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

Optimal. Leaf size=76 \[ \frac{2 d \sin (a+b x) \sqrt{d \tan (a+b x)}}{b}-\frac{2 d^2 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.0940729, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2593, 2601, 2572, 2639} \[ \frac{2 d \sin (a+b x) \sqrt{d \tan (a+b x)}}{b}-\frac{2 d^2 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^2*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(b*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]]) + (2*d*Sin[
a + b*x]*Sqrt[d*Tan[a + b*x]])/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 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 2572

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

Rule 2639

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

Rubi steps

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

Mathematica [C]  time = 0.273229, size = 61, normalized size = 0.8 \[ -\frac{2 \cos (a+b x) (d \tan (a+b x))^{3/2} \left (2 \sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )-3\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[a + b*x]*(-3 + 2*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*(d*Tan[a + b*
x])^(3/2))/(3*b)

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Maple [B]  time = 0.148, size = 519, normalized size = 6.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*2^(1/2)*(cos(b*x+a)-1)^2*(2*cos(b*x+a)*EllipticE((-(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))/s
in(b*x+a))^(1/2)-cos(b*x+a)*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
)+2*EllipticE((-(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)*((c
os(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)-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))/si
n(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^(1/2)+2^(1/2))*cos(b*x+a)*(cos(b*x+
a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(3/2)/sin(b*x+a)^6

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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{3}{2}} \csc \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*tan(b*x + a))^(3/2)*csc(b*x + a), 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 \csc \left (b x + a\right ) \tan \left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*tan(b*x + a))*d*csc(b*x + a)*tan(b*x + a), 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)*(d*tan(b*x+a))**(3/2),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{3}{2}} \csc \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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