3.3.54 \(\int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^3} \, dx\) [254]

Optimal. Leaf size=138 \[ \frac {22 e^3 (e \cos (c+d x))^{7/2}}{21 a^3 d}+\frac {22 e^6 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt {\cos (c+d x)}}+\frac {22 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^3 d}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a+a \sin (c+d x))^2} \]

[Out]

22/21*e^3*(e*cos(d*x+c))^(7/2)/a^3/d+22/15*e^5*(e*cos(d*x+c))^(3/2)*sin(d*x+c)/a^3/d+4/3*e*(e*cos(d*x+c))^(11/
2)/a/d/(a+a*sin(d*x+c))^2+22/5*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^3/d/cos(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 138, 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, 2761, 2715, 2721, 2719} \begin {gather*} \frac {22 e^6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 a^3 d \sqrt {\cos (c+d x)}}+\frac {22 e^5 \sin (c+d x) (e \cos (c+d x))^{3/2}}{15 a^3 d}+\frac {22 e^3 (e \cos (c+d x))^{7/2}}{21 a^3 d}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a \sin (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(13/2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(22*e^3*(e*Cos[c + d*x])^(7/2))/(21*a^3*d) + (22*e^6*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*a^3*d*
Sqrt[Cos[c + d*x]]) + (22*e^5*(e*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(15*a^3*d) + (4*e*(e*Cos[c + d*x])^(11/2))/
(3*a*d*(a + a*Sin[c + d*x])^2)

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]

Rule 2761

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*((g*Cos[e
+ f*x])^(p - 1)/(b*f*(p - 1))), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^3} \, dx &=\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a+a \sin (c+d x))^2}+\frac {\left (11 e^2\right ) \int \frac {(e \cos (c+d x))^{9/2}}{a+a \sin (c+d x)} \, dx}{3 a^2}\\ &=\frac {22 e^3 (e \cos (c+d x))^{7/2}}{21 a^3 d}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a+a \sin (c+d x))^2}+\frac {\left (11 e^4\right ) \int (e \cos (c+d x))^{5/2} \, dx}{3 a^3}\\ &=\frac {22 e^3 (e \cos (c+d x))^{7/2}}{21 a^3 d}+\frac {22 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^3 d}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a+a \sin (c+d x))^2}+\frac {\left (11 e^6\right ) \int \sqrt {e \cos (c+d x)} \, dx}{5 a^3}\\ &=\frac {22 e^3 (e \cos (c+d x))^{7/2}}{21 a^3 d}+\frac {22 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^3 d}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a+a \sin (c+d x))^2}+\frac {\left (11 e^6 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a^3 \sqrt {\cos (c+d x)}}\\ &=\frac {22 e^3 (e \cos (c+d x))^{7/2}}{21 a^3 d}+\frac {22 e^6 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt {\cos (c+d x)}}+\frac {22 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^3 d}+\frac {4 e (e \cos (c+d x))^{11/2}}{3 a d (a+a \sin (c+d x))^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.22, size = 66, normalized size = 0.48 \begin {gather*} -\frac {2\ 2^{3/4} (e \cos (c+d x))^{15/2} \, _2F_1\left (\frac {1}{4},\frac {15}{4};\frac {19}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{15 a^3 d e (1+\sin (c+d x))^{15/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(13/2)/(a + a*Sin[c + d*x])^3,x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 2.81, size = 216, normalized size = 1.57

method	result	size
default	\(\frac {2 e^{7} \left (-240 \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-504 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+480 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+504 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+200 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+231 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-126 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-440 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+125 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(216\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

2/105/a^3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^7*(-240*sin(1/2*d*x+1/2*c)^9-504*sin(1/2*d*
x+1/2*c)^6*cos(1/2*d*x+1/2*c)+480*sin(1/2*d*x+1/2*c)^7+504*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+200*sin(1/2
*d*x+1/2*c)^5+231*(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))-126*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-440*sin(1/2*d*x+1/2*c)^3+125*sin(1/2*d*x+1/2*c))/d

________________________________________________________________________________________

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))^(13/2)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

e^(13/2)*integrate(cos(d*x + c)^(13/2)/(a*sin(d*x + c) + a)^3, x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 114, normalized size = 0.83 \begin {gather*} \frac {231 i \, \sqrt {2} e^{\frac {13}{2}} {\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 i \, \sqrt {2} e^{\frac {13}{2}} {\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 (15 \, \cos \left (d x + c\right )^{3} e^{\frac {13}{2}} + 63 \, \cos \left (d x + c\right ) e^{\frac {13}{2}} \sin \left (d x + c\right ) - 140 \, \cos \left (d x + c\right ) e^{\frac {13}{2}}\right )} \sqrt {\cos \left (d x + c\right )}}{105 \, a^{3} d} \end {gather*}

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))^3,x, algorithm="fricas")

[Out]

1/105*(231*I*sqrt(2)*e^(13/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))
) - 231*I*sqrt(2)*e^(13/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) -
2*(15*cos(d*x + c)^3*e^(13/2) + 63*cos(d*x + c)*e^(13/2)*sin(d*x + c) - 140*cos(d*x + c)*e^(13/2))*sqrt(cos(d
*x + c)))/(a^3*d)

________________________________________________________________________________________

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))**(13/2)/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

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))^(13/2)/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(13/2)*e^(13/2)/(a*sin(d*x + c) + a)^3, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{13/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \end {gather*}

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))^3,x)

[Out]

int((e*cos(c + d*x))^(13/2)/(a + a*sin(c + d*x))^3, x)

________________________________________________________________________________________