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

Optimal. Leaf size=112 \[ -\frac{2 \sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{21 b d^2}-\frac{2 \csc ^3(a+b x)}{7 b d \sqrt{d \tan (a+b x)}}+\frac{2 \csc (a+b x)}{21 b d \sqrt{d \tan (a+b x)}} \]

[Out]

(2*Csc[a + b*x])/(21*b*d*Sqrt[d*Tan[a + b*x]]) - (2*Csc[a + b*x]^3)/(7*b*d*Sqrt[d*Tan[a + b*x]]) - (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]])/(21*b*d^2)

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Rubi [A]  time = 0.145929, antiderivative size = 112, 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 = {2597, 2599, 2601, 2573, 2641} \[ -\frac{2 \sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{21 b d^2}-\frac{2 \csc ^3(a+b x)}{7 b d \sqrt{d \tan (a+b x)}}+\frac{2 \csc (a+b x)}{21 b d \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Csc[a + b*x])/(21*b*d*Sqrt[d*Tan[a + b*x]]) - (2*Csc[a + b*x]^3)/(7*b*d*Sqrt[d*Tan[a + b*x]]) - (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]])/(21*b*d^2)

Rule 2597

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sin[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n + 1)), x] - Dist[(n + 1)/(b^2*(m + 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] && LtQ[n, -1] && NeQ[m + n + 1, 0] && Integer
sQ[2*m, 2*n] &&  !(EqQ[n, -3/2] && EqQ[m, 1])

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 \frac{\csc ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=-\frac{2 \csc ^3(a+b x)}{7 b d \sqrt{d \tan (a+b x)}}-\frac{\int \csc ^3(a+b x) \sqrt{d \tan (a+b x)} \, dx}{7 d^2}\\ &=\frac{2 \csc (a+b x)}{21 b d \sqrt{d \tan (a+b x)}}-\frac{2 \csc ^3(a+b x)}{7 b d \sqrt{d \tan (a+b x)}}-\frac{2 \int \csc (a+b x) \sqrt{d \tan (a+b x)} \, dx}{21 d^2}\\ &=\frac{2 \csc (a+b x)}{21 b d \sqrt{d \tan (a+b x)}}-\frac{2 \csc ^3(a+b x)}{7 b d \sqrt{d \tan (a+b x)}}-\frac{\left (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}{21 d^2 \sqrt{\sin (a+b x)}}\\ &=\frac{2 \csc (a+b x)}{21 b d \sqrt{d \tan (a+b x)}}-\frac{2 \csc ^3(a+b x)}{7 b d \sqrt{d \tan (a+b x)}}-\frac{\left (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}{21 d^2}\\ &=\frac{2 \csc (a+b x)}{21 b d \sqrt{d \tan (a+b x)}}-\frac{2 \csc ^3(a+b x)}{7 b d \sqrt{d \tan (a+b x)}}-\frac{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)}}{21 b d^2}\\ \end{align*}

Mathematica [C]  time = 1.67093, size = 136, normalized size = 1.21 \[ \frac{\csc ^3(a+b x) \left ((10 \cos (2 (a+b x))+\cos (4 (a+b x))+1) \sec ^2(a+b x)^{3/2}-8 \sqrt [4]{-1} \cos (2 (a+b x)) \tan ^{\frac{7}{2}}(a+b x) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (a+b x)}\right )\right |-1\right )\right )}{42 b d \left (\tan ^2(a+b x)-1\right ) \sqrt{\sec ^2(a+b x)} \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[a + b*x]^3*((1 + 10*Cos[2*(a + b*x)] + Cos[4*(a + b*x)])*(Sec[a + b*x]^2)^(3/2) - 8*(-1)^(1/4)*Cos[2*(a +
b*x)]*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Tan[a + b*x]^(7/2)))/(42*b*d*Sqrt[Sec[a + b*x]^
2]*Sqrt[d*Tan[a + b*x]]*(-1 + Tan[a + b*x]^2))

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Maple [B]  time = 0.169, size = 566, normalized size = 5.1 \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)^3/(d*tan(b*x+a))^(3/2),x)

[Out]

1/21/b*2^(1/2)*(cos(b*x+a)-1)^2*(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))/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)^3*sin(b*x+a)+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))/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)-2*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))/
sin(b*x+a))^(1/2),1/2*2^(1/2))-2*sin(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)-cos(b*x+a)^3*2^(1/2)-2*cos(b*x+a)*2^(1/2))*(cos(b*x+a)+1)^2/cos(b*x+a)^2/sin(b*x+a)^6/(d*sin
(b*x+a)/cos(b*x+a))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )^{3}}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \tan \left (b x + a\right )} \csc \left (b x + a\right )^{3}}{d^{2} \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)^3/(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{3}{\left (a + b x \right )}}{\left (d \tan{\left (a + b x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

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]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )^{3}}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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