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

Optimal. Leaf size=384 \[ \frac{e^{5/2} \left (b^2-a^2\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{5/2} d}-\frac{e^{5/2} \left (b^2-a^2\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{5/2} d}-\frac{a e^3 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \cos (c+d x)}}-\frac{a e^3 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \cos (c+d x)}}+\frac{2 a e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{b^2 d \sqrt{\cos (c+d x)}}+\frac{2 e (e \cos (c+d x))^{3/2}}{3 b d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.869977, antiderivative size = 384, 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, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac{e^{5/2} \left (b^2-a^2\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{5/2} d}-\frac{e^{5/2} \left (b^2-a^2\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{5/2} d}-\frac{a e^3 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \cos (c+d x)}}-\frac{a e^3 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \cos (c+d x)}}+\frac{2 a e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{b^2 d \sqrt{\cos (c+d x)}}+\frac{2 e (e \cos (c+d x))^{3/2}}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-a^2 + b^2)^(3/4)*e^(5/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(b^(5/2)*d) -
((-a^2 + b^2)^(3/4)*e^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(b^(5/2)*d)
+ (2*e*(e*Cos[c + d*x])^(3/2))/(3*b*d) + (2*a*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(b^2*d*Sqrt
[Cos[c + d*x]]) - (a*(a^2 - b^2)*e^3*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2,
2])/(b^3*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]]) - (a*(a^2 - b^2)*e^3*Sqrt[Cos[c + d*x]]*EllipticPi[(2*
b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(b^3*(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 2640

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

Rule 2639

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

Rule 2701

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 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 298

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 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]

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]

Rubi steps

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

Mathematica [C]  time = 21.4337, size = 709, normalized size = 1.85 \[ \frac{(e \cos (c+d x))^{5/2} \left (-\frac{a \left (a+b \sqrt{\sin ^2(c+d x)}\right ) \left (8 b^{5/2} \cos ^{\frac{3}{2}}(c+d x) F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right )+3 \sqrt{2} a \left (a^2-b^2\right )^{3/4} \left (-\log \left (-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (c+d x)}+\sqrt{a^2-b^2}+b \cos (c+d x)\right )+\log \left (\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (c+d x)}+\sqrt{a^2-b^2}+b \cos (c+d x)\right )+2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )\right )\right )}{4 b^{3/2} \left (b^2-a^2\right ) (a+b \sin (c+d x))}-\frac{6 b \sin (c+d x) \left (a+b \sqrt{\sin ^2(c+d x)}\right ) \left (\frac{a \cos ^{\frac{3}{2}}(c+d x) F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\cos ^2(c+d x),\frac{b^2 \cos ^2(c+d x)}{b^2-a^2}\right )}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (-\log \left (-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (c+d x)}+\sqrt{b^2-a^2}+i b \cos (c+d x)\right )+\log \left ((1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (c+d x)}+\sqrt{b^2-a^2}+i b \cos (c+d x)\right )+2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (1+\frac{(1+i) \sqrt{b} \sqrt{\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{\sin ^2(c+d x)} (a+b \sin (c+d x))}+2 \cos ^{\frac{3}{2}}(c+d x)\right )}{3 b d \cos ^{\frac{5}{2}}(c+d x)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((e*Cos[c + d*x])^(5/2)*(2*Cos[c + d*x]^(3/2) - (a*(8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Cos[c + d*x]^2, (b^2
*Cos[c + d*x]^2)/(-a^2 + b^2)]*Cos[c + d*x]^(3/2) + 3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(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]]))*(a + b*Sqrt[Sin[c + d*x]^
2]))/(4*b^(3/2)*(-a^2 + b^2)*(a + b*Sin[c + d*x])) - (6*b*((a*AppellF1[3/4, 1/2, 1, 7/4, Cos[c + d*x]^2, (b^2*
Cos[c + d*x]^2)/(-a^2 + b^2)]*Cos[c + d*x]^(3/2))/(3*(a^2 - b^2)) + ((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sqrt[b
]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/
4)] - Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[c + d*x]] + I*b*Cos[c + d*x]] + Log[S
qrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[c + d*x]] + I*b*Cos[c + d*x]]))/(Sqrt[b]*(-a^2 +
b^2)^(1/4)))*Sin[c + d*x]*(a + b*Sqrt[Sin[c + d*x]^2]))/(Sqrt[Sin[c + d*x]^2]*(a + b*Sin[c + d*x]))))/(3*b*d*
Cos[c + d*x]^(5/2))

________________________________________________________________________________________

Maple [C]  time = 2.72, size = 1131, normalized size = 3. \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))^(5/2)/(a+b*sin(d*x+c)),x)

[Out]

4/3/d*e^2/b*cos(1/2*d*x+1/2*c)^2*(2*cos(1/2*d*x+1/2*c)^2*e-e)^(1/2)+4/3/d*e^2/b*(2*cos(1/2*d*x+1/2*c)^2*e-e)^(
1/2)-2/d*e^2/b*(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)-1/2/d*e^3/b*sum((_R^6-_R^4*e-_R^2*e^2+e^3)/(_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))*a^2+1/
2/d*e^3*b*sum((_R^6-_R^4*e-_R^2*e^2+e^3)/(_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^3*a/b^2/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*
c)^2-1))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(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^3/a/b^4/sin(1/2*d*x+1/2*c)/(e*(2*cos(1
/2*d*x+1/2*c)^2-1))^(1/2)*sum((a^2-b^2)/_alpha*(8*(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-4*b^2/a^2*(_alpha^2-1),2^(1/2))*_al
pha^3*b^2-8*b^2*_alpha*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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))*(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)+a^2*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))*(-sin(1
/2*d*x+1/2*c)^2*e*(2*sin(1/2*d*x+1/2*c)^2-1))^(1/2))/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(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),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{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))^(5/2)/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{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))^(5/2)/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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