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

Optimal. Leaf size=444 \[ \frac{2 e^2 \left (a^2-3 b^2\right ) \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{3 a^3 d \sqrt{e \sin (c+d x)}}-\frac{b e^{3/2} \sqrt [4]{a^2-b^2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{5/2} d}-\frac{b e^{3/2} \sqrt [4]{a^2-b^2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{5/2} d}+\frac{b^2 e^2 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a-\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^3 d \left (-a \sqrt{a^2-b^2}+a^2-b^2\right ) \sqrt{e \sin (c+d x)}}+\frac{b^2 e^2 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a+\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^3 d \left (a \sqrt{a^2-b^2}+a^2-b^2\right ) \sqrt{e \sin (c+d x)}}+\frac{2 e \sqrt{e \sin (c+d x)} (3 b-a \cos (c+d x))}{3 a^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.04423, antiderivative size = 444, 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, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ -\frac{b e^{3/2} \sqrt [4]{a^2-b^2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{5/2} d}-\frac{b e^{3/2} \sqrt [4]{a^2-b^2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{5/2} d}+\frac{2 e^2 \left (a^2-3 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 a^3 d \sqrt{e \sin (c+d x)}}+\frac{b^2 e^2 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a-\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^3 d \left (-a \sqrt{a^2-b^2}+a^2-b^2\right ) \sqrt{e \sin (c+d x)}}+\frac{b^2 e^2 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a+\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^3 d \left (a \sqrt{a^2-b^2}+a^2-b^2\right ) \sqrt{e \sin (c+d x)}}+\frac{2 e \sqrt{e \sin (c+d x)} (3 b-a \cos (c+d x))}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 16.4583, size = 1959, normalized size = 4.41 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*(b + a*Cos[c + d*x])*Csc[c + d*x]*(e*Sin[c + d*x])^(3/2))/(3*a*d*(a + b*Sec[c + d*x])) + ((b + a*Cos[c + d
*x])*Sec[c + d*x]*(e*Sin[c + d*x])^(3/2)*((4*a*Cos[c + d*x]^2*(b + a*Sqrt[1 - Sin[c + d*x]^2])*((b*(-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)*Sqrt[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]]))/(
4*Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(3/4)) - (5*a*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (a^2*Sin[
c + d*x]^2)/(a^2 - b^2)]*Sqrt[Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]^2])/((5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5
/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)] + 2*(2*a^2*AppellF1[5/4, -1/2, 2, 9/4, Sin[c + d*x]^2, (
a^2*Sin[c + d*x]^2)/(a^2 - b^2)] + (-a^2 + b^2)*AppellF1[5/4, 1/2, 1, 9/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2
)/(a^2 - b^2)])*Sin[c + d*x]^2)*(b^2 + a^2*(-1 + Sin[c + d*x]^2)))))/((b + a*Cos[c + d*x])*(1 - Sin[c + d*x]^2
)) - (2*b*Cos[c + d*x]*(b + a*Sqrt[1 - Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[a]*(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]]))/(a^2 - b^2)^(3/4) + (5*b
*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Sqrt[Sin[c + d*x]])/
(Sqrt[1 - Sin[c + d*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2
- b^2)] + 2*(2*a^2*AppellF1[5/4, 1/2, 2, 9/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)] + (a^2 - b^2)*
AppellF1[5/4, 3/2, 1, 9/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)])*Sin[c + d*x]^2)*(b^2 + a^2*(-1 +
Sin[c + d*x]^2)))))/((b + a*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2]) + (3*b*Cos[c + d*x]*Cos[2*(c + d*x)]*(b +
a*Sqrt[1 - Sin[c + d*x]^2])*(((1/2 - I/2)*(a^2 - 2*b^2)*ArcTan[1 - ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2
- b^2)^(1/4)])/(a^(3/2)*(a^2 - b^2)^(3/4)) - ((1/2 - I/2)*(a^2 - 2*b^2)*ArcTan[1 + ((1 + I)*Sqrt[a]*Sqrt[Sin[c
+ d*x]])/(a^2 - b^2)^(1/4)])/(a^(3/2)*(a^2 - b^2)^(3/4)) + ((1/4 - I/4)*(a^2 - 2*b^2)*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]])/(a^(3/2)*(a^2 - b^2)^(3/4)) - ((1/4 -
I/4)*(a^2 - 2*b^2)*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]])/(a^(3/2)*(a^2 - b^2)^(3/4)) + (4*Sqrt[Sin[c + d*x]])/a + (4*b*AppellF1[5/4, 1/2, 1, 9/4, Sin[c + d*x]^2,
(a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Sin[c + d*x]^(5/2))/(5*(a^2 - b^2)) + (10*b*(a^2 - b^2)*AppellF1[1/4, 1/2,
1, 5/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Sqrt[Sin[c + d*x]])/(Sqrt[1 - Sin[c + d*x]^2]*(5*(a^
2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)] + 2*(2*a^2*AppellF1[5/4,
1/2, 2, 9/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)] + (a^2 - b^2)*AppellF1[5/4, 3/2, 1, 9/4, Sin[c
+ d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)])*Sin[c + d*x]^2)*(b^2 + a^2*(-1 + Sin[c + d*x]^2)))))/((b + a*Cos
[c + d*x])*(1 - 2*Sin[c + d*x]^2)*Sqrt[1 - Sin[c + d*x]^2])))/(6*a*d*(a + b*Sec[c + d*x])*Sin[c + d*x]^(3/2))

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Maple [B]  time = 4.273, size = 1120, normalized size = 2.5 \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))^(3/2)/(a+b*sec(d*x+c)),x)

[Out]

2/d*b*e/a^2*(e*sin(d*x+c))^(1/2)+1/d*b*e^3*(e^2*(a^2-b^2)/a^2)^(1/4)/(-a^2*e^2+b^2*e^2)*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^2*(e^2*(a^2-b^2)/a^2)^(1/4)/(-a^2*e^2+b^2*e^2)*arctan((e*sin(d*
x+c))^(1/2)/(e^2*(a^2-b^2)/a^2)^(1/4))+1/2/d*b*e^3*(e^2*(a^2-b^2)/a^2)^(1/4)/(-a^2*e^2+b^2*e^2)*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^2*(e^2
*(a^2-b^2)/a^2)^(1/4)/(-a^2*e^2+b^2*e^2)*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)))-1/3/d/a*e^2/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))-2/3/d/a*e^2*cos(d*x+c)/(e*sin(d*x+c)
)^(1/2)*sin(d*x+c)+1/d/a^3*e^2/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^2*e^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*
b^2/(a^2-b^2)^(1/2)*(-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)*Ellipt
icPi((-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^4*e^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*
b^4/(a^2-b^2)^(1/2)*(-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)*Ellipt
icPi((-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^2*e^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*
b^2/(a^2-b^2)^(1/2)*(-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)*Ellipt
icPi((-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^4*e^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*
b^4/(a^2-b^2)^(1/2)*(-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)*Ellipt
icPi((-sin(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{3}{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))^(3/2)/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

[Out]

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