\(\int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx\) [578]

Optimal result

Integrand size = 25, antiderivative size = 397 \[ \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx=-\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}-\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {2 a e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d \sqrt {e \cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}} \]

[Out]

-(-a^2+b^2)^(1/4)*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)/d-(-a^2+b^2)^(
1/4)*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)/d+2*a*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)-a*(a^2-b^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)-a*(a^2-b^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)+2*e*(e*cos(d*x+c))^(1/2)/b/d

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 397, 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 = {2774, 2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \[ \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx=-\frac {e^{3/2} \sqrt [4]{b^2-a^2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}-\frac {e^{3/2} \sqrt [4]{b^2-a^2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}-\frac {a e^2 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b^2 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {a e^2 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b^2 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}+\frac {2 a e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d} \]

[In]

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

[Out]

-(((-a^2 + b^2)^(1/4)*e^(3/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(b^(3/2)*d)
) - ((-a^2 + b^2)^(1/4)*e^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(b^(3/2)
*d) + (2*e*Sqrt[e*Cos[c + d*x]])/(b*d) + (2*a*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(b^2*d*Sqrt[e*
Cos[c + d*x]]) - (a*(a^2 - b^2)*e^2*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2
])/(b^2*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*d*Sqrt[e*Cos[c + d*x]]) - (a*(a^2 - b^2)*e^2*Sqrt[Cos[c + d*x]]*Ellip
ticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(b^2*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*d*Sqrt[e*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 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 2774

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 + p))), x] + Dist[g^2*((p - 1)/(b*(m + p))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && 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 {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {e^2 \int \frac {b+a \sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{b} \\ & = \frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {\left (a e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{b^2}+\frac {\left (\left (-a^2+b^2\right ) e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{b^2} \\ & = \frac {2 e \sqrt {e \cos (c+d x)}}{b d}-\frac {\left (a \sqrt {-a^2+b^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}{2 b^2}-\frac {\left (a \sqrt {-a^2+b^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}{2 b^2}-\frac {\left (\left (a^2-b^2\right ) 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 )}{b d}+\frac {\left (a e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{b^2 \sqrt {e \cos (c+d x)}} \\ & = \frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {2 a 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 {\left (2 \left (a^2-b^2\right ) 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 \sqrt {-a^2+b^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}{2 b^2 \sqrt {e \cos (c+d x)}}-\frac {\left (a \sqrt {-a^2+b^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}{2 b^2 \sqrt {e \cos (c+d x)}} \\ & = \frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {2 a 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 \sqrt {-a^2+b^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 )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {a \sqrt {-a^2+b^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 )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {\left (\sqrt {-a^2+b^2} 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 )}{b d}-\frac {\left (\sqrt {-a^2+b^2} 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 )}{b d} \\ & = -\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}-\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {2 a 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 \sqrt {-a^2+b^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 )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {a \sqrt {-a^2+b^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 )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 11.92 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.55 \[ \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx=-\frac {e \sqrt {e \cos (c+d x)} \left (a^2-b^2+b^2 \cos ^2(c+d x)\right ) \csc ^2(c+d x) \left (2 b \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},\left (1-\frac {a^2}{b^2}\right ) \sec ^2(c+d x)\right ) \tan (c+d x)+a \left (\operatorname {EllipticPi}\left (-\frac {\sqrt {-a^2+b^2}}{b},\arcsin \left (\sqrt [4]{\sec ^2(c+d x)}\right ),-1\right )+\operatorname {EllipticPi}\left (\frac {\sqrt {-a^2+b^2}}{b},\arcsin \left (\sqrt [4]{\sec ^2(c+d x)}\right ),-1\right )\right ) \sqrt [4]{\sec ^2(c+d x)} \sqrt {-\tan ^2(c+d x)}\right )}{b^2 d (b+a \csc (c+d x)) \left (-a \sqrt {\sec ^2(c+d x)}+b \tan (c+d x)\right )} \]

[In]

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

[Out]

-((e*Sqrt[e*Cos[c + d*x]]*(a^2 - b^2 + b^2*Cos[c + d*x]^2)*Csc[c + d*x]^2*(2*b*Hypergeometric2F1[-1/4, 1, 3/4,
(1 - a^2/b^2)*Sec[c + d*x]^2]*Tan[c + d*x] + a*(EllipticPi[-(Sqrt[-a^2 + b^2]/b), ArcSin[(Sec[c + d*x]^2)^(1/
4)], -1] + EllipticPi[Sqrt[-a^2 + b^2]/b, ArcSin[(Sec[c + d*x]^2)^(1/4)], -1])*(Sec[c + d*x]^2)^(1/4)*Sqrt[-Ta
n[c + d*x]^2]))/(b^2*d*(b + a*Csc[c + d*x])*(-(a*Sqrt[Sec[c + d*x]^2]) + b*Tan[c + d*x])))

Maple [C] (warning: unable to verify)

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

Time = 2.93 (sec) , antiderivative size = 844, normalized size of antiderivative = 2.13

method	result	size
default	\(\text {Expression too large to display}\)	\(844\)

[In]

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

[Out]

(4*e^2*b*(1/2/b^2/e*(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)-(a^2-b^2)/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*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)))/(16*a^2-16*b^2)/e)-2*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*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*(-a^2+b^2)/b^4*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)))/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. \[ \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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