3.3.68 \(\int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx\) [268]

Optimal. Leaf size=154 \[ \frac {2 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^4 d \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a+a \sin (c+d x))^3}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^4+a^4 \sin (c+d x)\right )} \]

[Out]

-4/9*e*(e*cos(d*x+c))^(3/2)/a/d/(a+a*sin(d*x+c))^3+2/15*e*(e*cos(d*x+c))^(3/2)/d/(a^2+a^2*sin(d*x+c))^2+2/15*e
*(e*cos(d*x+c))^(3/2)/d/(a^4+a^4*sin(d*x+c))+2/15*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
E(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/a^4/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 154, 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 = {2759, 2760, 2762, 2721, 2719} \begin {gather*} \frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{15 a^4 d \sqrt {\cos (c+d x)}}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^4 \sin (c+d x)+a^4\right )}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a \sin (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(5/2)/(a + a*Sin[c + d*x])^4,x]

[Out]

(2*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*a^4*d*Sqrt[Cos[c + d*x]]) - (4*e*(e*Cos[c + d*x])^(
3/2))/(9*a*d*(a + a*Sin[c + d*x])^3) + (2*e*(e*Cos[c + d*x])^(3/2))/(15*d*(a^2 + a^2*Sin[c + d*x])^2) + (2*e*(
e*Cos[c + d*x])^(3/2))/(15*d*(a^4 + a^4*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]

Rule 2760

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2762

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))^{5/2}}{(a+a \sin (c+d x))^4} \, dx &=-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a+a \sin (c+d x))^3}-\frac {e^2 \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^2} \, dx}{3 a^2}\\ &=-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a+a \sin (c+d x))^3}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {e^2 \int \frac {\sqrt {e \cos (c+d x)}}{a+a \sin (c+d x)} \, dx}{15 a^3}\\ &=-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a+a \sin (c+d x))^3}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^4+a^4 \sin (c+d x)\right )}+\frac {e^2 \int \sqrt {e \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a+a \sin (c+d x))^3}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^4+a^4 \sin (c+d x)\right )}+\frac {\left (e^2 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 a^4 \sqrt {\cos (c+d x)}}\\ &=\frac {2 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^4 d \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a+a \sin (c+d x))^3}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^4+a^4 \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 = 0.09, size = 66, normalized size = 0.43 \begin {gather*} -\frac {(e \cos (c+d x))^{7/2} \, _2F_1\left (\frac {7}{4},\frac {13}{4};\frac {11}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{14 \sqrt [4]{2} a^4 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])^4,x]

[Out]

-1/14*((e*Cos[c + d*x])^(7/2)*Hypergeometric2F1[7/4, 13/4, 11/4, (1 - Sin[c + d*x])/2])/(2^(1/4)*a^4*d*e*(1 +
Sin[c + d*x])^(7/4))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(513\) vs. \(2(162)=324\).
time = 10.38, size = 514, normalized size = 3.34

method	result	size
default	\(\frac {2 \left (48 \EllipticE \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 ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-96 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-96 \EllipticE \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 ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+192 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+72 \EllipticE \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 ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-272 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-24 \EllipticE \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 ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+176 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+144 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+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 )+42 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-144 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{3}}{45 \left (16 \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-32 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) a^{4} \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}\)	\(514\)

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))^4,x,method=_RETURNVERBOSE)

[Out]

2/45/(16*sin(1/2*d*x+1/2*c)^8-32*sin(1/2*d*x+1/2*c)^6+24*sin(1/2*d*x+1/2*c)^4-8*sin(1/2*d*x+1/2*c)^2+1)/a^4/si
n(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(48*EllipticE(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)^8-96*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c
)-96*EllipticE(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)^6+192*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+72*EllipticE(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)^4-272*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x
+1/2*c)-24*EllipticE(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)^2+176*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+144*sin(1/2*d*x+1/2*c)^5+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))+42*sin(1/2*d*x+1/2*c)^2*cos
(1/2*d*x+1/2*c)-144*sin(1/2*d*x+1/2*c)^3-4*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))^4,x, algorithm="maxima")

