3.3.34 \(\int \frac {(e \sin (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx\) [234]

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

[Out]

b*(a^2-b^2)^(3/4)*e^(5/2)*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(7/2)/d-b*(a^2-b^2)^(
3/4)*e^(5/2)*arctanh(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(7/2)/d+2/15*e*(5*b-3*a*cos(d*x+c
))*(e*sin(d*x+c))^(3/2)/a^2/d+b^2*(a^2-b^2)*e^3*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*
EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*a/(a-(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/a^4/d/(a-(a^2-b^2)^(1/2
))/(e*sin(d*x+c))^(1/2)+b^2*(a^2-b^2)*e^3*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*Ellipt
icPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*a/(a+(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/a^4/d/(a+(a^2-b^2)^(1/2))/(e*
sin(d*x+c))^(1/2)-2/5*(3*a^2-5*b^2)*e^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*Elliptic
E(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/a^3/d/sin(d*x+c)^(1/2)

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Rubi [A]
time = 0.79, antiderivative size = 430, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3957, 2944, 2946, 2721, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \begin {gather*} \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}+\frac {b e^{5/2} \left (a^2-b^2\right )^{3/4} \text {ArcTan}\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 {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^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)}} \end {gather*}

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 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 304

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 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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 2780

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 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 2944

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 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]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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]

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 33.87, size = 853, normalized size = 1.98 \begin {gather*} -\frac {(b+a \cos (c+d x)) \sec (c+d x) (e \sin (c+d x))^{5/2} \left (\frac {\left (-3 a^2+5 b^2\right ) \cos ^2(c+d x) \left (3 \sqrt {2} b \left (-a^2+b^2\right )^{3/4} \left (2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )\right )+8 a^{5/2} 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)\right ) \left (b+a \sqrt {1-\sin ^2(c+d x)}\right )}{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 \cos (c+d x) \left (\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \text {ArcTan}\left (1-\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \text {ArcTan}\left (1+\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )\right )}{\sqrt {a} \sqrt [4]{a^2-b^2}}+\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 (-a^2+b^2\right )}\right ) \left (b+a \sqrt {1-\sin ^2(c+d x)}\right )}{(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)}+\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))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/5*((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
)*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]]) + 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])))/(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 [A]
time = 0.38, size = 852, normalized size = 1.98

method	result	size
default	\(\frac {\frac {2 e b \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3 a^{2}}-\frac {e^{3} b \ln \left (\frac {\sqrt {e \sin \left (d x +c \right )}+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}{\sqrt {e \sin \left (d x +c \right )}-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{2 a^{2} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}+\frac {e^{3} b^{3} \ln \left (\frac {\sqrt {e \sin \left (d x +c \right )}+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}{\sqrt {e \sin \left (d x +c \right )}-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{2 a^{4} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}+\frac {e^{3} b \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{a^{2} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}-\frac {e^{3} b^{3} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{a^{4} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}+\frac {\sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) e \sin \left (d x +c \right )}\, a \,e^{3} \left (-\frac {6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\cos ^{4}\left (d x +c \right )\right )+2 \left (\cos ^{2}\left (d x +c \right )\right )}{5 a^{2} \sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) e \sin \left (d x +c \right )}}+\frac {b^{2} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \left (2 \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{a^{4} \sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) e \sin \left (d x +c \right )}}-\frac {b^{2} \left (a^{2}-b^{2}\right ) \left (-\frac {\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {1}{1-\frac {\sqrt {a^{2}-b^{2}}}{a}}, \frac {\sqrt {2}}{2}\right )}{2 a^{2} \sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) e \sin \left (d x +c \right )}\, \left (1-\frac {\sqrt {a^{2}-b^{2}}}{a}\right )}-\frac {\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {1}{1+\frac {\sqrt {a^{2}-b^{2}}}{a}}, \frac {\sqrt {2}}{2}\right )}{2 a^{2} \sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) e \sin \left (d x +c \right )}\, \left (1+\frac {\sqrt {a^{2}-b^{2}}}{a}\right )}\right )}{a^{4}}\right )}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(852\)

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,method=_RETURNVERBOSE)

[Out]

(2/3*e*b/a^2*(e*sin(d*x+c))^(3/2)-1/2*e^3*b/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*e^3*b^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)))+e^3*b/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))-e^3*b^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))+(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*a*e^3*(-1/5/
a^2/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*(6*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellipti
cE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-3*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Elliptic
F((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-2*cos(d*x+c)^4+2*cos(d*x+c)^2)+b^2/a^4*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c
)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*(2*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-
EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2)))-b^2*(a^2-b^2)/a^4*(-1/2/a^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+
2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*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/a^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2
)/(cos(d*x+c)^2*e*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))))/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}}{b+a\,\cos \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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