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

Optimal. Leaf size=430 \[ \frac{b e^{5/2} \left (a^2-b^2\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{7/2} d}-\frac{b e^{5/2} \left (a^2-b^2\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{7/2} d}+\frac{2 e^2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 a^3 d \sqrt{\sin (c+d x)}}-\frac{b^2 e^3 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a-\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^4 d \left (a-\sqrt{a^2-b^2}\right ) \sqrt{e \sin (c+d x)}}-\frac{b^2 e^3 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a+\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^4 d \left (\sqrt{a^2-b^2}+a\right ) \sqrt{e \sin (c+d x)}}+\frac{2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d} \]

[Out]

(b*(a^2 - b^2)^(3/4)*e^(5/2)*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^(7/2)*d) -
(b*(a^2 - b^2)^(3/4)*e^(5/2)*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^(7/2)*d)
- (b^2*(a^2 - b^2)*e^3*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a^
4*(a - Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) - (b^2*(a^2 - b^2)*e^3*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]),
(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a^4*(a + Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (2*(3*a^2 - 5
*b^2)*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*a^3*d*Sqrt[Sin[c + d*x]]) + (2*e*(5*b - 3*
a*Cos[c + d*x])*(e*Sin[c + d*x])^(3/2))/(15*a^2*d)

________________________________________________________________________________________

Rubi [A]  time = 1.10903, antiderivative size = 430, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {3872, 2865, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac{b e^{5/2} \left (a^2-b^2\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{7/2} d}-\frac{b e^{5/2} \left (a^2-b^2\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{7/2} d}+\frac{2 e^2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 a^3 d \sqrt{\sin (c+d x)}}-\frac{b^2 e^3 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a-\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^4 d \left (a-\sqrt{a^2-b^2}\right ) \sqrt{e \sin (c+d x)}}-\frac{b^2 e^3 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a+\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^4 d \left (\sqrt{a^2-b^2}+a\right ) \sqrt{e \sin (c+d x)}}+\frac{2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(a^2 - b^2)^(3/4)*e^(5/2)*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^(7/2)*d) -
(b*(a^2 - b^2)^(3/4)*e^(5/2)*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/(a^(7/2)*d)
- (b^2*(a^2 - b^2)*e^3*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a^
4*(a - Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) - (b^2*(a^2 - b^2)*e^3*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]),
(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a^4*(a + Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) + (2*(3*a^2 - 5
*b^2)*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*a^3*d*Sqrt[Sin[c + d*x]]) + (2*e*(5*b - 3*
a*Cos[c + d*x])*(e*Sin[c + d*x])^(3/2))/(15*a^2*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2865

