\(\int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx\) [71]

Optimal result

Integrand size = 25, antiderivative size = 404 \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx=-\frac {3 a e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}+\frac {3 a e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}+\frac {3 a^2 e^3 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^3 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a^2 e^3 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^3 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {3 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b^2 d \sqrt {\sin (c+d x)}}+\frac {e (e \sin (c+d x))^{3/2}}{b d (a+b \cos (c+d x))} \]

[Out]

-3/2*a*e^(5/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(5/2)/(-a^2+b^2)^(1/4)/d+3/2*a*
e^(5/2)*arctanh(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(5/2)/(-a^2+b^2)^(1/4)/d+e*(e*sin(d*x
+c))^(3/2)/b/d/(a+b*cos(d*x+c))-3/2*a^2*e^3*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*Elli
pticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/b^3/d/(b-(-a^2+b^2)^(1/2))
/(e*sin(d*x+c))^(1/2)-3/2*a^2*e^3*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticPi(cos
(1/2*c+1/4*Pi+1/2*d*x),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/b^3/d/(b+(-a^2+b^2)^(1/2))/(e*sin(d*
x+c))^(1/2)+3*e^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2
*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/b^2/d/sin(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {2772, 2946, 2721, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx=-\frac {3 a e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{5/2} d \sqrt [4]{b^2-a^2}}+\frac {3 a e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{5/2} d \sqrt [4]{b^2-a^2}}+\frac {3 a^2 e^3 \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{2 b^3 d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}+\frac {3 a^2 e^3 \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{2 b^3 d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}+\frac {e (e \sin (c+d x))^{3/2}}{b d (a+b \cos (c+d x))}-\frac {3 e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b^2 d \sqrt {\sin (c+d x)}} \]

[In]

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

[Out]

(-3*a*e^(5/2)*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*b^(5/2)*(-a^2 + b^2)^(1/
4)*d) + (3*a*e^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*b^(5/2)*(-a^2 +
b^2)^(1/4)*d) + (3*a^2*e^3*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])
/(2*b^3*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (3*a^2*e^3*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (
c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*b^3*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) - (3*e^2*Ellip
ticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(b^2*d*Sqrt[Sin[c + d*x]]) + (e*(e*Sin[c + d*x])^(3/2))/(b*d
*(a + b*Cos[c + d*x]))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, 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 2772

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

Rule 2780

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> With[{q = Rt[-a^2
+ b^2, 2]}, Dist[a*(g/(2*b)), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (-Dist[a*(g/(2*b)),
Int[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x] + Dist[b*(g/f), Subst[Int[Sqrt[x]/(g^2*(a^2 - b^2)
+ b^2*x^2), x], x, g*Cos[e + f*x]], x])] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 2886

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/
(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2946

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]))/((a_) + (b_.)*sin[(e_.) + (
f_.)*(x_)]), x_Symbol] :> Dist[d/b, Int[(g*Cos[e + f*x])^p, x], x] + Dist[(b*c - a*d)/b, Int[(g*Cos[e + f*x])^
p/(a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {e (e \sin (c+d x))^{3/2}}{b d (a+b \cos (c+d x))}-\frac {\left (3 e^2\right ) \int \frac {\cos (c+d x) \sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{2 b} \\ & = \frac {e (e \sin (c+d x))^{3/2}}{b d (a+b \cos (c+d x))}-\frac {\left (3 e^2\right ) \int \sqrt {e \sin (c+d x)} \, dx}{2 b^2}+\frac {\left (3 a e^2\right ) \int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{2 b^2} \\ & = \frac {e (e \sin (c+d x))^{3/2}}{b d (a+b \cos (c+d x))}-\frac {\left (3 a^2 e^3\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{4 b^3}+\frac {\left (3 a^2 e^3\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{4 b^3}-\frac {\left (3 a e^3\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \sin (c+d x)\right )}{2 b d}-\frac {\left (3 e^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{2 b^2 \sqrt {\sin (c+d x)}} \\ & = -\frac {3 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b^2 d \sqrt {\sin (c+d x)}}+\frac {e (e \sin (c+d x))^{3/2}}{b d (a+b \cos (c+d x))}-\frac {\left (3 a e^3\right ) \text {Subst}\left (\int \frac {x^2}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{b d}-\frac {\left (3 a^2 e^3 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{4 b^3 \sqrt {e \sin (c+d x)}}+\frac {\left (3 a^2 e^3 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{4 b^3 \sqrt {e \sin (c+d x)}} \\ & = \frac {3 a^2 e^3 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^3 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a^2 e^3 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^3 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {3 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b^2 d \sqrt {\sin (c+d x)}}+\frac {e (e \sin (c+d x))^{3/2}}{b d (a+b \cos (c+d x))}+\frac {\left (3 a e^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{2 b^2 d}-\frac {\left (3 a e^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{2 b^2 d} \\ & = -\frac {3 a e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}+\frac {3 a e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}+\frac {3 a^2 e^3 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^3 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a^2 e^3 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^3 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {3 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b^2 d \sqrt {\sin (c+d x)}}+\frac {e (e \sin (c+d x))^{3/2}}{b d (a+b \cos (c+d x))} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 5.90 (sec) , antiderivative size = 366, normalized size of antiderivative = 0.91 \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx=\frac {(e \sin (c+d x))^{5/2} \left (8 b^{3/2} \csc (c+d x)+\frac {\left (a+b \sqrt {\cos ^2(c+d x)}\right ) \left (3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )\right )+8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{\left (a^2-b^2\right ) \sin ^{\frac {5}{2}}(c+d x)}\right )}{8 b^{5/2} d (a+b \cos (c+d x))} \]

