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

Optimal. Leaf size=112 \[ \frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 a d e^2 \sqrt{e \cos (c+d x)}}+\frac{10 \sin (c+d x)}{21 a d e (e \cos (c+d x))^{3/2}}-\frac{2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}} \]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*a*d*e^2*Sqrt[e*Cos[c + d*x]]) + (10*Sin[c + d*x])/(21*a*
d*e*(e*Cos[c + d*x])^(3/2)) - 2/(7*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.0943606, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2683, 2636, 2642, 2641} \[ \frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 a d e^2 \sqrt{e \cos (c+d x)}}+\frac{10 \sin (c+d x)}{21 a d e (e \cos (c+d x))^{3/2}}-\frac{2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*a*d*e^2*Sqrt[e*Cos[c + d*x]]) + (10*Sin[c + d*x])/(21*a*
d*e*(e*Cos[c + d*x])^(3/2)) - 2/(7*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x]))

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx &=-\frac{2}{7 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))}+\frac{5 \int \frac{1}{(e \cos (c+d x))^{5/2}} \, dx}{7 a}\\ &=\frac{10 \sin (c+d x)}{21 a d e (e \cos (c+d x))^{3/2}}-\frac{2}{7 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))}+\frac{5 \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{21 a e^2}\\ &=\frac{10 \sin (c+d x)}{21 a d e (e \cos (c+d x))^{3/2}}-\frac{2}{7 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))}+\frac{\left (5 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 a 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 )}{21 a d e^2 \sqrt{e \cos (c+d x)}}+\frac{10 \sin (c+d x)}{21 a d e (e \cos (c+d x))^{3/2}}-\frac{2}{7 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.0609495, size = 66, normalized size = 0.59 \[ \frac{(\sin (c+d x)+1)^{3/4} \, _2F_1\left (-\frac{3}{4},\frac{11}{4};\frac{1}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{3\ 2^{3/4} a d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[-3/4, 11/4, 1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(3/4))/(3*2^(3/4)*a*d*e*(e*Cos[c
+ d*x])^(3/2))

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Maple [B]  time = 1.511, size = 375, normalized size = 3.4 \begin{align*} -{\frac{2}{21\,{e}^{2}ad} \left ( 40\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-60\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+40\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +30\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-40\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -5\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +16\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -3\,\sin \left ( 1/2\,dx+c/2 \right ) \right ) \left ( 8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-12\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+6\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) ^{-1} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

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)),x)

[Out]

-2/21/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/a/sin(1/2*d*x+1/2*c)/(-2*sin(1
/2*d*x+1/2*c)^2*e+e)^(1/2)/e^2*(40*(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)*sin(1/2*d*x+1/2*c)^6-60*(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)*sin(1/2*d*x+1/2*c)^4+40*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+30*(2*s
in(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)^2-40*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-5*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*sin(1/2*d*x+1/2*c))/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}}\,{d x} \end{align*}

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)),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e \cos \left (d x + c\right )}}{a e^{3} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a e^{3} \cos \left (d x + c\right )^{3}}, x\right ) \end{align*}

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)),x, algorithm="fricas")

[Out]

integral(sqrt(e*cos(d*x + c))/(a*e^3*cos(d*x + c)^3*sin(d*x + c) + a*e^3*cos(d*x + c)^3), 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(1/(e*cos(d*x+c))**(5/2)/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}}\,{d x} \end{align*}

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)),x, algorithm="giac")

[Out]

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