3.1.80 \(\int \csc (a+b x) (d \tan (a+b x))^{5/2} \, dx\) [80]

Optimal. Leaf size=80 \[ -\frac {d^2 \csc (a+b x) F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{3 b}+\frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b} \]

[Out]

1/3*d^2*csc(b*x+a)*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))*sin(2*b*
x+2*a)^(1/2)*(d*tan(b*x+a))^(1/2)/b+2/3*d*csc(b*x+a)*(d*tan(b*x+a))^(3/2)/b

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Rubi [A]
time = 0.08, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2674, 2681, 2653, 2720} \begin {gather*} \frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(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]])/b + (2*d*Csc[
a + b*x]*(d*Tan[a + b*x])^(3/2))/(3*b)

Rule 2653

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 2674

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

Rule 2681

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 2720

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

Rubi steps

\begin {align*} \int \csc (a+b x) (d \tan (a+b x))^{5/2} \, dx &=\frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {1}{3} d^2 \int \csc (a+b x) \sqrt {d \tan (a+b x)} \, dx\\ &=\frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {\left (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 {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {1}{3} \left (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 {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 (a+b x) (d \tan (a+b x))^{3/2}}{3 b}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.44, size = 71, normalized size = 0.89 \begin {gather*} \frac {2 d^2 \cos (a+b x) \left (\sec ^2(a+b x)-\, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sqrt {d \tan (a+b x)}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^2*Cos[a + b*x]*(Sec[a + b*x]^2 - Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*
Sqrt[d*Tan[a + b*x]])/(3*b)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(95)=190\).
time = 0.32, size = 192, normalized size = 2.40

method	result	size
default	\(\frac {\left (-1+\cos \left (b x +a \right )\right ) \left (\EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sin \left (b x +a \right ) \cos \left (b x +a \right )+\cos \left (b x +a \right ) \sqrt {2}-\sqrt {2}\right ) \cos \left (b x +a \right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}} \sqrt {2}}{3 b \sin \left (b x +a \right )^{6}}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/b*(-1+cos(b*x+a))*(EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((-1+cos(b*x+a))/si
n(b*x+a))^(1/2)*((1-cos(b*x+a)+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)*cos(b*x+a)+cos(b*x+a)*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)^6*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains complex when optimal does not.
time = 0.12, size = 101, normalized size = 1.26 \begin {gather*} \frac {\sqrt {i \, d} d^{2} \cos \left (b x + a\right ) {\rm ellipticF}\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ), -1\right ) + \sqrt {-i \, d} d^{2} \cos \left (b x + a\right ) {\rm ellipticF}\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ), -1\right ) + 2 \, d^{2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{3 \, b \cos \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(I*d)*d^2*cos(b*x + a)*ellipticF(cos(b*x + a) + I*sin(b*x + a), -1) + sqrt(-I*d)*d^2*cos(b*x + a)*ell
ipticF(cos(b*x + a) - I*sin(b*x + a), -1) + 2*d^2*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(b*cos(b*x + a))

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6190 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2}}{\sin \left (a+b\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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