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

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

[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 = 0.16, 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 = {2681, 2683, 2636, 2642, 2641} \[ \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{33 a^2 d e^2 \sqrt {e \cos (c+d x)}}+\frac {10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac {2}{11 d e \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{3/2}}-\frac {2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}} \]

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 2636

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 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]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2681

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[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 2683

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]  time = 0.08, size = 66, normalized size = 0.44 \[ \frac {(\sin (c+d x)+1)^{3/4} \, _2F_1\left (-\frac {3}{4},\frac {15}{4};\frac {1}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{6\ 2^{3/4} a^2 d e (e \cos (c+d x))^{3/2}} \]

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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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {e \cos \left (d x + c\right )}}{a^{2} e^{3} \cos \left (d x + c\right )^{5} - 2 \, a^{2} e^{3} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} e^{3} \cos \left (d x + c\right )^{3}}, x\right ) \]

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]

integral(-sqrt(e*cos(d*x + c))/(a^2*e^3*cos(d*x + c)^5 - 2*a^2*e^3*cos(d*x + c)^3*sin(d*x + c) - 2*a^2*e^3*cos
(d*x + c)^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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

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maple [B]  time = 3.65, size = 557, normalized size = 3.71 \[ -\frac {2 \left (160 \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \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}\, \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 {\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}\, \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 {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 ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \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 {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 ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \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 \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 ^{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}}\, \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}+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} \]

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)

[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))*(sin(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)^10-400*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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)^8+160*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+400*(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)^(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*(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)^(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*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)^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)*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)+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 \[ \int \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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]

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

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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