3.1424 \(\int \frac {(g \cos (e+f x))^{5/2}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=572 \[ \frac {a g^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {d \sin (e+f x)}}\right )}{\sqrt {2} b^2 \sqrt {d} f}-\frac {a g^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {d \sin (e+f x)}}+1\right )}{\sqrt {2} b^2 \sqrt {d} f}+\frac {2 \sqrt {2} g^{5/2} \sqrt {b-a} \sqrt {a+b} \sqrt {\sin (e+f x)} \Pi \left (-\frac {\sqrt {b-a}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{b^2 f \sqrt {d \sin (e+f x)}}-\frac {2 \sqrt {2} g^{5/2} \sqrt {b-a} \sqrt {a+b} \sqrt {\sin (e+f x)} \Pi \left (\frac {\sqrt {b-a}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{b^2 f \sqrt {d \sin (e+f x)}}-\frac {a g^{5/2} \log \left (-\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)}}+\sqrt {g} \cot (e+f x)+\sqrt {g}\right )}{2 \sqrt {2} b^2 \sqrt {d} f}+\frac {a g^{5/2} \log \left (\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)}}+\sqrt {g} \cot (e+f x)+\sqrt {g}\right )}{2 \sqrt {2} b^2 \sqrt {d} f}-\frac {g^2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{b d f \sqrt {\sin (2 e+2 f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.02, antiderivative size = 572, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 15, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {2901, 2838, 2575, 297, 1162, 617, 204, 1165, 628, 2572, 2639, 2906, 2905, 490, 1218} \[ \frac {a g^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {d \sin (e+f x)}}\right )}{\sqrt {2} b^2 \sqrt {d} f}-\frac {a g^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {d \sin (e+f x)}}+1\right )}{\sqrt {2} b^2 \sqrt {d} f}+\frac {2 \sqrt {2} g^{5/2} \sqrt {b-a} \sqrt {a+b} \sqrt {\sin (e+f x)} \Pi \left (-\frac {\sqrt {b-a}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{b^2 f \sqrt {d \sin (e+f x)}}-\frac {2 \sqrt {2} g^{5/2} \sqrt {b-a} \sqrt {a+b} \sqrt {\sin (e+f x)} \Pi \left (\frac {\sqrt {b-a}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{b^2 f \sqrt {d \sin (e+f x)}}-\frac {a g^{5/2} \log \left (-\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)}}+\sqrt {g} \cot (e+f x)+\sqrt {g}\right )}{2 \sqrt {2} b^2 \sqrt {d} f}+\frac {a g^{5/2} \log \left (\frac {\sqrt {2} \sqrt {d} \sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)}}+\sqrt {g} \cot (e+f x)+\sqrt {g}\right )}{2 \sqrt {2} b^2 \sqrt {d} f}-\frac {g^2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{b d f \sqrt {\sin (2 e+2 f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*g^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d]*Sqrt[g*Cos[e + f*x]])/(Sqrt[g]*Sqrt[d*Sin[e + f*x]])])/(Sqrt[2]*b^2*Sqr
t[d]*f) - (a*g^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d]*Sqrt[g*Cos[e + f*x]])/(Sqrt[g]*Sqrt[d*Sin[e + f*x]])])/(Sqrt[
2]*b^2*Sqrt[d]*f) - (a*g^(5/2)*Log[Sqrt[g] + Sqrt[g]*Cot[e + f*x] - (Sqrt[2]*Sqrt[d]*Sqrt[g*Cos[e + f*x]])/Sqr
t[d*Sin[e + f*x]]])/(2*Sqrt[2]*b^2*Sqrt[d]*f) + (a*g^(5/2)*Log[Sqrt[g] + Sqrt[g]*Cot[e + f*x] + (Sqrt[2]*Sqrt[
d]*Sqrt[g*Cos[e + f*x]])/Sqrt[d*Sin[e + f*x]]])/(2*Sqrt[2]*b^2*Sqrt[d]*f) + (2*Sqrt[2]*Sqrt[-a + b]*Sqrt[a + b
]*g^(5/2)*EllipticPi[-(Sqrt[-a + b]/Sqrt[a + b]), ArcSin[Sqrt[g*Cos[e + f*x]]/(Sqrt[g]*Sqrt[1 + Sin[e + f*x]])
], -1]*Sqrt[Sin[e + f*x]])/(b^2*f*Sqrt[d*Sin[e + f*x]]) - (2*Sqrt[2]*Sqrt[-a + b]*Sqrt[a + b]*g^(5/2)*Elliptic
Pi[Sqrt[-a + b]/Sqrt[a + b], ArcSin[Sqrt[g*Cos[e + f*x]]/(Sqrt[g]*Sqrt[1 + Sin[e + f*x]])], -1]*Sqrt[Sin[e + f
*x]])/(b^2*f*Sqrt[d*Sin[e + f*x]]) - (g^2*Sqrt[g*Cos[e + f*x]]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Sin[e + f*x
]])/(b*d*f*Sqrt[Sin[2*e + 2*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 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(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)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 2572

