3.82 \(\int \csc ^5(a+b x) (d \tan (a+b x))^{5/2} \, dx\)

Optimal. Leaf size=110 \[ -\frac{4 d^3 \csc (a+b x)}{3 b \sqrt{d \tan (a+b x)}}+\frac{4 d^2 \sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{3 b}+\frac{2 d \csc ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b} \]

[Out]

(-4*d^3*Csc[a + b*x])/(3*b*Sqrt[d*Tan[a + b*x]]) + (4*d^2*Csc[a + b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2
*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(3*b) + (2*d*Csc[a + b*x]^3*(d*Tan[a + b*x])^(3/2))/(3*b)

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Rubi [A]  time = 0.146866, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2593, 2599, 2601, 2573, 2641} \[ -\frac{4 d^3 \csc (a+b x)}{3 b \sqrt{d \tan (a+b x)}}+\frac{4 d^2 \sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{3 b}+\frac{2 d \csc ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*d^3*Csc[a + b*x])/(3*b*Sqrt[d*Tan[a + b*x]]) + (4*d^2*Csc[a + b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2
*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(3*b) + (2*d*Csc[a + b*x]^3*(d*Tan[a + b*x])^(3/2))/(3*b)

Rule 2593

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sin[e +
f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 1))/(a^2*f*(n - 1)), x] - Dist[(b^2*(m + 2))/(a^2*(n - 1)), Int[(a*Sin[e
+ f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ
[m, -1] && EqQ[n, 3/2])) && IntegersQ[2*m, 2*n]

Rule 2599

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(b*(a*Sin[e
+ f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 1))/(a^2*f*(m + n + 1)), x] + Dist[(m + 2)/(a^2*(m + n + 1)), Int[(a*Sin
[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n + 1, 0]
&& IntegersQ[2*m, 2*n]

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \csc ^5(a+b x) (d \tan (a+b x))^{5/2} \, dx &=\frac{2 d \csc ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}+\left (2 d^2\right ) \int \csc ^3(a+b x) \sqrt{d \tan (a+b x)} \, dx\\ &=-\frac{4 d^3 \csc (a+b x)}{3 b \sqrt{d \tan (a+b x)}}+\frac{2 d \csc ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}+\frac{1}{3} \left (4 d^2\right ) \int \csc (a+b x) \sqrt{d \tan (a+b x)} \, dx\\ &=-\frac{4 d^3 \csc (a+b x)}{3 b \sqrt{d \tan (a+b x)}}+\frac{2 d \csc ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}+\frac{\left (4 d^2 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)}} \, dx}{3 \sqrt{\sin (a+b x)}}\\ &=-\frac{4 d^3 \csc (a+b x)}{3 b \sqrt{d \tan (a+b x)}}+\frac{2 d \csc ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}+\frac{1}{3} \left (4 d^2 \csc (a+b x) \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx\\ &=-\frac{4 d^3 \csc (a+b x)}{3 b \sqrt{d \tan (a+b x)}}+\frac{4 d^2 \csc (a+b x) F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}{3 b}+\frac{2 d \csc ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\\ \end{align*}

Mathematica [C]  time = 0.469204, size = 110, normalized size = 1. \[ -\frac{2 d \csc ^3(a+b x) (d \tan (a+b x))^{3/2} \left (\cos (2 (a+b x)) \sqrt{\sec ^2(a+b x)}+2 \sqrt [4]{-1} \sin (2 (a+b x)) \sqrt{\tan (a+b x)} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (a+b x)}\right )\right |-1\right )\right )}{3 b \sqrt{\sec ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*Csc[a + b*x]^3*(Cos[2*(a + b*x)]*Sqrt[Sec[a + b*x]^2] + 2*(-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt
[Tan[a + b*x]]], -1]*Sin[2*(a + b*x)]*Sqrt[Tan[a + b*x]])*(d*Tan[a + b*x])^(3/2))/(3*b*Sqrt[Sec[a + b*x]^2])

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Maple [B]  time = 0.158, size = 320, normalized size = 2.9 \begin{align*}{\frac{\sqrt{2} \left ( \cos \left ( bx+a \right ) -1 \right ) ^{2}\cos \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{3\,b \left ( \sin \left ( bx+a \right ) \right ) ^{8}} \left ( 4\,{\it EllipticF} \left ( \sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}} \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sin \left ( bx+a \right ) +4\,\cos \left ( bx+a \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sin \left ( bx+a \right ){\it EllipticF} \left ( \sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) -2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}+\sqrt{2} \right ) \left ({\frac{d\sin \left ( bx+a \right ) }{\cos \left ( bx+a \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/b*2^(1/2)*(cos(b*x+a)-1)^2*(4*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b
*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a)
)^(1/2)*cos(b*x+a)^2*sin(b*x+a)+4*cos(b*x+a)*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(
b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*sin(b*x+a)*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/s
in(b*x+a))^(1/2),1/2*2^(1/2))-2*cos(b*x+a)^2*2^(1/2)+2^(1/2))*cos(b*x+a)*(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*
x+a))^(5/2)/sin(b*x+a)^8

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{d \tan \left (b x + a\right )} d^{2} \csc \left (b x + a\right )^{5} \tan \left (b x + a\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*tan(b*x + a))*d^2*csc(b*x + a)^5*tan(b*x + a)^2, x)

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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(b*x+a)**5*(d*tan(b*x+a))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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