3.1414 \(\int \frac {\sqrt {g \cos (e+f x)}}{(d \sin (e+f x))^{7/2} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=513 \[ -\frac {2 \sqrt {2} b^3 \sqrt {g} \sqrt {\sin (e+f x)} \Pi \left (-\frac {\sqrt {b-a}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{a^3 d^3 f \sqrt {b-a} \sqrt {a+b} \sqrt {d \sin (e+f x)}}+\frac {2 \sqrt {2} b^3 \sqrt {g} \sqrt {\sin (e+f x)} \Pi \left (\frac {\sqrt {b-a}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{a^3 d^3 f \sqrt {b-a} \sqrt {a+b} \sqrt {d \sin (e+f x)}}-\frac {2 b^2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{a^3 d^4 f \sqrt {\sin (2 e+2 f x)}}-\frac {2 b^2 (g \cos (e+f x))^{3/2}}{a^3 d^3 f g \sqrt {d \sin (e+f x)}}+\frac {2 b (g \cos (e+f x))^{3/2}}{3 a^2 d^2 f g (d \sin (e+f x))^{3/2}}-\frac {4 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{5 a d^4 f \sqrt {\sin (2 e+2 f x)}}-\frac {4 (g \cos (e+f x))^{3/2}}{5 a d^3 f g \sqrt {d \sin (e+f x)}}-\frac {2 (g \cos (e+f x))^{3/2}}{5 a d f g (d \sin (e+f x))^{5/2}} \]

[Out]

-2/5*(g*cos(f*x+e))^(3/2)/a/d/f/g/(d*sin(f*x+e))^(5/2)+2/3*b*(g*cos(f*x+e))^(3/2)/a^2/d^2/f/g/(d*sin(f*x+e))^(
3/2)-4/5*(g*cos(f*x+e))^(3/2)/a/d^3/f/g/(d*sin(f*x+e))^(1/2)-2*b^2*(g*cos(f*x+e))^(3/2)/a^3/d^3/f/g/(d*sin(f*x
+e))^(1/2)-2*b^3*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)*g^(1/2)*sin(f*x+e)^(1/2)/a^3/d^3/f/(-a+b)^(1/2)/(a+b)^(1/2)/(d*sin(f*x+e))^(1/2)+2*b^3*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)*g^(1/2)*sin(f*x+e)^(1/2)/a^3/d^
3/f/(-a+b)^(1/2)/(a+b)^(1/2)/(d*sin(f*x+e))^(1/2)+4/5*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*EllipticE(
cos(e+1/4*Pi+f*x),2^(1/2))*(g*cos(f*x+e))^(1/2)*(d*sin(f*x+e))^(1/2)/a/d^4/f/sin(2*f*x+2*e)^(1/2)+2*b^2*(sin(e
+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*EllipticE(cos(e+1/4*Pi+f*x),2^(1/2))*(g*cos(f*x+e))^(1/2)*(d*sin(f*x+e
))^(1/2)/a^3/d^4/f/sin(2*f*x+2*e)^(1/2)

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Rubi [A]  time = 1.45, antiderivative size = 513, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {2910, 2570, 2572, 2639, 2563, 2906, 2905, 490, 1218} \[ -\frac {2 b^2 (g \cos (e+f x))^{3/2}}{a^3 d^3 f g \sqrt {d \sin (e+f x)}}-\frac {2 b^2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{a^3 d^4 f \sqrt {\sin (2 e+2 f x)}}-\frac {2 \sqrt {2} b^3 \sqrt {g} \sqrt {\sin (e+f x)} \Pi \left (-\frac {\sqrt {b-a}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{a^3 d^3 f \sqrt {b-a} \sqrt {a+b} \sqrt {d \sin (e+f x)}}+\frac {2 \sqrt {2} b^3 \sqrt {g} \sqrt {\sin (e+f x)} \Pi \left (\frac {\sqrt {b-a}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right )\right |-1\right )}{a^3 d^3 f \sqrt {b-a} \sqrt {a+b} \sqrt {d \sin (e+f x)}}+\frac {2 b (g \cos (e+f x))^{3/2}}{3 a^2 d^2 f g (d \sin (e+f x))^{3/2}}-\frac {4 (g \cos (e+f x))^{3/2}}{5 a d^3 f g \sqrt {d \sin (e+f x)}}-\frac {4 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{5 a d^4 f \sqrt {\sin (2 e+2 f x)}}-\frac {2 (g \cos (e+f x))^{3/2}}{5 a d f g (d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[g*Cos[e + f*x]]/((d*Sin[e + f*x])^(7/2)*(a + b*Sin[e + f*x])),x]

