3.81 \(\int \csc ^3(a+b x) (d \tan (a+b x))^{5/2} \, dx\)

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

[Out]

-2/3*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 = 0.11, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2593, 2601, 2573, 2641} \[ \frac {2 d^2 \sqrt {\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (a+b x)}}{3 b}+\frac {2 d \csc (a+b x) (d \tan (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3*(d*Tan[a + b*x])^(5/2),x]

[Out]

(2*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]*(d*Tan[a + b*x])^(3/2))/(3*b)

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 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 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 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 ^3(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} \left (2 d^2\right ) \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 (2 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 (2 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 {2 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]  time = 0.35, size = 71, normalized size = 0.89 \[ \frac {2 d^2 \cos (a+b x) \sqrt {d \tan (a+b x)} \left (2 \sqrt {\sec ^2(a+b x)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2(a+b x)\right )+\sec ^2(a+b x)\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3*(d*Tan[a + b*x])^(5/2),x]

[Out]

(2*d^2*Cos[a + b*x]*(Sec[a + b*x]^2 + 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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \tan \left (b x + a\right )} d^{2} \csc \left (b x + a\right )^{3} \tan \left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3*(d*tan(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(b*x + a))*d^2*csc(b*x + a)^3*tan(b*x + a)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3*(d*tan(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate((d*tan(b*x + a))^(5/2)*csc(b*x + a)^3, x)

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maple [B]  time = 0.51, size = 192, normalized size = 2.40 \[ -\frac {\left (-1+\cos \left (b x +a \right )\right ) \left (2 \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \cos \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sin \left (b x +a \right )-\cos \left (b x +a \right ) \sqrt {2}+\sqrt {2}\right ) \cos \left (b x +a \right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}} \sqrt {2}}{3 b \sin \left (b x +a \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^3*(d*tan(b*x+a))^(5/2),x)

[Out]

-1/3/b*(-1+cos(b*x+a))*(2*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*cos(b*x+a)*((-1+
cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x
+a))^(1/2)*sin(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 \[ \int \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} \csc \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3*(d*tan(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*tan(b*x + a))^(5/2)*csc(b*x + a)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(a + b*x))^(5/2)/sin(a + b*x)^3,x)

[Out]

int((d*tan(a + b*x))^(5/2)/sin(a + b*x)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**3*(d*tan(b*x+a))**(5/2),x)

[Out]

Timed out

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