[In]

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

[Out]

((e*Sin[c + d*x])^(5/2)*(8*b^(3/2)*Csc[c + d*x] + ((a + b*Sqrt[Cos[c + d*x]^2])*(3*Sqrt[2]*a*(a^2 - b^2)^(3/4)
*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Si
n[c + d*x]])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] +
b*Sin[c + d*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]
]) + 8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(3/
2)))/((a^2 - b^2)*Sin[c + d*x]^(5/2))))/(8*b^(5/2)*d*(a + b*Cos[c + d*x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1667\) vs. \(2(436)=872\).

Time = 15.83 (sec) , antiderivative size = 1668, normalized size of antiderivative = 4.13

method	result	size
default	\(\text {Expression too large to display}\)	\(1668\)

[In]

int((e*sin(d*x+c))^(5/2)/(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)

[Out]

(-2*e^3*a*b*(-1/2*(e*sin(d*x+c))^(3/2)/b^2/(-b^2*cos(d*x+c)^2*e^2+a^2*e^2)+3/16/b^4/(e^2*(a^2-b^2)/b^2)^(1/4)*
2^(1/2)*(ln((e*sin(d*x+c)-(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(e
*sin(d*x+c)+(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan(2^(1/2
)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)+1)+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^
(1/2)-1)))+1/4*e^3*a^2*(3*(-a^2+b^2)^(1/2)*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*Ellipt
icPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*b^2-3*(-a^2+b^2)^(1/2)*(1-sin(d*x+c))^(1/2)*(2
*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticPi((1-sin(d*x+c))^(1/2),1/(b+(-a^2+b^2)^(1/2))*b,1/2*2^(1/2))*b^
2-12*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*
b^3+6*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))
*b^3+3*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a
^2+b^2)^(1/2)),1/2*2^(1/2))*b^3+3*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticPi((1-s
in(d*x+c))^(1/2),1/(b+(-a^2+b^2)^(1/2))*b,1/2*2^(1/2))*b^3+3*(-a^2+b^2)^(1/2)*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+
c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*a^2-3*(-a^2
+b^2)^(1/2)*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticPi((1-sin(d*x+c))^(1/2),-b/(-
b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*b^2-3*(-a^2+b^2)^(1/2)*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)
^(1/2)*EllipticPi((1-sin(d*x+c))^(1/2),1/(b+(-a^2+b^2)^(1/2))*b,1/2*2^(1/2))*a^2+3*(-a^2+b^2)^(1/2)*(1-sin(d*x
+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticPi((1-sin(d*x+c))^(1/2),1/(b+(-a^2+b^2)^(1/2))*b,1/
2*2^(1/2))*b^2-12*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),
1/2*2^(1/2))*a^2*b+12*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((1-sin(d*x+c))^(1
/2),1/2*2^(1/2))*b^3+6*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(
1/2),1/2*2^(1/2))*a^2*b-6*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c)
)^(1/2),1/2*2^(1/2))*b^3+3*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticPi((1-sin(d*x+
c))^(1/2),-b/(-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*a^2*b-3*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)
^(1/2)*EllipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*b^3+3*(1-sin(d*x+c))^(1/2)*(2*sin
(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticPi((1-sin(d*x+c))^(1/2),1/(b+(-a^2+b^2)^(1/2))*b,1/2*2^(1/2))*a^2*b-
3*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticPi((1-sin(d*x+c))^(1/2),1/(b+(-a^2+b^2)
^(1/2))*b,1/2*2^(1/2))*b^3-4*b^3*sin(d*x+c)^4+4*b^3*sin(d*x+c)^2)/b^3/(b+(-a^2+b^2)^(1/2))/(-b+(-a^2+b^2)^(1/2
))/(-b^2*cos(d*x+c)^2+a^2)/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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