[Out]

e^(5/2)*integrate(cos(d*x + c)^(5/2)/(a*sin(d*x + c) + a)^4, x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 407, normalized size = 2.64 \begin {gather*} -\frac {3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} e^{\frac {5}{2}} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} e^{\frac {5}{2}} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) e^{\frac {5}{2}} + {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} e^{\frac {5}{2}} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) e^{\frac {5}{2}} + 4 i \, \sqrt {2} e^{\frac {5}{2}}\right )} \sin \left (d x + c\right ) + 4 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 )^{3} e^{\frac {5}{2}} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} e^{\frac {5}{2}} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) e^{\frac {5}{2}} + {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} e^{\frac {5}{2}} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) e^{\frac {5}{2}} - 4 i \, \sqrt {2} e^{\frac {5}{2}}\right )} \sin \left (d x + c\right ) - 4 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 ) - 2 \, {\left (3 \, \cos \left (d x + c\right )^{3} e^{\frac {5}{2}} - 6 \, \cos \left (d x + c\right )^{2} e^{\frac {5}{2}} + \cos \left (d x + c\right ) e^{\frac {5}{2}} - {\left (3 \, \cos \left (d x + c\right )^{2} e^{\frac {5}{2}} + 9 \, \cos \left (d x + c\right ) e^{\frac {5}{2}} + 10 \, e^{\frac {5}{2}}\right )} \sin \left (d x + c\right ) + 10 \, e^{\frac {5}{2}}\right )} \sqrt {\cos \left (d x + c\right )}}{45 \, {\left (a^{4} d \cos \left (d x + c\right )^{3} + 3 \, a^{4} d \cos \left (d x + c\right )^{2} - 2 \, a^{4} d \cos \left (d x + c\right ) - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 2 \, a^{4} d \cos \left (d x + c\right ) - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \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))^4,x, algorithm="fricas")

[Out]

-1/45*(3*(-I*sqrt(2)*cos(d*x + c)^3*e^(5/2) - 3*I*sqrt(2)*cos(d*x + c)^2*e^(5/2) + 2*I*sqrt(2)*cos(d*x + c)*e^
(5/2) + (-I*sqrt(2)*cos(d*x + c)^2*e^(5/2) + 2*I*sqrt(2)*cos(d*x + c)*e^(5/2) + 4*I*sqrt(2)*e^(5/2))*sin(d*x +
c) + 4*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)^3*e^(5/2) + 3*I*sqrt(2)*cos(d*x + c)^2*e^(5/2) - 2*I*sqrt(2)*cos(d*x + c)*e^(5/2) +
(I*sqrt(2)*cos(d*x + c)^2*e^(5/2) - 2*I*sqrt(2)*cos(d*x + c)*e^(5/2) - 4*I*sqrt(2)*e^(5/2))*sin(d*x + c) - 4*
I*sqrt(2)*e^(5/2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*co
s(d*x + c)^3*e^(5/2) - 6*cos(d*x + c)^2*e^(5/2) + cos(d*x + c)*e^(5/2) - (3*cos(d*x + c)^2*e^(5/2) + 9*cos(d*x
+ c)*e^(5/2) + 10*e^(5/2))*sin(d*x + c) + 10*e^(5/2))*sqrt(cos(d*x + c)))/(a^4*d*cos(d*x + c)^3 + 3*a^4*d*cos
(d*x + c)^2 - 2*a^4*d*cos(d*x + c) - 4*a^4*d + (a^4*d*cos(d*x + c)^2 - 2*a^4*d*cos(d*x + c) - 4*a^4*d)*sin(d*x
+ c))

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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))**4,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))^4,x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(5/2)*e^(5/2)/(a*sin(d*x + c) + a)^4, 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 )}^4} \,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))^4,x)

[Out]

int((e*cos(c + d*x))^(5/2)/(a + a*sin(c + d*x))^4, x)

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