3.2.26 \(\int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx\) [126]

Optimal. Leaf size=135 \[ -\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}+\frac {2 e \cos (c+d x)}{7 a d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{21 a d e (e \sin (c+d x))^{3/2}}+\frac {4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{21 a d e^2 \sqrt {e \sin (c+d x)}} \]

[Out]

-2/7*e/a/d/(e*sin(d*x+c))^(7/2)+2/7*e*cos(d*x+c)/a/d/(e*sin(d*x+c))^(7/2)-4/21*cos(d*x+c)/a/d/e/(e*sin(d*x+c))
^(3/2)-4/21*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),
2^(1/2))*sin(d*x+c)^(1/2)/a/d/e^2/(e*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3957, 2918, 2644, 30, 2647, 2716, 2721, 2720} \begin {gather*} \frac {4 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{21 a d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}+\frac {2 e \cos (c+d x)}{7 a d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{21 a d e (e \sin (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[c + d*x])*(e*Sin[c + d*x])^(5/2)),x]

[Out]

(-2*e)/(7*a*d*(e*Sin[c + d*x])^(7/2)) + (2*e*Cos[c + d*x])/(7*a*d*(e*Sin[c + d*x])^(7/2)) - (4*Cos[c + d*x])/(
21*a*d*e*(e*Sin[c + d*x])^(3/2)) + (4*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*a*d*e^2*Sqrt[e*
Sin[c + d*x]])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2647

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a*Cos[e +
f*x])^(m - 1)*((b*Sin[e + f*x])^(n + 1)/(b*f*(n + 1))), x] + Dist[a^2*((m - 1)/(b^2*(n + 1))), Int[(a*Cos[e +
f*x])^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Intege
rsQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2716

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

Rule 2720

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

Rule 2721

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

Rule 2918

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx &=-\int \frac {\cos (c+d x)}{(-a-a \cos (c+d x)) (e \sin (c+d x))^{5/2}} \, dx\\ &=\frac {e^2 \int \frac {\cos (c+d x)}{(e \sin (c+d x))^{9/2}} \, dx}{a}-\frac {e^2 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{9/2}} \, dx}{a}\\ &=\frac {2 e \cos (c+d x)}{7 a d (e \sin (c+d x))^{7/2}}+\frac {2 \int \frac {1}{(e \sin (c+d x))^{5/2}} \, dx}{7 a}+\frac {e \text {Subst}\left (\int \frac {1}{x^{9/2}} \, dx,x,e \sin (c+d x)\right )}{a d}\\ &=-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}+\frac {2 e \cos (c+d x)}{7 a d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{21 a d e (e \sin (c+d x))^{3/2}}+\frac {2 \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{21 a e^2}\\ &=-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}+\frac {2 e \cos (c+d x)}{7 a d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{21 a d e (e \sin (c+d x))^{3/2}}+\frac {\left (2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{21 a e^2 \sqrt {e \sin (c+d x)}}\\ &=-\frac {2 e}{7 a d (e \sin (c+d x))^{7/2}}+\frac {2 e \cos (c+d x)}{7 a d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{21 a d e (e \sin (c+d x))^{3/2}}+\frac {4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{21 a d e^2 \sqrt {e \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.85, size = 91, normalized size = 0.67 \begin {gather*} -\frac {2 \left (4+2 \cos (c+d x)+\cos (2 (c+d x))+\csc ^2\left (\frac {1}{2} (c+d x)\right ) F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sin ^{\frac {7}{2}}(c+d x)\right )}{21 a d e (1+\cos (c+d x)) (e \sin (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sec[c + d*x])*(e*Sin[c + d*x])^(5/2)),x]

[Out]

(-2*(4 + 2*Cos[c + d*x] + Cos[2*(c + d*x)] + Csc[(c + d*x)/2]^2*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*
x]^(7/2)))/(21*a*d*e*(1 + Cos[c + d*x])*(e*Sin[c + d*x])^(3/2))

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Maple [A]
time = 0.19, size = 136, normalized size = 1.01

method	result	size
default	\(\frac {-\frac {2 e}{7 a \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}-\frac {2 \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {9}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{5}\left (d x +c \right )\right )+5 \left (\sin ^{3}\left (d x +c \right )\right )-3 \sin \left (d x +c \right )\right )}{21 e^{2} a \sin \left (d x +c \right )^{4} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(136\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

(-2/7/a*e/(e*sin(d*x+c))^(7/2)-2/21/e^2*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(9/2)*Ellipti
cF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^5+5*sin(d*x+c)^3-3*sin(d*x+c))/a/sin(d*x+c)^4/cos(d*x+c)/(e
*sin(d*x+c))^(1/2))/d

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.79, size = 205, normalized size = 1.52 \begin {gather*} \frac {2 \, {\left (\sqrt {-i} {\left (\sqrt {2} \cos \left (d x + c\right )^{3} + \sqrt {2} \cos \left (d x + c\right )^{2} - \sqrt {2} \cos \left (d x + c\right ) - \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {i} {\left (\sqrt {2} \cos \left (d x + c\right )^{3} + \sqrt {2} \cos \left (d x + c\right )^{2} - \sqrt {2} \cos \left (d x + c\right ) - \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (2 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {\sin \left (d x + c\right )}\right )}}{21 \, {\left (a d \cos \left (d x + c\right )^{3} e^{\frac {5}{2}} + a d \cos \left (d x + c\right )^{2} e^{\frac {5}{2}} - a d \cos \left (d x + c\right ) e^{\frac {5}{2}} - a d e^{\frac {5}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/21*(sqrt(-I)*(sqrt(2)*cos(d*x + c)^3 + sqrt(2)*cos(d*x + c)^2 - sqrt(2)*cos(d*x + c) - sqrt(2))*weierstrassP
Inverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(I)*(sqrt(2)*cos(d*x + c)^3 + sqrt(2)*cos(d*x + c)^2 - sqrt
(2)*cos(d*x + c) - sqrt(2))*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + (2*cos(d*x + c)^2 + 2*c
os(d*x + c) + 3)*sqrt(sin(d*x + c)))/(a*d*cos(d*x + c)^3*e^(5/2) + a*d*cos(d*x + c)^2*e^(5/2) - a*d*cos(d*x +
c)*e^(5/2) - a*d*e^(5/2))

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))/(e*sin(d*x+c))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5010 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)*(e*sin(d*x + c))^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\cos \left (c+d\,x\right )}{a\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*sin(c + d*x))^(5/2)*(a + a/cos(c + d*x))),x)

[Out]

int(cos(c + d*x)/(a*(e*sin(c + d*x))^(5/2)*(cos(c + d*x) + 1)), x)

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