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

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [B] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 154 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {2 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^4 d \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a+a \sin (c+d x))^3}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^4+a^4 \sin (c+d x)\right )} \]

[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)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2759, 2760, 2762, 2721, 2719} \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{15 a^4 d \sqrt {\cos (c+d x)}}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^4 \sin (c+d x)+a^4\right )}+\frac {2 e (e \cos (c+d x))^{3/2}}{15 d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{9 a d (a \sin (c+d x)+a)^3} \]

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

Mathematica [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.43 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {(e \cos (c+d x))^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {13}{4},\frac {11}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{14 \sqrt [4]{2} a^4 d e (1+\sin (c+d x))^{7/4}} \]

[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))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(513\) vs. \(2(162)=324\).

Time = 708.68 (sec) , antiderivative size = 514, normalized size of antiderivative = 3.34

method result size
default \(-\frac {2 \left (96 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-48 E\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 )-192 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+96 E\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 )+272 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-72 \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 ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\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 )+24 \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 ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-144 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )-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}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {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}\) \(514\)

[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/s
in(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(96*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-48*Elliptic
E(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)
^8-192*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+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+272*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-72*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+
1/2*c)^4-176*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+24*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-144*sin(1/2*d*x+1/2*c)^5-42*sin(1/2*d*x+1/2
*c)^2*cos(1/2*d*x+1/2*c)-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))+144*sin(1/2*d*x+1/2*c)^3+4*sin(1/2*d*x+1/2*c))*e^3/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 = 436, normalized size of antiderivative = 2.83 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {3 \, {\left (-i \, \sqrt {2} e^{2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} e^{2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} e^{2} \cos \left (d x + c\right ) + 4 i \, \sqrt {2} e^{2} + {\left (-i \, \sqrt {2} e^{2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} e^{2} \cos \left (d x + c\right ) + 4 i \, \sqrt {2} e^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (i \, \sqrt {2} e^{2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} e^{2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} e^{2} \cos \left (d x + c\right ) - 4 i \, \sqrt {2} e^{2} + {\left (i \, \sqrt {2} e^{2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} e^{2} \cos \left (d x + c\right ) - 4 i \, \sqrt {2} e^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, e^{2} \cos \left (d x + c\right )^{3} - 6 \, e^{2} \cos \left (d x + c\right )^{2} + e^{2} \cos \left (d x + c\right ) + 10 \, e^{2} - {\left (3 \, e^{2} \cos \left (d x + c\right )^{2} + 9 \, e^{2} \cos \left (d x + c\right ) + 10 \, e^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{45 \, {\left (a^{4} d \cos \left (d x + c\right )^{3} + 3 \, a^{4} d \cos \left (d x + c\right )^{2} - 2 \, a^{4} d \cos \left (d x + c\right ) - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 2 \, a^{4} d \cos \left (d x + c\right ) - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

[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)*e^2*cos(d*x + c)^3 - 3*I*sqrt(2)*e^2*cos(d*x + c)^2 + 2*I*sqrt(2)*e^2*cos(d*x + c) + 4*I*
sqrt(2)*e^2 + (-I*sqrt(2)*e^2*cos(d*x + c)^2 + 2*I*sqrt(2)*e^2*cos(d*x + c) + 4*I*sqrt(2)*e^2)*sin(d*x + c))*s
qrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(I*sqrt(2)*e^2*co
s(d*x + c)^3 + 3*I*sqrt(2)*e^2*cos(d*x + c)^2 - 2*I*sqrt(2)*e^2*cos(d*x + c) - 4*I*sqrt(2)*e^2 + (I*sqrt(2)*e^
2*cos(d*x + c)^2 - 2*I*sqrt(2)*e^2*cos(d*x + c) - 4*I*sqrt(2)*e^2)*sin(d*x + c))*sqrt(e)*weierstrassZeta(-4, 0
, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*e^2*cos(d*x + c)^3 - 6*e^2*cos(d*x + c)^2
+ e^2*cos(d*x + c) + 10*e^2 - (3*e^2*cos(d*x + c)^2 + 9*e^2*cos(d*x + c) + 10*e^2)*sin(d*x + c))*sqrt(e*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))

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]

[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)