3.396 \(\int \csc ^4(e+f x) (b \sec (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=124 \[ -\frac{7 b^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{2 f \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{b \csc ^3(e+f x) \sqrt{b \sec (e+f x)}}{3 f}-\frac{7 b \csc (e+f x) \sqrt{b \sec (e+f x)}}{6 f}+\frac{7 b \sin (e+f x) \sqrt{b \sec (e+f x)}}{2 f} \]

[Out]

(-7*b^2*EllipticE[(e + f*x)/2, 2])/(2*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (7*b*Csc[e + f*x]*Sqrt[b*Se
c[e + f*x]])/(6*f) - (b*Csc[e + f*x]^3*Sqrt[b*Sec[e + f*x]])/(3*f) + (7*b*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x])/(
2*f)

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Rubi [A]  time = 0.128983, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2625, 3768, 3771, 2639} \[ -\frac{7 b^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{2 f \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{b \csc ^3(e+f x) \sqrt{b \sec (e+f x)}}{3 f}-\frac{7 b \csc (e+f x) \sqrt{b \sec (e+f x)}}{6 f}+\frac{7 b \sin (e+f x) \sqrt{b \sec (e+f x)}}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4*(b*Sec[e + f*x])^(3/2),x]

[Out]

(-7*b^2*EllipticE[(e + f*x)/2, 2])/(2*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (7*b*Csc[e + f*x]*Sqrt[b*Se
c[e + f*x]])/(6*f) - (b*Csc[e + f*x]^3*Sqrt[b*Sec[e + f*x]])/(3*f) + (7*b*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x])/(
2*f)

Rule 2625

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> -Simp[(a*b*(a*Csc
[e + f*x])^(m - 1)*(b*Sec[e + f*x])^(n - 1))/(f*(m - 1)), x] + Dist[(a^2*(m + n - 2))/(m - 1), Int[(a*Csc[e +
f*x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && IntegersQ[2*m, 2*n] &&
!GtQ[n, m]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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 ^4(e+f x) (b \sec (e+f x))^{3/2} \, dx &=-\frac{b \csc ^3(e+f x) \sqrt{b \sec (e+f x)}}{3 f}+\frac{7}{6} \int \csc ^2(e+f x) (b \sec (e+f x))^{3/2} \, dx\\ &=-\frac{7 b \csc (e+f x) \sqrt{b \sec (e+f x)}}{6 f}-\frac{b \csc ^3(e+f x) \sqrt{b \sec (e+f x)}}{3 f}+\frac{7}{4} \int (b \sec (e+f x))^{3/2} \, dx\\ &=-\frac{7 b \csc (e+f x) \sqrt{b \sec (e+f x)}}{6 f}-\frac{b \csc ^3(e+f x) \sqrt{b \sec (e+f x)}}{3 f}+\frac{7 b \sqrt{b \sec (e+f x)} \sin (e+f x)}{2 f}-\frac{1}{4} \left (7 b^2\right ) \int \frac{1}{\sqrt{b \sec (e+f x)}} \, dx\\ &=-\frac{7 b \csc (e+f x) \sqrt{b \sec (e+f x)}}{6 f}-\frac{b \csc ^3(e+f x) \sqrt{b \sec (e+f x)}}{3 f}+\frac{7 b \sqrt{b \sec (e+f x)} \sin (e+f x)}{2 f}-\frac{\left (7 b^2\right ) \int \sqrt{\cos (e+f x)} \, dx}{4 \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=-\frac{7 b^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{2 f \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{7 b \csc (e+f x) \sqrt{b \sec (e+f x)}}{6 f}-\frac{b \csc ^3(e+f x) \sqrt{b \sec (e+f x)}}{3 f}+\frac{7 b \sqrt{b \sec (e+f x)} \sin (e+f x)}{2 f}\\ \end{align*}

Mathematica [A]  time = 0.206314, size = 77, normalized size = 0.62 \[ -\frac{b \sin (e+f x) \sqrt{b \sec (e+f x)} \left (2 \csc ^4(e+f x)+7 \csc ^2(e+f x)+21 \sqrt{\cos (e+f x)} \csc (e+f x) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )-21\right )}{6 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4*(b*Sec[e + f*x])^(3/2),x]

[Out]

-(b*(-21 + 7*Csc[e + f*x]^2 + 2*Csc[e + f*x]^4 + 21*Sqrt[Cos[e + f*x]]*Csc[e + f*x]*EllipticE[(e + f*x)/2, 2])
*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x])/(6*f)

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Maple [C]  time = 0.158, size = 622, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4*(b*sec(f*x+e))^(3/2),x)

[Out]

1/6/f*(cos(f*x+e)+1)^2*(-1+cos(f*x+e))^2*(21*I*cos(f*x+e)^3*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1
))^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)-21*I*cos(f*x+e)^3*(1/(cos(f*x+e)+1))^(1/2)*(cos(
f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)+21*I*cos(f*x+e)^2*(1/(cos(f*
x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)-21*I*cos
(f*x+e)^2*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)
*sin(f*x+e)-21*I*cos(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(co
s(f*x+e)+1))^(1/2)*sin(f*x+e)+21*I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*(1/(cos(f*x
+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-21*I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(cos(f*x+e)+
1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+21*I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+
e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+21*cos(f*x+e)^3-14*cos(f*x+e)^2-21*cos(f*x+e)+12
)*cos(f*x+e)*(b/cos(f*x+e))^(3/2)/sin(f*x+e)^7

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}} \csc \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^(3/2)*csc(f*x + e)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \sec \left (f x + e\right )} b \csc \left (f x + e\right )^{4} \sec \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(f*x + e))*b*csc(f*x + e)^4*sec(f*x + e), 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(f*x+e)**4*(b*sec(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}} \csc \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^(3/2)*csc(f*x + e)^4, x)