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

Optimal. Leaf size=397 \[ -\frac{e^{3/2} \sqrt [4]{b^2-a^2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}-\frac{e^{3/2} \sqrt [4]{b^2-a^2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}-\frac{a e^2 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \cos (c+d x)}}-\frac{a e^2 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \cos (c+d x)}}+\frac{2 a e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt{e \cos (c+d x)}}+\frac{2 e \sqrt{e \cos (c+d x)}}{b d} \]

[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]])

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Rubi [A]  time = 0.881962, antiderivative size = 397, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {2695, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ -\frac{e^{3/2} \sqrt [4]{b^2-a^2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}-\frac{e^{3/2} \sqrt [4]{b^2-a^2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}-\frac{a e^2 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \cos (c+d x)}}-\frac{a e^2 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \cos (c+d x)}}+\frac{2 a e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt{e \cos (c+d x)}}+\frac{2 e \sqrt{e \cos (c+d x)}}{b d} \]

Antiderivative was successfully verified.

[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 2695

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[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 2867

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]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2702

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 2807

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*
Sin[e + f*x])/(c + d)]), 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 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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 329

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 212

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] &&
!GtQ[a/b, 0]

Rule 208

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx &=\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 ) \operatorname{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)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt{e \cos (c+d x)}}-\frac{\left (2 \left (a^2-b^2\right ) e^3\right ) \operatorname{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)} F\left (\left .\frac{1}{2} (c+d x)\right |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)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |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)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |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 ) \operatorname{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 ) \operatorname{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} \tan ^{-1}\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} \tanh ^{-1}\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)} F\left (\left .\frac{1}{2} (c+d x)\right |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)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |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)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 4.70315, size = 233, normalized size = 0.59 \[ \frac{\sec ^4(c+d x) (e \cos (c+d x))^{3/2} \left (a^2+b^2 \cos ^2(c+d x)-b^2\right ) \left (\frac{2 b \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};\left (1-\frac{a^2}{b^2}\right ) \sec ^2(c+d x)\right )}{\sqrt [4]{\sec ^2(c+d x)}}+\frac{a \tan (c+d x) \left (\Pi \left (-\frac{\sqrt{b^2-a^2}}{b};\left .-\sin ^{-1}\left (\sqrt [4]{\sec ^2(c+d x)}\right )\right |-1\right )+\Pi \left (\frac{\sqrt{b^2-a^2}}{b};\left .-\sin ^{-1}\left (\sqrt [4]{\sec ^2(c+d x)}\right )\right |-1\right )\right )}{\sqrt{-\tan ^2(c+d x)}}\right )}{b^2 d \sec ^2(c+d x)^{3/4} (a+b \sin (c+d x)) \left (a \sqrt{\sec ^2(c+d x)}-b \tan (c+d x)\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 2.537, size = 1266, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*e/b*(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)-2/d*e^3/b*sum((_R^4+_R^2*e)/(_R^7*b^2-3*_R^5*b^2*e+8*_R^3*a^2*e^2
-5*_R^3*b^2*e^2-_R*b^2*e^3)*ln((-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)-e^(1/2)*cos(1/2*d*x+1/2*c)*2^(1/2)-_R),_R=R
ootOf(b^2*_Z^8-4*b^2*e*_Z^6+(16*a^2*e^2-10*b^2*e^2)*_Z^4-4*b^2*e^3*_Z^2+b^2*e^4))*a^2+2/d*e^3*b*sum((_R^4+_R^2
*e)/(_R^7*b^2-3*_R^5*b^2*e+8*_R^3*a^2*e^2-5*_R^3*b^2*e^2-_R*b^2*e^3)*ln((-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)-e^
(1/2)*cos(1/2*d*x+1/2*c)*2^(1/2)-_R),_R=RootOf(b^2*_Z^8-4*b^2*e*_Z^6+(16*a^2*e^2-10*b^2*e^2)*_Z^4-4*b^2*e^3*_Z
^2+b^2*e^4))-2/d*(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*a/sin(1/2*d*x+1/2*c)/(e*(2*cos(
1/2*d*x+1/2*c)^2-1))^(1/2)/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))+1/8/d*(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*a^3/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/b^4*sum(1/_al
pha/(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*cos(1/2*d*x+1/2*c)^2*a^2-3*b^2*cos(1/2*d*x+1/2*c)^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)/(-sin(1/2*d*x+1/2*c)^2*e*(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))-1/8/d*(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*a/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2
*d*x+1/2*c)^2-1))^(1/2)/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)*arct
anh(1/2*e*(4*_alpha^2-3)/(4*a^2-3*b^2)*(4*cos(1/2*d*x+1/2*c)^2*a^2-3*b^2*cos(1/2*d*x+1/2*c)^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)/(-sin(1/2*
d*x+1/2*c)^2*e*(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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{b \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{b \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)