3.264 \(\int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=149 \[ -\frac {154 e^6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 a^4 d \sqrt {\cos (c+d x)}}-\frac {154 e^5 \sin (c+d x) (e \cos (c+d x))^{3/2}}{15 a^4 d}-\frac {44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {4 e (e \cos (c+d x))^{11/2}}{a d (a \sin (c+d x)+a)^3} \]

[Out]

-154/15*e^5*(e*cos(d*x+c))^(3/2)*sin(d*x+c)/a^4/d-4*e*(e*cos(d*x+c))^(11/2)/a/d/(a+a*sin(d*x+c))^3-44/3*e^3*(e
*cos(d*x+c))^(7/2)/d/(a^4+a^4*sin(d*x+c))-154/5*e^6*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(
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 = 0.15, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2680, 2635, 2640, 2639} \[ -\frac {154 e^5 \sin (c+d x) (e \cos (c+d x))^{3/2}}{15 a^4 d}-\frac {44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {154 e^6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 a^4 d \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{11/2}}{a d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(13/2)/(a + a*Sin[c + d*x])^4,x]

[Out]

(-154*e^6*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*a^4*d*Sqrt[Cos[c + d*x]]) - (154*e^5*(e*Cos[c + d
*x])^(3/2)*Sin[c + d*x])/(15*a^4*d) - (4*e*(e*Cos[c + d*x])^(11/2))/(a*d*(a + a*Sin[c + d*x])^3) - (44*e^3*(e*
Cos[c + d*x])^(7/2))/(3*d*(a^4 + a^4*Sin[c + d*x]))

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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[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]

Rubi steps

\begin {align*} \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx &=-\frac {4 e (e \cos (c+d x))^{11/2}}{a d (a+a \sin (c+d x))^3}-\frac {\left (11 e^2\right ) \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^2} \, dx}{a^2}\\ &=-\frac {4 e (e \cos (c+d x))^{11/2}}{a d (a+a \sin (c+d x))^3}-\frac {44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac {\left (77 e^4\right ) \int (e \cos (c+d x))^{5/2} \, dx}{3 a^4}\\ &=-\frac {154 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^4 d}-\frac {4 e (e \cos (c+d x))^{11/2}}{a d (a+a \sin (c+d x))^3}-\frac {44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac {\left (77 e^6\right ) \int \sqrt {e \cos (c+d x)} \, dx}{5 a^4}\\ &=-\frac {154 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^4 d}-\frac {4 e (e \cos (c+d x))^{11/2}}{a d (a+a \sin (c+d x))^3}-\frac {44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac {\left (77 e^6 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a^4 \sqrt {\cos (c+d x)}}\\ &=-\frac {154 e^6 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt {\cos (c+d x)}}-\frac {154 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^4 d}-\frac {4 e (e \cos (c+d x))^{11/2}}{a d (a+a \sin (c+d x))^3}-\frac {44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4+a^4 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 66, normalized size = 0.44 \[ -\frac {2^{3/4} (e \cos (c+d x))^{15/2} \, _2F_1\left (\frac {5}{4},\frac {15}{4};\frac {19}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{15 a^4 d e (\sin (c+d x)+1)^{15/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(13/2)/(a + a*Sin[c + d*x])^4,x]

[Out]

-1/15*(2^(3/4)*(e*Cos[c + d*x])^(15/2)*Hypergeometric2F1[5/4, 15/4, 19/4, (1 - Sin[c + d*x])/2])/(a^4*d*e*(1 +
Sin[c + d*x])^(15/4))

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \cos \left (d x + c\right )} e^{6} \cos \left (d x + c\right )^{6}}{a^{4} \cos \left (d x + c\right )^{4} - 8 \, a^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{4} - 4 \, {\left (a^{4} \cos \left (d x + c\right )^{2} - 2 \, a^{4}\right )} \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(13/2)/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

integral(sqrt(e*cos(d*x + c))*e^6*cos(d*x + c)^6/(a^4*cos(d*x + c)^4 - 8*a^4*cos(d*x + c)^2 + 8*a^4 - 4*(a^4*c
os(d*x + c)^2 - 2*a^4)*sin(d*x + c)), x)

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giac [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))^(13/2)/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.65, size = 190, normalized size = 1.28 \[ -\frac {2 \left (-24 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+80 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+231 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-246 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-80 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+140 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{7}}{15 \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(13/2)/(a+a*sin(d*x+c))^4,x)

[Out]

-2/15/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/sin(1/2*d*x+1/2*c)/a^4*(-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+
24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+80*sin(1/2*d*x+1/2*c)^5+231*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)-246*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-80*si
n(1/2*d*x+1/2*c)^3+140*sin(1/2*d*x+1/2*c))*e^7/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {13}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(13/2)/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(13/2)/(a*sin(d*x + c) + a)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{13/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(13/2)/(a + a*sin(c + d*x))^4,x)

[Out]

int((e*cos(c + d*x))^(13/2)/(a + a*sin(c + d*x))^4, x)

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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))**(13/2)/(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

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