3.1430 \(\int \frac {(d \sin (e+f x))^{5/2}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=616 \[ -\frac {2 \sqrt {2} a^2 d^{5/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{b^2 f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} a^2 d^{5/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{b^2 f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {a d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}\right )}{\sqrt {2} b^2 f \sqrt {g}}-\frac {a d^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}+1\right )}{\sqrt {2} b^2 f \sqrt {g}}-\frac {a d^{5/2} \log \left (-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)+\sqrt {d}\right )}{2 \sqrt {2} b^2 f \sqrt {g}}+\frac {a d^{5/2} \log \left (\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)+\sqrt {d}\right )}{2 \sqrt {2} b^2 f \sqrt {g}}+\frac {d^3 \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{2 b f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {d^2 \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{b f g} \]

[Out]

1/2*a*d^(5/2)*arctan(1-2^(1/2)*g^(1/2)*(d*sin(f*x+e))^(1/2)/d^(1/2)/(g*cos(f*x+e))^(1/2))/b^2/f*2^(1/2)/g^(1/2
)-1/2*a*d^(5/2)*arctan(1+2^(1/2)*g^(1/2)*(d*sin(f*x+e))^(1/2)/d^(1/2)/(g*cos(f*x+e))^(1/2))/b^2/f*2^(1/2)/g^(1
/2)-1/4*a*d^(5/2)*ln(d^(1/2)-2^(1/2)*g^(1/2)*(d*sin(f*x+e))^(1/2)/(g*cos(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/b^2
/f*2^(1/2)/g^(1/2)+1/4*a*d^(5/2)*ln(d^(1/2)+2^(1/2)*g^(1/2)*(d*sin(f*x+e))^(1/2)/(g*cos(f*x+e))^(1/2)+d^(1/2)*
tan(f*x+e))/b^2/f*2^(1/2)/g^(1/2)-2*a^2*d^(5/2)*EllipticPi((d*sin(f*x+e))^(1/2)/d^(1/2)/(1+cos(f*x+e))^(1/2),-
a/(b-(-a^2+b^2)^(1/2)),I)*2^(1/2)*cos(f*x+e)^(1/2)/b^2/f/(-a^2+b^2)^(1/2)/(g*cos(f*x+e))^(1/2)+2*a^2*d^(5/2)*E
llipticPi((d*sin(f*x+e))^(1/2)/d^(1/2)/(1+cos(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)),I)*2^(1/2)*cos(f*x+e)^(1/2
)/b^2/f/(-a^2+b^2)^(1/2)/(g*cos(f*x+e))^(1/2)-d^2*(g*cos(f*x+e))^(1/2)*(d*sin(f*x+e))^(1/2)/b/f/g-1/2*d^3*(sin
(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*EllipticF(cos(e+1/4*Pi+f*x),2^(1/2))*sin(2*f*x+2*e)^(1/2)/b/f/(g*cos
(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.14, antiderivative size = 616, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 14, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.378, Rules used = {2909, 2568, 2573, 2641, 2574, 297, 1162, 617, 204, 1165, 628, 2908, 2907, 1218} \[ -\frac {2 \sqrt {2} a^2 d^{5/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{b^2 f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} a^2 d^{5/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{b^2 f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {a d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}\right )}{\sqrt {2} b^2 f \sqrt {g}}-\frac {a d^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}+1\right )}{\sqrt {2} b^2 f \sqrt {g}}-\frac {a d^{5/2} \log \left (-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)+\sqrt {d}\right )}{2 \sqrt {2} b^2 f \sqrt {g}}+\frac {a d^{5/2} \log \left (\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)+\sqrt {d}\right )}{2 \sqrt {2} b^2 f \sqrt {g}}-\frac {d^2 \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{b f g}+\frac {d^3 \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{2 b f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[g]*Sqrt[d*Sin[e + f*x]])/(Sqrt[d]*Sqrt[g*Cos[e + f*x]])])/(Sqrt[2]*b^2*f*S
qrt[g]) - (a*d^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[g]*Sqrt[d*Sin[e + f*x]])/(Sqrt[d]*Sqrt[g*Cos[e + f*x]])])/(Sqrt[
2]*b^2*f*Sqrt[g]) - (2*Sqrt[2]*a^2*d^(5/2)*Sqrt[Cos[e + f*x]]*EllipticPi[-(a/(b - Sqrt[-a^2 + b^2])), ArcSin[S
qrt[d*Sin[e + f*x]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x]])], -1])/(b^2*Sqrt[-a^2 + b^2]*f*Sqrt[g*Cos[e + f*x]]) + (2
*Sqrt[2]*a^2*d^(5/2)*Sqrt[Cos[e + f*x]]*EllipticPi[-(a/(b + Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(S
qrt[d]*Sqrt[1 + Cos[e + f*x]])], -1])/(b^2*Sqrt[-a^2 + b^2]*f*Sqrt[g*Cos[e + f*x]]) - (a*d^(5/2)*Log[Sqrt[d] -
(Sqrt[2]*Sqrt[g]*Sqrt[d*Sin[e + f*x]])/Sqrt[g*Cos[e + f*x]] + Sqrt[d]*Tan[e + f*x]])/(2*Sqrt[2]*b^2*f*Sqrt[g]
) + (a*d^(5/2)*Log[Sqrt[d] + (Sqrt[2]*Sqrt[g]*Sqrt[d*Sin[e + f*x]])/Sqrt[g*Cos[e + f*x]] + Sqrt[d]*Tan[e + f*x
]])/(2*Sqrt[2]*b^2*f*Sqrt[g]) - (d^2*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]])/(b*f*g) + (d^3*EllipticF[e - P
i/4 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]])/(2*b*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1218

