\(\int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx\) [588]

Optimal result

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

[Out]

-1/2*a*e^(3/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(3/2)/(-a^2+b^2)^(3/4)/d-1/2*a*
e^(3/2)*arctanh(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(3/2)/(-a^2+b^2)^(3/4)/d-e^2*(cos(1/2
*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/b^2/d/(e*cos(d*
x+c))^(1/2)+1/2*a^2*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(b-(
-a^2+b^2)^(1/2)),2^(1/2))*cos(d*x+c)^(1/2)/b^2/d/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2)+1/2*a^2*e^2
*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2
))*cos(d*x+c)^(1/2)/b^2/d/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2)-e*(e*cos(d*x+c))^(1/2)/b/d/(a+b*si
n(d*x+c))

Rubi [A] (verified)

Time = 0.59 (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, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx=-\frac {a e^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} d \left (b^2-a^2\right )^{3/4}}-\frac {a e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} d \left (b^2-a^2\right )^{3/4}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{2 b^2 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{2 b^2 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d \sqrt {e \cos (c+d x)}} \]

[In]

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

[Out]

-1/2*(a*e^(3/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(b^(3/2)*(-a^2 + b^2)^(3/
4)*d) - (a*e^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*b^(3/2)*(-a^2 + b^
2)^(3/4)*d) - (e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(b^2*d*Sqrt[e*Cos[c + d*x]]) + (a^2*e^2*Sqrt[
Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(2*b^2*(a^2 - b*(b - Sqrt[-a^2 + b^2])
)*d*Sqrt[e*Cos[c + d*x]]) + (a^2*e^2*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2,
2])/(2*b^2*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*d*Sqrt[e*Cos[c + d*x]]) - (e*Sqrt[e*Cos[c + d*x]])/(b*d*(a + b*Sin
[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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 2781

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[
-a^2 + b^2, 2]}, Dist[-a/(2*q), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Dist[b*(g/f), Sub
st[Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e
+ f*x]]*(q - b*Cos[e + f*x])), 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 \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {e^2 \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{2 b} \\ & = -\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{2 b^2}+\frac {\left (a e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{2 b^2} \\ & = -\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {\left (a^2 e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{4 b^2 \sqrt {-a^2+b^2}}-\frac {\left (a^2 e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{4 b^2 \sqrt {-a^2+b^2}}+\frac {\left (a e^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \cos (c+d x)\right )}{2 b d}-\frac {\left (e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 b^2 \sqrt {e \cos (c+d x)}} \\ & = -\frac {e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d \sqrt {e \cos (c+d x)}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}+\frac {\left (a e^3\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{b d}-\frac {\left (a^2 e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{4 b^2 \sqrt {-a^2+b^2} \sqrt {e \cos (c+d x)}}-\frac {\left (a^2 e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{4 b^2 \sqrt {-a^2+b^2} \sqrt {e \cos (c+d x)}} \\ & = -\frac {e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {a^2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^2 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {\left (a e^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{2 b \sqrt {-a^2+b^2} d}-\frac {\left (a e^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{2 b \sqrt {-a^2+b^2} d} \\ & = -\frac {a e^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{3/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {a e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{3/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {a^2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^2 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (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 = 8.82 (sec) , antiderivative size = 562, normalized size of antiderivative = 1.39 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx=\frac {(e \cos (c+d x))^{3/2} \left (-\sqrt {\cos (c+d x)}+\left (a+b \sqrt {\sin ^2(c+d x)}\right ) \left (\frac {a \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}+\frac {5 b \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\cos (c+d x)} \sqrt {\sin ^2(c+d x)}}{\left (a^2-b^2+b^2 \cos ^2(c+d x)\right ) \left (-5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )+2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},2,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )+\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )\right ) \cos ^2(c+d x)\right )}\right )\right )}{b d \cos ^{\frac {3}{2}}(c+d x) (a+b \sin (c+d x))} \]

[In]

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

[Out]

((e*Cos[c + d*x])^(3/2)*(-Sqrt[Cos[c + d*x]] + (a + b*Sqrt[Sin[c + d*x]^2])*((a*(-2*ArcTan[1 - (Sqrt[2]*Sqrt[b
]*Sqrt[Cos[c + d*x]])/(a^2 - b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[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[Cos[c + d*x]] + b*Cos[c + d*x]] + Log[Sqrt[a^
2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[c + d*x]] + b*Cos[c + d*x]]))/(4*Sqrt[2]*Sqrt[b]*(a^2 -
b^2)^(3/4)) + (5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]*
Sqrt[Cos[c + d*x]]*Sqrt[Sin[c + d*x]^2])/((a^2 - b^2 + b^2*Cos[c + d*x]^2)*(-5*(a^2 - b^2)*AppellF1[1/4, -1/2,
1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Cos[c + d*x
]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Cos[c + d*x]^2, (b^2*Cos[c +
d*x]^2)/(-a^2 + b^2)])*Cos[c + d*x]^2)))))/(b*d*Cos[c + d*x]^(3/2)*(a + b*Sin[c + d*x]))

