(e*4i + f*x
*4i) + 4*b*exp(e*2i + f*x*2i))^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**2/(a+b*sec(f*x+e)**2)**(5/2),x)

[Out]

Integral(csc(e + f*x)**2/(a + b*sec(e + f*x)**2)**(5/2), x)

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