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

Rule 2575

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 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*Cos[e + f*x])^(1/k)/(b*Si
n[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

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 2838

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

Rule 2901

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

Rule 2905

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

Rule 2906

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[(d_)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x
_)])), x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]], Int[Sqrt[g*Cos[e + f*x]]/(Sqrt[Sin[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]

Rubi steps

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

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Mathematica [C]  time = 25.81, size = 1399, normalized size = 2.45 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((g*Cos[e + f*x])^(5/2)*Sqrt[Sin[e + f*x]]*((Sqrt[Tan[e + f*x]]*((3*Sqrt[2]*a^(3/2)*(-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]])/Sqr
t[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]]))/(a^2 - b^2)^(1/4) - 8
*b*AppellF1[3/4, 1/2, 1, 7/4, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Tan[e + f*x]^(3/2))*(b*Tan[e
+ f*x] + a*Sqrt[1 + Tan[e + f*x]^2]))/(12*a^2*Cos[e + f*x]^(3/2)*Sqrt[Sin[e + f*x]]*(a + b*Sin[e + f*x])*(1 +
Tan[e + f*x]^2)^(3/2)) + (Cos[2*(e + f*x)]*Sqrt[Tan[e + f*x]]*(b*Tan[e + f*x] + a*Sqrt[1 + Tan[e + f*x]^2])*(
56*b*(-3*a^2 + b^2)*AppellF1[3/4, 1/2, 1, 7/4, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x]^(3
/2) + 24*b*(-a^2 + b^2)*AppellF1[7/4, 1/2, 1, 11/4, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*
x]^(7/2) + 21*a^(3/2)*(4*Sqrt[2]*a^(3/2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] - 4*Sqrt[2]*a^(3/2)*ArcTan[1 +
Sqrt[2]*Sqrt[Tan[e + f*x]]] - (4*Sqrt[2]*a^2*ArcTan[1 - (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a
]])/(a^2 - b^2)^(1/4) + (2*Sqrt[2]*b^2*ArcTan[1 - (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]])/(a^
2 - b^2)^(1/4) + (4*Sqrt[2]*a^2*ArcTan[1 + (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]])/(a^2 - b^2
)^(1/4) - (2*Sqrt[2]*b^2*ArcTan[1 + (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]])/(a^2 - b^2)^(1/4)
+ 2*Sqrt[2]*a^(3/2)*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - 2*Sqrt[2]*a^(3/2)*Log[1 + Sqrt[2]*Sq
rt[Tan[e + f*x]] + Tan[e + f*x]] - (2*Sqrt[2]*a^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)^(1/4) + (Sqrt[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)^(1/4) + (2*Sqrt[2]*a^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)^(1/4) - (Sqrt[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)^(1/4) + (
8*Sqrt[a]*b*Tan[e + f*x]^(3/2))/Sqrt[1 + Tan[e + f*x]^2])))/(84*a^2*b^2*Cos[e + f*x]^(3/2)*Sqrt[Sin[e + f*x]]*
(a + b*Sin[e + f*x])*(-1 + Tan[e + f*x]^2)*Sqrt[1 + Tan[e + f*x]^2])))/(2*f*Cos[e + f*x]^(5/2)*Sqrt[d*Sin[e +
f*x]])

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.69, size = 5224, normalized size = 9.13 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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