3.93 \(\int \frac{\csc ^3(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\)

Optimal. Leaf size=102 \[ -\frac{4 \cos (a+b x)}{5 b \sqrt{d \tan (a+b x)}}-\frac{2 d \csc (a+b x)}{5 b (d \tan (a+b x))^{3/2}}-\frac{4 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{5 b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}} \]

[Out]

(-2*d*Csc[a + b*x])/(5*b*(d*Tan[a + b*x])^(3/2)) - (4*Cos[a + b*x])/(5*b*Sqrt[d*Tan[a + b*x]]) - (4*EllipticE[
a - Pi/4 + b*x, 2]*Sin[a + b*x])/(5*b*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])

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Rubi [A]  time = 0.133896, antiderivative size = 102, 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 = {2599, 2601, 2570, 2572, 2639} \[ -\frac{4 \cos (a+b x)}{5 b \sqrt{d \tan (a+b x)}}-\frac{2 d \csc (a+b x)}{5 b (d \tan (a+b x))^{3/2}}-\frac{4 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{5 b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3/Sqrt[d*Tan[a + b*x]],x]

[Out]

(-2*d*Csc[a + b*x])/(5*b*(d*Tan[a + b*x])^(3/2)) - (4*Cos[a + b*x])/(5*b*Sqrt[d*Tan[a + b*x]]) - (4*EllipticE[
a - Pi/4 + b*x, 2]*Sin[a + b*x])/(5*b*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])

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 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 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 \frac{\csc ^3(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx &=-\frac{2 d \csc (a+b x)}{5 b (d \tan (a+b x))^{3/2}}+\frac{2}{5} \int \frac{\csc (a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=-\frac{2 d \csc (a+b x)}{5 b (d \tan (a+b x))^{3/2}}+\frac{\left (2 \sqrt{\sin (a+b x)}\right ) \int \frac{\sqrt{\cos (a+b x)}}{\sin ^{\frac{3}{2}}(a+b x)} \, dx}{5 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{2 d \csc (a+b x)}{5 b (d \tan (a+b x))^{3/2}}-\frac{4 \cos (a+b x)}{5 b \sqrt{d \tan (a+b x)}}-\frac{\left (4 \sqrt{\sin (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)} \, dx}{5 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{2 d \csc (a+b x)}{5 b (d \tan (a+b x))^{3/2}}-\frac{4 \cos (a+b x)}{5 b \sqrt{d \tan (a+b x)}}-\frac{(4 \sin (a+b x)) \int \sqrt{\sin (2 a+2 b x)} \, dx}{5 \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{2 d \csc (a+b x)}{5 b (d \tan (a+b x))^{3/2}}-\frac{4 \cos (a+b x)}{5 b \sqrt{d \tan (a+b x)}}-\frac{4 E\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sin (a+b x)}{5 b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.673756, size = 104, normalized size = 1.02 \[ \frac{6 (\cos (2 (a+b x))-2) \cot (a+b x) \csc (a+b x) \sqrt{\sec ^2(a+b x)}-8 \tan ^2(a+b x) \sec (a+b x) \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )}{15 b \sqrt{\sec ^2(a+b x)} \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3/Sqrt[d*Tan[a + b*x]],x]

[Out]

(6*(-2 + Cos[2*(a + b*x)])*Cot[a + b*x]*Csc[a + b*x]*Sqrt[Sec[a + b*x]^2] - 8*Hypergeometric2F1[3/4, 3/2, 7/4,
-Tan[a + b*x]^2]*Sec[a + b*x]*Tan[a + b*x]^2)/(15*b*Sqrt[Sec[a + b*x]^2]*Sqrt[d*Tan[a + b*x]])

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Maple [B]  time = 0.186, size = 972, normalized size = 9.5 \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))^(1/2),x)

[Out]

-1/5/b*2^(1/2)*(4*cos(b*x+a)^3*((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)*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2
))-2*cos(b*x+a)^3*((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)*((
cos(b*x+a)-1)/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+4*cos(b*x+
a)^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)*((cos(b*x+a)-1)
/sin(b*x+a))^(1/2)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-2*cos(b*x+a)^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)*((cos(b*x+a)-1)/sin(b*x+a))^
(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-4*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))*((cos(b*x+a)-1)/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)-1+sin(b*x+a))/sin(b*x+a))^(1/2)+2*cos(b*x+a)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/s
in(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/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)-1+sin(b*x+a))/sin(b*x+a))^(1/2)-2*cos(b*x+a)^3*2^(1/2)-4*EllipticE(((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)/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((co
s(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)+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)/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)-1+sin(b*x+a))/sin
(b*x+a))^(1/2)+cos(b*x+a)^2*2^(1/2)+2*cos(b*x+a)*2^(1/2))/sin(b*x+a)^2/(d*sin(b*x+a)/cos(b*x+a))^(1/2)/cos(b*x
+a)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )^{3}}{\sqrt{d \tan \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)^3/(d*tan(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^3/sqrt(d*tan(b*x + a)), 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 \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)^3/(d*tan(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(b*x + a))*csc(b*x + a)^3/(d*tan(b*x + a)), 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 )}}{\sqrt{d \tan{\left (a + b x \right )}}}\, 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))**(1/2),x)

[Out]

Integral(csc(a + b*x)**3/sqrt(d*tan(a + b*x)), 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}}{\sqrt{d \tan \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)^3/(d*tan(b*x+a))^(1/2),x, algorithm="giac")

[Out]

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