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

Optimal. Leaf size=97 \[ -\frac {8 b \tan (e+f x)}{3 a^3 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {4 b \tan (e+f x)}{3 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot (e+f x)}{a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 97, 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 = {3663, 271, 192, 191} \[ -\frac {8 b \tan (e+f x)}{3 a^3 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {4 b \tan (e+f x)}{3 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot (e+f x)}{a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 1.05, size = 133, normalized size = 1.37 \[ -\frac {\cot (e+f x) \left (4 \left (3 a^2-8 b^2\right ) \cos (2 (e+f x))+\left (3 a^2-12 a b+8 b^2\right ) \cos (4 (e+f x))+3 \left (3 a^2+4 a b+8 b^2\right )\right ) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{6 \sqrt {2} a^3 f ((a-b) \cos (2 (e+f x))+a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*((3*(3*a^2 + 4*a*b + 8*b^2) + 4*(3*a^2 - 8*b^2)*Cos[2*(e + f*x)] + (3*a^2 - 12*a*b + 8*b^2)*Cos[4*(e + f*
x)])*Cot[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(Sqrt[2]*a^3*f*(a + b + (a - b)*Cos
[2*(e + f*x)])^2)

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fricas [A]  time = 10.47, size = 156, normalized size = 1.61 \[ -\frac {{\left ({\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 4 \, {\left (3 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \, b^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a^{3} b^{2} f + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.12, size = 153, normalized size = 1.58 \[ -\frac {\left (3 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2}-12 \left (\cos ^{4}\left (f x +e \right )\right ) a b +8 \left (\cos ^{4}\left (f x +e \right )\right ) b^{2}+12 \left (\cos ^{2}\left (f x +e \right )\right ) a b -16 b^{2} \left (\cos ^{2}\left (f x +e \right )\right )+8 b^{2}\right ) \left (\cos ^{5}\left (f x +e \right )\right ) \left (\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}\right )^{\frac {5}{2}}}{3 f \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{4} \sin \left (f x +e \right ) a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/f/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^4*(3*cos(f*x+e)^4*a^2-12*cos(f*x+e)^4*a*b+8*cos(f*x+e)^4*b^2+12*cos(f
*x+e)^2*a*b-16*b^2*cos(f*x+e)^2+8*b^2)*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)/a^3

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maxima [A]  time = 0.79, size = 85, normalized size = 0.88 \[ -\frac {\frac {8 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3}} + \frac {4 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 27.44, size = 324, normalized size = 3.34 \[ -\frac {\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {a+\frac {b\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}}\,\left (-a\,b\,12{}\mathrm {i}+a^2\,3{}\mathrm {i}+b^2\,8{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,12{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,18{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,12{}\mathrm {i}+a^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,3{}\mathrm {i}-b^2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,32{}\mathrm {i}+b^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,48{}\mathrm {i}-b^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,32{}\mathrm {i}+b^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,8{}\mathrm {i}+a\,b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,24{}\mathrm {i}-a\,b\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,12{}\mathrm {i}\right )}{3\,a^3\,f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )\,{\left (a-b+2\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}+2\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((exp(e*2i + f*x*2i) + 1)*(a + (b*(exp(e*2i + f*x*2i)*1i - 1i)^2)/(exp(e*2i + f*x*2i) + 1)^2)^(1/2)*(a^2*3i -
a*b*12i + b^2*8i + 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)*32i + b^2*exp(e*4i + f*x*4i)*48i - b^2*exp(e*6i + f*x*6i)*32i
+ b^2*exp(e*8i + f*x*8i)*8i + a*b*exp(e*4i + f*x*4i)*24i - a*b*exp(e*8i + f*x*8i)*12i))/(3*a^3*f*(exp(e*2i + f
*x*2i) - 1)*(a - b + 2*a*exp(e*2i + f*x*2i) + a*exp(e*4i + f*x*4i) + 2*b*exp(e*2i + f*x*2i) - b*exp(e*4i + f*x
*4i))^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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