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2574

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{k = Denomina
tor[m]}, Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Sin[e + f*x])^(1/k)/(b*Cos
[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

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 2907

Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]))
, x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Dist[(2*Sqrt[2]*d*(b + q))/(f*q), Subst[Int[1/((d*(b + q) + a*x^2
)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x] - Dist[(2*Sqrt[2]*d*(b - q))/(f*
q), Subst[Int[1/((d*(b - q) + a*x^2)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]],
x]] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2908

Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(
x_)])), x_Symbol] :> Dist[Sqrt[Cos[e + f*x]]/Sqrt[g*Cos[e + f*x]], Int[Sqrt[d*Sin[e + f*x]]/(Sqrt[Cos[e + f*x]
]*(a + b*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2909

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[d/b, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a*d)/b, Int[(
(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1))/(a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && N
eQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[-1, p, 1] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(d \sin (e+f x))^{5/2}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx &=\frac {d \int \frac {(d \sin (e+f x))^{3/2}}{\sqrt {g \cos (e+f x)}} \, dx}{b}-\frac {(a d) \int \frac {(d \sin (e+f x))^{3/2}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b}\\ &=-\frac {d^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{b f g}-\frac {\left (a d^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}} \, dx}{b^2}+\frac {\left (a^2 d^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b^2}+\frac {d^3 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \, dx}{2 b}\\ &=-\frac {d^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{b f g}-\frac {\left (2 a d^3 g\right ) \operatorname {Subst}\left (\int \frac {x^2}{d^2+g^2 x^4} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}\right )}{b^2 f}+\frac {\left (a^2 d^2 \sqrt {\cos (e+f x)}\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b^2 \sqrt {g \cos (e+f x)}}+\frac {\left (d^3 \sqrt {\sin (2 e+2 f x)}\right ) \int \frac {1}{\sqrt {\sin (2 e+2 f x)}} \, dx}{2 b \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ &=-\frac {d^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{b f g}+\frac {d^3 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{2 b f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}+\frac {\left (a d^3\right ) \operatorname {Subst}\left (\int \frac {d-g x^2}{d^2+g^2 x^4} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}\right )}{b^2 f}-\frac {\left (a d^3\right ) \operatorname {Subst}\left (\int \frac {d+g x^2}{d^2+g^2 x^4} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}\right )}{b^2 f}+\frac {\left (2 \sqrt {2} a^2 \left (1-\frac {b}{\sqrt {-a^2+b^2}}\right ) d^3 \sqrt {\cos (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (b-\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {1+\cos (e+f x)}}\right )}{b^2 f \sqrt {g \cos (e+f x)}}+\frac {\left (2 \sqrt {2} a^2 \left (1+\frac {b}{\sqrt {-a^2+b^2}}\right ) d^3 \sqrt {\cos (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (b+\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {1+\cos (e+f x)}}\right )}{b^2 f \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 \sqrt {2} a^2 d^{5/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{b^2 \sqrt {-a^2+b^2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} a^2 d^{5/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{b^2 \sqrt {-a^2+b^2} f \sqrt {g \cos (e+f x)}}-\frac {d^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{b f g}+\frac {d^3 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{2 b f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}-\frac {\left (a d^3\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {d}{g}-\frac {\sqrt {2} \sqrt {d} x}{\sqrt {g}}+x^2} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}\right )}{2 b^2 f g}-\frac {\left (a d^3\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {d}{g}+\frac {\sqrt {2} \sqrt {d} x}{\sqrt {g}}+x^2} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}\right )}{2 b^2 f g}-\frac {\left (a d^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {d}}{\sqrt {g}}+2 x}{-\frac {d}{g}-\frac {\sqrt {2} \sqrt {d} x}{\sqrt {g}}-x^2} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}\right )}{2 \sqrt {2} b^2 f \sqrt {g}}-\frac {\left (a d^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {d}}{\sqrt {g}}-2 x}{-\frac {d}{g}+\frac {\sqrt {2} \sqrt {d} x}{\sqrt {g}}-x^2} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}\right )}{2 \sqrt {2} b^2 f \sqrt {g}}\\ &=-\frac {2 \sqrt {2} a^2 d^{5/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{b^2 \sqrt {-a^2+b^2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} a^2 d^{5/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{b^2 \sqrt {-a^2+b^2} f \sqrt {g \cos (e+f x)}}-\frac {a d^{5/2} \log \left (\sqrt {d}-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)\right )}{2 \sqrt {2} b^2 f \sqrt {g}}+\frac {a d^{5/2} \log \left (\sqrt {d}+\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)\right )}{2 \sqrt {2} b^2 f \sqrt {g}}-\frac {d^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{b f g}+\frac {d^3 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{2 b f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}-\frac {\left (a d^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}\right )}{\sqrt {2} b^2 f \sqrt {g}}+\frac {\left (a d^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}\right )}{\sqrt {2} b^2 f \sqrt {g}}\\ &=\frac {a d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}\right )}{\sqrt {2} b^2 f \sqrt {g}}-\frac {a d^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}\right )}{\sqrt {2} b^2 f \sqrt {g}}-\frac {2 \sqrt {2} a^2 d^{5/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{b^2 \sqrt {-a^2+b^2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} a^2 d^{5/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{b^2 \sqrt {-a^2+b^2} f \sqrt {g \cos (e+f x)}}-\frac {a d^{5/2} \log \left (\sqrt {d}-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)\right )}{2 \sqrt {2} b^2 f \sqrt {g}}+\frac {a d^{5/2} \log \left (\sqrt {d}+\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)\right )}{2 \sqrt {2} b^2 f \sqrt {g}}-\frac {d^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{b f g}+\frac {d^3 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{2 b f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 27.08, size = 1318, normalized size = 2.14 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[Cos[e + f*x]]*(d*Sin[e + f*x])^(5/2)*((2*Sqrt[Sin[e + f*x]]*((Sqrt[a]*(-2*ArcTan[1 - (Sqrt[2]*(a^2 - b^2
)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]] + 2*ArcTan[1 + (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]] +
Log[-a + Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]] - Sqrt[a^2 - b^2]*Tan[e + f*x]] - Log[a + Sqrt[2
]*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]] + Sqrt[a^2 - b^2]*Tan[e + f*x]]))/(4*Sqrt[2]*(a^2 - b^2)^(3/4))
- (b*AppellF1[5/4, 1/2, 1, 9/4, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Tan[e + f*x]^(5/2))/(5*a^
2))*(b*Tan[e + f*x] + a*Sqrt[1 + Tan[e + f*x]^2]))/(Cos[e + f*x]^(5/2)*(a + b*Sin[e + f*x])*Sqrt[Tan[e + f*x]]
*(1 + Tan[e + f*x]^2)^(3/2)) + (Cos[2*(e + f*x)]*Sqrt[Sin[e + f*x]]*(b*Tan[e + f*x] + a*Sqrt[1 + Tan[e + f*x]^
2])*(-20*Sqrt[2]*a*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] + 20*Sqrt[2]*a*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]
] + (10*Sqrt[2]*Sqrt[a]*(2*a^2 - b^2)*ArcTan[1 - (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]])/(a^2
- b^2)^(3/4) - (10*Sqrt[2]*Sqrt[a]*(2*a^2 - b^2)*ArcTan[1 + (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sq
rt[a]])/(a^2 - b^2)^(3/4) + 10*Sqrt[2]*a*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - 10*Sqrt[2]*a*Log
[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - (5*Sqrt[2]*Sqrt[a]*(2*a^2 - b^2)*Log[-a + Sqrt[2]*Sqrt[a]*(a
^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]] - Sqrt[a^2 - b^2]*Tan[e + f*x]])/(a^2 - b^2)^(3/4) + (5*Sqrt[2]*Sqrt[a]*(2*
a^2 - b^2)*Log[a + Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]] + Sqrt[a^2 - b^2]*Tan[e + f*x]])/(a^2
- b^2)^(3/4) + 8*b*AppellF1[5/4, 1/2, 1, 9/4, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x]^(5/
2) + (40*b*Sqrt[Tan[e + f*x]])/Sqrt[1 + Tan[e + f*x]^2] + (200*a^4*b*AppellF1[1/4, 1/2, 1, 5/4, -Tan[e + f*x]^
2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sqrt[Tan[e + f*x]])/(Sqrt[1 + Tan[e + f*x]^2]*(-5*a^2*AppellF1[1/4, 1/2, 1,
5/4, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] + 2*(2*(a^2 - b^2)*AppellF1[5/4, 1/2, 2, 9/4, -Tan[e + f*
x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] + a^2*AppellF1[5/4, 3/2, 1, 9/4, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e +
f*x]^2])*Tan[e + f*x]^2)*(-(b^2*Tan[e + f*x]^2) + a^2*(1 + Tan[e + f*x]^2)))))/(20*b^2*Cos[e + f*x]^(5/2)*(a +
b*Sin[e + f*x])*Sqrt[Tan[e + f*x]]*(-1 + Tan[e + f*x]^2)*Sqrt[1 + Tan[e + f*x]^2])))/(2*f*Sqrt[g*Cos[e + f*x]
]*Sin[e + f*x]^(5/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sin \left (f x + e\right )\right )^{\frac {5}{2}}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.89, size = 2344, normalized size = 3.81 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x)

