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

Optimal. Leaf size=102 \[ -\frac {4 d^2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {4 d^2 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 102, normalized size of antiderivative = 1.00, 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, 2601, 2570, 2572, 2639} \[ -\frac {4 d^2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {4 d^2 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \csc (a+b x) \sqrt {d \tan (a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*d^2*Cos[a + b*x])/(b*Sqrt[d*Tan[a + b*x]]) - (4*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*Csc[a + b*x]*Sqrt[d*Tan[a + b*x]])/b

Rule 2570

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

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

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Mathematica [C]  time = 0.59, size = 71, normalized size = 0.70 \[ -\frac {2 \cos (a+b x) (d \tan (a+b x))^{3/2} \left (4 \sqrt {\sec ^2(a+b x)} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(a+b x)\right )+3 \csc ^2(a+b x)-6\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[a + b*x]*(-6 + 3*Csc[a + b*x]^2 + 4*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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \tan \left (b x + a\right )} d \csc \left (b x + a\right )^{3} \tan \left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.57, size = 491, normalized size = 4.81 \[ \frac {\left (4 \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 )}}\, \cos \left (b x +a \right ) \EllipticE \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 )-2 \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 )}}\, \cos \left (b x +a \right ) \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 )+4 \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 )}}\, \EllipticE \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 )-2 \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 )}}\, \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 )-2 \cos \left (b x +a \right ) \sqrt {2}+\sqrt {2}\right ) \cos \left (b x +a \right ) \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sqrt {2}}{b \sin \left (b x +a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(4*((-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)*cos(b*x+a)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-2*((-1+c
os(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)*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))+4*((-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)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*Ell
ipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-2*((-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)*EllipticF(((1-cos(b*x+a)+si
n(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-2*cos(b*x+a)*2^(1/2)+2^(1/2))*cos(b*x+a)*(d*sin(b*x+a)/cos(b*x+a))^(3
/2)/sin(b*x+a)^2*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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