\(\int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx\) [264]

Optimal result

Integrand size = 25, antiderivative size = 149 \[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {154 e^6 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt {\cos (c+d x)}}-\frac {154 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^4 d}-\frac {4 e (e \cos (c+d x))^{11/2}}{a d (a+a \sin (c+d x))^3}-\frac {44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4+a^4 \sin (c+d x)\right )} \]

[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)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2759, 2715, 2721, 2719} \[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {154 e^6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 a^4 d \sqrt {\cos (c+d x)}}-\frac {154 e^5 \sin (c+d x) (e \cos (c+d x))^{3/2}}{15 a^4 d}-\frac {44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {4 e (e \cos (c+d x))^{11/2}}{a d (a \sin (c+d x)+a)^3} \]

[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 2715

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] && Integ
erQ[2*n]

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*} \text {integral}& = -\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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.44 \[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {2^{3/4} (e \cos (c+d x))^{15/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {15}{4},\frac {19}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{15 a^4 d e (1+\sin (c+d x))^{15/4}} \]

[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))

Maple [A] (verified)

Time = 4.76 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.28

\[\frac {2 \left (24 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-80 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+246 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-231 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+80 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-140 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{7}}{15 \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{4} d}\]

[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+246*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2
31*(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))+80*sin(
1/2*d*x+1/2*c)^3-140*sin(1/2*d*x+1/2*c))*e^7/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.69 \[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {231 \, {\left (i \, \sqrt {2} e^{6} \cos \left (d x + c\right ) + i \, \sqrt {2} e^{6} \sin \left (d x + c\right ) + i \, \sqrt {2} e^{6}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 231 \, {\left (-i \, \sqrt {2} e^{6} \cos \left (d x + c\right ) - i \, \sqrt {2} e^{6} \sin \left (d x + c\right ) - i \, \sqrt {2} e^{6}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, e^{6} \cos \left (d x + c\right )^{3} + 20 \, e^{6} \cos \left (d x + c\right )^{2} + 137 \, e^{6} \cos \left (d x + c\right ) + 120 \, e^{6} - {\left (3 \, e^{6} \cos \left (d x + c\right )^{2} - 17 \, e^{6} \cos \left (d x + c\right ) + 120 \, e^{6}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{15 \, {\left (a^{4} d \cos \left (d x + c\right ) + a^{4} d \sin \left (d x + c\right ) + a^{4} d\right )}} \]

[In]

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

[Out]

-1/15*(231*(I*sqrt(2)*e^6*cos(d*x + c) + I*sqrt(2)*e^6*sin(d*x + c) + I*sqrt(2)*e^6)*sqrt(e)*weierstrassZeta(-
4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 231*(-I*sqrt(2)*e^6*cos(d*x + c) - I*sqrt(2
)*e^6*sin(d*x + c) - I*sqrt(2)*e^6)*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I
*sin(d*x + c))) + 2*(3*e^6*cos(d*x + c)^3 + 20*e^6*cos(d*x + c)^2 + 137*e^6*cos(d*x + c) + 120*e^6 - (3*e^6*co
s(d*x + c)^2 - 17*e^6*cos(d*x + c) + 120*e^6)*sin(d*x + c))*sqrt(e*cos(d*x + c)))/(a^4*d*cos(d*x + c) + a^4*d*
sin(d*x + c) + a^4*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {13}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {13}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{13/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]

[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)