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

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 \sin (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx &=-\int \frac{\cos (c+d x) (e \sin (c+d x))^{5/2}}{-b-a \cos (c+d x)} \, dx\\ &=\frac{2 e (5 b-3 a \cos (c+d x)) (e \sin (c+d x))^{3/2}}{15 a^2 d}-\frac{\left (2 e^2\right ) \int \frac{\left (-a b+\frac{1}{2} \left (3 a^2-5 b^2\right ) \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{-b-a \cos (c+d x)} \, dx}{5 a^2}\\ &=\frac{2 e (5 b-3 a \cos (c+d x)) (e \sin (c+d x))^{3/2}}{15 a^2 d}+\frac{\left (\left (3 a^2-5 b^2\right ) e^2\right ) \int \sqrt{e \sin (c+d x)} \, dx}{5 a^3}+\frac{\left (b \left (a^2-b^2\right ) e^2\right ) \int \frac{\sqrt{e \sin (c+d x)}}{-b-a \cos (c+d x)} \, dx}{a^3}\\ &=\frac{2 e (5 b-3 a \cos (c+d x)) (e \sin (c+d x))^{3/2}}{15 a^2 d}+\frac{\left (b^2 \left (a^2-b^2\right ) e^3\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{a^2-b^2}-a \sin (c+d x)\right )} \, dx}{2 a^4}-\frac{\left (b^2 \left (a^2-b^2\right ) e^3\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{a^2-b^2}+a \sin (c+d x)\right )} \, dx}{2 a^4}+\frac{\left (b \left (a^2-b^2\right ) e^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{\left (-a^2+b^2\right ) e^2+a^2 x^2} \, dx,x,e \sin (c+d x)\right )}{a^2 d}+\frac{\left (\left (3 a^2-5 b^2\right ) e^2 \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{5 a^3 \sqrt{\sin (c+d x)}}\\ &=\frac{2 \left (3 a^2-5 b^2\right ) 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)}}{5 a^3 d \sqrt{\sin (c+d x)}}+\frac{2 e (5 b-3 a \cos (c+d x)) (e \sin (c+d x))^{3/2}}{15 a^2 d}+\frac{\left (2 b \left (a^2-b^2\right ) e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (-a^2+b^2\right ) e^2+a^2 x^4} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{a^2 d}+\frac{\left (b^2 \left (a^2-b^2\right ) e^3 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)} \left (\sqrt{a^2-b^2}-a \sin (c+d x)\right )} \, dx}{2 a^4 \sqrt{e \sin (c+d x)}}-\frac{\left (b^2 \left (a^2-b^2\right ) e^3 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)} \left (\sqrt{a^2-b^2}+a \sin (c+d x)\right )} \, dx}{2 a^4 \sqrt{e \sin (c+d x)}}\\ &=-\frac{b^2 \left (a^2-b^2\right ) e^3 \Pi \left (\frac{2 a}{a-\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{a^4 \left (a-\sqrt{a^2-b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{b^2 \left (a^2-b^2\right ) e^3 \Pi \left (\frac{2 a}{a+\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{a^4 \left (a+\sqrt{a^2-b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{2 \left (3 a^2-5 b^2\right ) 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)}}{5 a^3 d \sqrt{\sin (c+d x)}}+\frac{2 e (5 b-3 a \cos (c+d x)) (e \sin (c+d x))^{3/2}}{15 a^2 d}-\frac{\left (b \left (a^2-b^2\right ) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a^2-b^2} e-a x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{a^3 d}+\frac{\left (b \left (a^2-b^2\right ) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a^2-b^2} e+a x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{a^3 d}\\ &=\frac{b \left (a^2-b^2\right )^{3/4} e^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt{e}}\right )}{a^{7/2} d}-\frac{b \left (a^2-b^2\right )^{3/4} e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt{e}}\right )}{a^{7/2} d}-\frac{b^2 \left (a^2-b^2\right ) e^3 \Pi \left (\frac{2 a}{a-\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{a^4 \left (a-\sqrt{a^2-b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{b^2 \left (a^2-b^2\right ) e^3 \Pi \left (\frac{2 a}{a+\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{a^4 \left (a+\sqrt{a^2-b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{2 \left (3 a^2-5 b^2\right ) 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)}}{5 a^3 d \sqrt{\sin (c+d x)}}+\frac{2 e (5 b-3 a \cos (c+d x)) (e \sin (c+d x))^{3/2}}{15 a^2 d}\\ \end{align*}

Mathematica [C]  time = 14.8852, size = 853, normalized size = 1.98 \[ \frac{(b+a \cos (c+d x)) \csc ^2(c+d x) \sec (c+d x) (e \sin (c+d x))^{5/2} \left (\frac{2 b \sin (c+d x)}{3 a^2}-\frac{\sin (2 (c+d x))}{5 a}\right )}{d (a+b \sec (c+d x))}-\frac{(b+a \cos (c+d x)) \sec (c+d x) (e \sin (c+d x))^{5/2} \left (\frac{\left (5 b^2-3 a^2\right ) \left (8 F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac{3}{2}}(c+d x) a^{5/2}+3 \sqrt{2} b \left (b^2-a^2\right )^{3/4} \left (2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{a} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )-\log \left (a \sin (c+d x)-\sqrt{2} \sqrt{a} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}\right )+\log \left (a \sin (c+d x)+\sqrt{2} \sqrt{a} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}\right )\right )\right ) \left (\sqrt{1-\sin ^2(c+d x)} a+b\right ) \cos ^2(c+d x)}{12 a^{3/2} \left (a^2-b^2\right ) (b+a \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac{4 a b \left (\frac{b F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac{3}{2}}(c+d x)}{3 \left (b^2-a^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{a} \sqrt{\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (i a \sin (c+d x)-(1+i) \sqrt{a} \sqrt [4]{a^2-b^2} \sqrt{\sin (c+d x)}+\sqrt{a^2-b^2}\right )+\log \left (i a \sin (c+d x)+(1+i) \sqrt{a} \sqrt [4]{a^2-b^2} \sqrt{\sin (c+d x)}+\sqrt{a^2-b^2}\right )\right )}{\sqrt{a} \sqrt [4]{a^2-b^2}}\right ) \left (\sqrt{1-\sin ^2(c+d x)} a+b\right ) \cos (c+d x)}{(b+a \cos (c+d x)) \sqrt{1-\sin ^2(c+d x)}}\right )}{5 a^2 d (a+b \sec (c+d x)) \sin ^{\frac{5}{2}}(c+d x)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((b + a*Cos[c + d*x])*Sec[c + d*x]*(e*Sin[c + d*x])^(5/2)*(((-3*a^2 + 5*b^2)*Cos[c + d*x]^2*(3*Sqrt[2]*b*(-a^
2 + b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*
Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sq
rt[Sin[c + d*x]] + a*Sin[c + d*x]] + Log[Sqrt[-a^2 + b^2] + Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*
x]] + a*Sin[c + d*x]]) + 8*a^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2
)]*Sin[c + d*x]^(3/2))*(b + a*Sqrt[1 - Sin[c + d*x]^2]))/(12*a^(3/2)*(a^2 - b^2)*(b + a*Cos[c + d*x])*(1 - Sin
[c + d*x]^2)) + (4*a*b*Cos[c + d*x]*(((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2 - b^
2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - (1 +
I)*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*a*Sin[c + d*x]] + Log[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[a]*(a
^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*a*Sin[c + d*x]]))/(Sqrt[a]*(a^2 - b^2)^(1/4)) + (b*AppellF1[3/4, 1/2, 1
, 7/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Sin[c + d*x]^(3/2))/(3*(-a^2 + b^2)))*(b + a*Sqrt[1 -
Sin[c + d*x]^2]))/((b + a*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2])))/(5*a^2*d*(a + b*Sec[c + d*x])*Sin[c + d*x
]^(5/2)) + ((b + a*Cos[c + d*x])*Csc[c + d*x]^2*Sec[c + d*x]*(e*Sin[c + d*x])^(5/2)*((2*b*Sin[c + d*x])/(3*a^2
) - Sin[2*(c + d*x)]/(5*a)))/(d*(a + b*Sec[c + d*x]))