[Out]

-1/f*(I*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2
)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2
^(1/2))*(-a^2+b^2)^(1/2)*a^2-I*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f
*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e)
)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a*b+I*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2
)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)
-sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a*b-I*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f
*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*Elli
pticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a^2+sin(f*x+e)*E
llipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-(-1+cos(f*x+
e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1
/2)*(-a^2+b^2)^(1/2)*a^2-sin(f*x+e)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)
^(1/2)-a),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^
(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*a^3+sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+co
s(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+
e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*a^2*b-sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin
(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(
-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a^2+sin(f*x+e)*(-(-1+cos(f
*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))
^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a*b-s
in(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+co
s(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(
-a^2+b^2)^(1/2)*a^2+sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(
f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2
+1/2*I,1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a*b-(-a^2+b^2)^(1/2)*sin(f*x+e)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/si
n(f*x+e))^(1/2),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*
x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*a*b+(-a^2+b^2)^(1/2)*sin(f*x+e)*EllipticF((-(-1+cos(f*x+e)-sin(
f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+
e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*b^2+sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e
))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos
(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a^2+sin(f*x+e)*(-
(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/si
n(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2)
)*a^3-sin(f*x+e)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2
))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e
))/sin(f*x+e))^(1/2)*a^2*b-cos(f*x+e)^2*(-a^2+b^2)^(1/2)*2^(1/2)*a*b+cos(f*x+e)^2*2^(1/2)*(-a^2+b^2)^(1/2)*b^2
+cos(f*x+e)*(-a^2+b^2)^(1/2)*2^(1/2)*a*b-cos(f*x+e)*2^(1/2)*(-a^2+b^2)^(1/2)*b^2)*(d*sin(f*x+e))^(5/2)/(-1+cos
(f*x+e))/sin(f*x+e)^2/(g*cos(f*x+e))^(1/2)*2^(1/2)*a/b^2/(-a^2+b^2)^(1/2)/(a-b+(-a^2+b^2)^(1/2))/(b+(-a^2+b^2)
^(1/2)-a)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sin \left (f x + e\right )\right )^{\frac {5}{2}}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^{5/2}}{\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sin(e + f*x))^(5/2)/((g*cos(e + f*x))^(1/2)*(a + b*sin(e + f*x))),x)

[Out]

int((d*sin(e + f*x))^(5/2)/((g*cos(e + f*x))^(1/2)*(a + b*sin(e + f*x))), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))**(5/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))**(1/2),x)

[Out]

Timed out

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