3.259 \(\int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=118 \[ -\frac {2 e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 a^3 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sqrt {e \cos (c+d x)}}{21 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {4 e \sqrt {e \cos (c+d x)}}{7 a d (a \sin (c+d x)+a)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2680, 2683, 2642, 2641} \[ -\frac {2 e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 a^3 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sqrt {e \cos (c+d x)}}{21 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {4 e \sqrt {e \cos (c+d x)}}{7 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(3/2)/(a + a*Sin[c + d*x])^3,x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [C]  time = 0.08, size = 66, normalized size = 0.56 \[ -\frac {(e \cos (c+d x))^{5/2} \, _2F_1\left (\frac {5}{4},\frac {11}{4};\frac {9}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{5\ 2^{3/4} a^3 d e (\sin (c+d x)+1)^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(3/2)/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/5*((e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[5/4, 11/4, 9/4, (1 - Sin[c + d*x])/2])/(2^(3/4)*a^3*d*e*(1 + Si
n[c + d*x])^(5/4))

________________________________________________________________________________________

fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {e \cos \left (d x + c\right )} e \cos \left (d x + c\right )}{3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(-sqrt(e*cos(d*x + c))*e*cos(d*x + c)/(3*a^3*cos(d*x + c)^2 - 4*a^3 + (a^3*cos(d*x + c)^2 - 4*a^3)*sin
(d*x + c)), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(3/2)/(a*sin(d*x + c) + a)^3, x)

________________________________________________________________________________________

maple [B]  time = 2.32, size = 401, normalized size = 3.40 \[ \frac {2 \left (8 \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 )-12 \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 )+8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+6 \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 )-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-28 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\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}-22 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+28 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{2}}{21 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) a^{3} \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}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^3,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^3/sin(1/2*d*x+1/2*c)/(-2*sin(
1/2*d*x+1/2*c)^2*e+e)^(1/2)*(8*(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)^6-12*(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+8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+6*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
-8*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-28*sin(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)-22*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+28*sin(1/2
*d*x+1/2*c)^3+5*sin(1/2*d*x+1/2*c))*e^2/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(3/2)/(a*sin(d*x + c) + a)^3, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(3/2)/(a + a*sin(c + d*x))^3,x)

[Out]

int((e*cos(c + d*x))^(3/2)/(a + a*sin(c + d*x))^3, x)

________________________________________________________________________________________

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((e*cos(d*x+c))**(3/2)/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________