3.3.46 \(\int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^2} \, dx\) [246]

Optimal. Leaf size=79 \[ -\frac {6 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

-4*e*(e*cos(d*x+c))^(3/2)/d/(a^2+a^2*sin(d*x+c))-6*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti
cE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/a^2/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2759, 2721, 2719} \begin {gather*} -\frac {6 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{a^2 d \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2 \sin (c+d x)+a^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 2759

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]

Rubi steps

\begin {align*} \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^2} \, dx &=-\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2+a^2 \sin (c+d x)\right )}-\frac {\left (3 e^2\right ) \int \sqrt {e \cos (c+d x)} \, dx}{a^2}\\ &=-\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2+a^2 \sin (c+d x)\right )}-\frac {\left (3 e^2 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{a^2 \sqrt {\cos (c+d x)}}\\ &=-\frac {6 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{3/2}}{d \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 = 0.84 \begin {gather*} -\frac {2^{3/4} (e \cos (c+d x))^{7/2} \, _2F_1\left (\frac {5}{4},\frac {7}{4};\frac {11}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{7 a^2 d e (1+\sin (c+d x))^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/7*(2^(3/4)*(e*Cos[c + d*x])^(7/2)*Hypergeometric2F1[5/4, 7/4, 11/4, (1 - Sin[c + d*x])/2])/(a^2*d*e*(1 + Si
n[c + d*x])^(7/4))

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Maple [A]
time = 3.32, size = 120, normalized size = 1.52

method	result	size
default	\(-\frac {2 \left (3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{3}}{\sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} d}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/sin(1/2*d*x+1/2*c)/a^2*(3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-4*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+2*sin(1/2*d*x
+1/2*c))*e^3/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((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 179, normalized size = 2.27 \begin {gather*} -\frac {3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right ) e^{\frac {5}{2}} + i \, \sqrt {2} e^{\frac {5}{2}} \sin \left (d x + c\right ) + i \, \sqrt {2} e^{\frac {5}{2}}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) e^{\frac {5}{2}} - i \, \sqrt {2} e^{\frac {5}{2}} \sin \left (d x + c\right ) - i \, \sqrt {2} e^{\frac {5}{2}}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 4 \, {\left (\cos \left (d x + c\right ) e^{\frac {5}{2}} - e^{\frac {5}{2}} \sin \left (d x + c\right ) + e^{\frac {5}{2}}\right )} \sqrt {\cos \left (d x + c\right )}}{a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*(I*sqrt(2)*cos(d*x + c)*e^(5/2) + I*sqrt(2)*e^(5/2)*sin(d*x + c) + I*sqrt(2)*e^(5/2))*weierstrassZeta(-4,
0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(-I*sqrt(2)*cos(d*x + c)*e^(5/2) - I*sqrt(2)
*e^(5/2)*sin(d*x + c) - I*sqrt(2)*e^(5/2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*
sin(d*x + c))) + 4*(cos(d*x + c)*e^(5/2) - e^(5/2)*sin(d*x + c) + e^(5/2))*sqrt(cos(d*x + c)))/(a^2*d*cos(d*x
+ c) + a^2*d*sin(d*x + c) + a^2*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

integrate(cos(d*x + c)^(5/2)*e^(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 \begin {gather*} \int \frac {{\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((e*cos(c + d*x))^(5/2)/(a + a*sin(c + d*x))^2,x)

[Out]

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

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