3.3.51 \(\int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2} \, dx\) [251]

Optimal. Leaf size=150 \[ \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{33 a^2 d e^2 \sqrt {e \cos (c+d x)}}+\frac {10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac {2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}-\frac {2}{11 d e (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

10/33*sin(d*x+c)/a^2/d/e/(e*cos(d*x+c))^(3/2)-2/11/d/e/(e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^2-2/11/d/e/(e*cos
(d*x+c))^(3/2)/(a^2+a^2*sin(d*x+c))+10/33*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*
x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/a^2/d/e^2/(e*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2760, 2762, 2716, 2721, 2720} \begin {gather*} \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{33 a^2 d e^2 \sqrt {e \cos (c+d x)}}+\frac {10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac {2}{11 d e \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{3/2}}-\frac {2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(33*a^2*d*e^2*Sqrt[e*Cos[c + d*x]]) + (10*Sin[c + d*x])/(33*
a^2*d*e*(e*Cos[c + d*x])^(3/2)) - 2/(11*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^2) - 2/(11*d*e*(e*Cos[
c + d*x])^(3/2)*(a^2 + a^2*Sin[c + d*x]))

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 2760

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1))), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rule 2762

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

Rubi steps

\begin {align*} \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2} \, dx &=-\frac {2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}+\frac {7 \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx}{11 a}\\ &=-\frac {2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}-\frac {2}{11 d e (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}+\frac {5 \int \frac {1}{(e \cos (c+d x))^{5/2}} \, dx}{11 a^2}\\ &=\frac {10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac {2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}-\frac {2}{11 d e (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}+\frac {5 \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{33 a^2 e^2}\\ &=\frac {10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac {2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}-\frac {2}{11 d e (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\left (5 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{33 a^2 e^2 \sqrt {e \cos (c+d x)}}\\ &=\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{33 a^2 d e^2 \sqrt {e \cos (c+d x)}}+\frac {10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac {2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}-\frac {2}{11 d e (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.08, size = 66, normalized size = 0.44 \begin {gather*} \frac {\, _2F_1\left (-\frac {3}{4},\frac {15}{4};\frac {1}{4};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{3/4}}{6\ 2^{3/4} a^2 d e (e \cos (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[-3/4, 15/4, 1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(3/4))/(6*2^(3/4)*a^2*d*e*(e*Cos[
c + d*x])^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(556\) vs. \(2(158)=316\).
time = 10.11, size = 557, normalized size = 3.71

method	result	size
default	\(-\frac {2 \left (160 \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-400 \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+160 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+400 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-320 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-200 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+264 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+50 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-104 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+28 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{33 \left (32 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-80 \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+80 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\)	\(557\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/33/(32*sin(1/2*d*x+1/2*c)^10-80*sin(1/2*d*x+1/2*c)^8+80*sin(1/2*d*x+1/2*c)^6-40*sin(1/2*d*x+1/2*c)^4+10*sin
(1/2*d*x+1/2*c)^2-1)/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^2*(160*EllipticF(cos(1/2*d*x
+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^10-400*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2
*c)^8+160*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+400*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)
^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^6-320*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-
200*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/
2*d*x+1/2*c)^4+264*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+50*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*
c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-104*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+
1/2*c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2
8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-6*sin(1/2*d*x+1/2*c))/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 232, normalized size = 1.55 \begin {gather*} \frac {5 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{4} + 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{4} - 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (10 \, \cos \left (d x + c\right )^{2} + {\left (5 \, \cos \left (d x + c\right )^{2} - 7\right )} \sin \left (d x + c\right ) - 4\right )} \sqrt {\cos \left (d x + c\right )}}{33 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} e^{\frac {5}{2}} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} e^{\frac {5}{2}} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{2} e^{\frac {5}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/33*(5*(-I*sqrt(2)*cos(d*x + c)^4 + 2*I*sqrt(2)*cos(d*x + c)^2*sin(d*x + c) + 2*I*sqrt(2)*cos(d*x + c)^2)*wei
erstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(I*sqrt(2)*cos(d*x + c)^4 - 2*I*sqrt(2)*cos(d*x + c
)^2*sin(d*x + c) - 2*I*sqrt(2)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(
10*cos(d*x + c)^2 + (5*cos(d*x + c)^2 - 7)*sin(d*x + c) - 4)*sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^4*e^(5/2)
- 2*a^2*d*cos(d*x + c)^2*e^(5/2)*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^2*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/(e*cos(d*x+c))**(5/2)/(a+a*sin(d*x+c))**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3881 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/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(e^(-5/2)/((a*sin(d*x + c) + a)^2*cos(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 {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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