Maple [C] (warning: unable to verify)

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

Time = 7.73 (sec) , antiderivative size = 1803, normalized size of antiderivative = 4.46

method	result	size
default	\(\text {Expression too large to display}\)	\(1803\)

[In]

int((e*cos(d*x+c))^(3/2)/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

(8*e^2*a*b*(1/b^2*(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((2*e*cos(1/2*d*x+1/2*c)^2-e+(e^2*(a^2-b^2)/b^2)^(1/4)*
(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(2*e*cos(1/2*d*x+1/2*c)^2-e-(e^2*(a^2-b^
2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2)*(2*e*co
s(1/2*d*x+1/2*c)^2-e)^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))+2*arctan((2^(1/2)*(2*e*cos(1
/2*d*x+1/2*c)^2-e)^(1/2)-(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4)))/(16*a^2-16*b^2)/e-1/64*(a^2-b^
2)/b^2*(3*(ln((2*e*cos(1/2*d*x+1/2*c)^2-e+(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)
+(e^2*(a^2-b^2)/b^2)^(1/2))/(2*e*cos(1/2*d*x+1/2*c)^2-e-(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)
^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)+(e^2*(a^2-b^2)
/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))+2*arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)-(e^2*(a^2-b^2)/b^
2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4)))*(4*cos(1/2*d*x+1/2*c)^4*b^2-4*cos(1/2*d*x+1/2*c)^2*b^2+a^2)*2^(1/2)*(e^2
*(a^2-b^2)/b^2)^(1/4)+(8*a^2-8*b^2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2))/e/(a-b)^2/(a+b)^2/(4*cos(1/2*d*x+1/2*c
)^4*b^2-4*cos(1/2*d*x+1/2*c)^2*b^2+a^2))-2*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*e^2*(-1/b
^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*
c)^2))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/16/b^4*(-3*a^2+b^2)*sum(1/_alpha/(2*_alpha^2-1)*(2^(1/2)/
(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*arctanh(1/2*e*(4*_alpha^2-3)/(4*a^2-3*b^2)*(4*a^2*cos(1/2*d*x+1/2*c)^
2-3*cos(1/2*d*x+1/2*c)^2*b^2+b^2*_alpha^2-3*a^2+2*b^2)*2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-e*(2
*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2))+8*b^2/a^2*_alpha*(_alpha^2-1)*(sin(1/2*d*x+1/2*c)^2)^(1/2)
*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-e*sin(1/2*d*x+1/2*c)^2*(2*sin(1/2*d*x+1/2*c)^2-1))^(1/2)*EllipticPi(cos(1
/2*d*x+1/2*c),-4*b^2/a^2*(_alpha^2-1),2^(1/2))),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2))-2*a^2*(a^2-b^2)/b^2*
(1/2*b^2/(a^2-b^2)/a^2/e*cos(1/2*d*x+1/2*c)*(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(4*cos(1/
2*d*x+1/2*c)^4*b^2-4*cos(1/2*d*x+1/2*c)^2*b^2+a^2)-1/4/(a^2-b^2)/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*
d*x+1/2*c)^2+1)^(1/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^
(1/2))-1/64/a^2/b^2*sum((-5*a^2+2*b^2)/(a-b)/(a+b)/(2*_alpha^2-1)/_alpha*(2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2
)/b^2)^(1/2)*arctanh(1/2*e*(4*_alpha^2-3)/(4*a^2-3*b^2)*(4*a^2*cos(1/2*d*x+1/2*c)^2-3*cos(1/2*d*x+1/2*c)^2*b^2
+b^2*_alpha^2-3*a^2+2*b^2)*2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/
2*d*x+1/2*c)^2))^(1/2))+8*b^2/a^2*_alpha*(_alpha^2-1)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)
^(1/2)/(-e*sin(1/2*d*x+1/2*c)^2*(2*sin(1/2*d*x+1/2*c)^2-1))^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-4*b^2/a^2*(_a
lpha^2-1),2^(1/2))),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2))))/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-
1))^(1/2))/d

Fricas [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]

[In]

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

[Out]

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