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Maple [B]  time = 4.602, size = 1195, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d*b*e*(e*sin(d*x+c))^(3/2)/a^2+1/d*b*e^3/a^2/(e^2*(a^2-b^2)/a^2)^(1/4)*arctan((e*sin(d*x+c))^(1/2)/(e^2*(a
^2-b^2)/a^2)^(1/4))-1/d*b^3*e^3/a^4/(e^2*(a^2-b^2)/a^2)^(1/4)*arctan((e*sin(d*x+c))^(1/2)/(e^2*(a^2-b^2)/a^2)^
(1/4))-1/2/d*b*e^3/a^2/(e^2*(a^2-b^2)/a^2)^(1/4)*ln(((e*sin(d*x+c))^(1/2)+(e^2*(a^2-b^2)/a^2)^(1/4))/((e*sin(d
*x+c))^(1/2)-(e^2*(a^2-b^2)/a^2)^(1/4)))+1/2/d*b^3*e^3/a^4/(e^2*(a^2-b^2)/a^2)^(1/4)*ln(((e*sin(d*x+c))^(1/2)+
(e^2*(a^2-b^2)/a^2)^(1/4))/((e*sin(d*x+c))^(1/2)-(e^2*(a^2-b^2)/a^2)^(1/4)))+2/5/d/a*e^3*cos(d*x+c)^3/(e*sin(d
*x+c))^(1/2)-6/5/d/a*e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+
c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+3/5/d/a*e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(-sin(d*x+c)
+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-2/5/d/a*e^3*cos
(d*x+c)/(e*sin(d*x+c))^(1/2)+2/d/a^3*e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^2*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*
x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-1/d/a^3*e^3/cos(d*x+c)/(e*sin(d*x+c)
)^(1/2)*b^2*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*
2^(1/2))+1/2/d/a^3*e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^2*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*
x+c)^(1/2)/(1-(a^2-b^2)^(1/2)/a)*EllipticPi((-sin(d*x+c)+1)^(1/2),1/(1-(a^2-b^2)^(1/2)/a),1/2*2^(1/2))-1/2/d/a
^5*e^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^4*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1-(a
^2-b^2)^(1/2)/a)*EllipticPi((-sin(d*x+c)+1)^(1/2),1/(1-(a^2-b^2)^(1/2)/a),1/2*2^(1/2))+1/2/d/a^3*e^3/cos(d*x+c
)/(e*sin(d*x+c))^(1/2)*b^2*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1+(a^2-b^2)^(1/2)/a)
*EllipticPi((-sin(d*x+c)+1)^(1/2),1/(1+(a^2-b^2)^(1/2)/a),1/2*2^(1/2))-1/2/d/a^5*e^3/cos(d*x+c)/(e*sin(d*x+c))
^(1/2)*b^4*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1+(a^2-b^2)^(1/2)/a)*EllipticPi((-si
n(d*x+c)+1)^(1/2),1/(1+(a^2-b^2)^(1/2)/a),1/2*2^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*sin(d*x + c))^(5/2)/(b*sec(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*sin(d*x+c))^(5/2)/(a+b*sec(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*sin(d*x+c))**(5/2)/(a+b*sec(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 \sin \left (d x + c\right )\right )^{\frac{5}{2}}}{b \sec \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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