3.245 \(\int \frac{(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=112 \[ \frac{10 e^3 \sin (c+d x) \sqrt{e \cos (c+d x)}}{3 a^2 d}+\frac{10 e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d \sqrt{e \cos (c+d x)}}+\frac{4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )} \]

[Out]

(10*e^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*a^2*d*Sqrt[e*Cos[c + d*x]]) + (10*e^3*Sqrt[e*Cos[c +
d*x]]*Sin[c + d*x])/(3*a^2*d) + (4*e*(e*Cos[c + d*x])^(5/2))/(d*(a^2 + a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.0970945, 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 = {2680, 2635, 2642, 2641} \[ \frac{10 e^3 \sin (c+d x) \sqrt{e \cos (c+d x)}}{3 a^2 d}+\frac{10 e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d \sqrt{e \cos (c+d x)}}+\frac{4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(7/2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(10*e^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*a^2*d*Sqrt[e*Cos[c + d*x]]) + (10*e^3*Sqrt[e*Cos[c +
d*x]]*Sin[c + d*x])/(3*a^2*d) + (4*e*(e*Cos[c + d*x])^(5/2))/(d*(a^2 + a^2*Sin[c + d*x]))

Rule 2680

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[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{(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^2} \, dx &=\frac{4 e (e \cos (c+d x))^{5/2}}{d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{\left (5 e^2\right ) \int (e \cos (c+d x))^{3/2} \, dx}{a^2}\\ &=\frac{10 e^3 \sqrt{e \cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{4 e (e \cos (c+d x))^{5/2}}{d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{\left (5 e^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{3 a^2}\\ &=\frac{10 e^3 \sqrt{e \cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{4 e (e \cos (c+d x))^{5/2}}{d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{\left (5 e^4 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a^2 \sqrt{e \cos (c+d x)}}\\ &=\frac{10 e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d \sqrt{e \cos (c+d x)}}+\frac{10 e^3 \sqrt{e \cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{4 e (e \cos (c+d x))^{5/2}}{d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.0765677, size = 66, normalized size = 0.59 \[ -\frac{2 \sqrt [4]{2} (e \cos (c+d x))^{9/2} \, _2F_1\left (\frac{3}{4},\frac{9}{4};\frac{13}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{9 a^2 d e (\sin (c+d x)+1)^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(7/2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-2*2^(1/4)*(e*Cos[c + d*x])^(9/2)*Hypergeometric2F1[3/4, 9/4, 13/4, (1 - Sin[c + d*x])/2])/(9*a^2*d*e*(1 + Si
n[c + d*x])^(9/4))

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Maple [A]  time = 0.434, size = 155, normalized size = 1.4 \begin{align*} -{\frac{2\,{e}^{4}}{3\,{a}^{2}d} \left ( -4\, \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 ) +2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +12\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-6\,\sin \left ( 1/2\,dx+c/2 \right ) \right ) \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((e*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^2,x)

[Out]

-2/3/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^4*(-4*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))+2*sin(
1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+12*sin(1/2*d*x+1/2*c)^3-6*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{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(7/2)/(a*sin(d*x + c) + a)^2, 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 )} e^{3} \cos \left (d x + c\right )^{3}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(-sqrt(e*cos(d*x + c))*e^3*cos(d*x + c)^3/(a^2*cos(d*x + c)^2 - 2*a^2*sin(d*x + c) - 2*a^2), 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((e*cos(d*x+c))**(7/2)/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(7/2)/(a*sin(d*x + c) + a)^2, x)