[Out]

(-2*(g*Cos[e + f*x])^(3/2))/(5*a*d*f*g*(d*Sin[e + f*x])^(5/2)) + (2*b*(g*Cos[e + f*x])^(3/2))/(3*a^2*d^2*f*g*(
d*Sin[e + f*x])^(3/2)) - (4*(g*Cos[e + f*x])^(3/2))/(5*a*d^3*f*g*Sqrt[d*Sin[e + f*x]]) - (2*b^2*(g*Cos[e + f*x
])^(3/2))/(a^3*d^3*f*g*Sqrt[d*Sin[e + f*x]]) - (2*Sqrt[2]*b^3*Sqrt[g]*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]])/(a^3*Sqrt[-a + b]*Sqrt[
a + b]*d^3*f*Sqrt[d*Sin[e + f*x]]) + (2*Sqrt[2]*b^3*Sqrt[g]*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]])/(a^3*Sqrt[-a + b]*Sqrt[a + b]*d^3*f*
Sqrt[d*Sin[e + f*x]]) - (4*Sqrt[g*Cos[e + f*x]]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Sin[e + f*x]])/(5*a*d^4*f*
Sqrt[Sin[2*e + 2*f*x]]) - (2*b^2*Sqrt[g*Cos[e + f*x]]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Sin[e + f*x]])/(a^3*
d^4*f*Sqrt[Sin[2*e + 2*f*x]])

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 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 2563

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

Rule 2570

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

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 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 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]

Rule 2910

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

Rubi steps

\begin {align*} \int \frac {\sqrt {g \cos (e+f x)}}{(d \sin (e+f x))^{7/2} (a+b \sin (e+f x))} \, dx &=\frac {\int \frac {\sqrt {g \cos (e+f x)}}{(d \sin (e+f x))^{7/2}} \, dx}{a}-\frac {b \int \frac {\sqrt {g \cos (e+f x)}}{(d \sin (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx}{a d}\\ &=-\frac {2 (g \cos (e+f x))^{3/2}}{5 a d f g (d \sin (e+f x))^{5/2}}+\frac {2 \int \frac {\sqrt {g \cos (e+f x)}}{(d \sin (e+f x))^{3/2}} \, dx}{5 a d^2}+\frac {b^2 \int \frac {\sqrt {g \cos (e+f x)}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx}{a^2 d^2}-\frac {b \int \frac {\sqrt {g \cos (e+f x)}}{(d \sin (e+f x))^{5/2}} \, dx}{a^2 d}\\ &=-\frac {2 (g \cos (e+f x))^{3/2}}{5 a d f g (d \sin (e+f x))^{5/2}}+\frac {2 b (g \cos (e+f x))^{3/2}}{3 a^2 d^2 f g (d \sin (e+f x))^{3/2}}-\frac {4 (g \cos (e+f x))^{3/2}}{5 a d^3 f g \sqrt {d \sin (e+f x)}}-\frac {4 \int \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)} \, dx}{5 a d^4}-\frac {b^3 \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx}{a^3 d^3}+\frac {b^2 \int \frac {\sqrt {g \cos (e+f x)}}{(d \sin (e+f x))^{3/2}} \, dx}{a^3 d^2}\\ &=-\frac {2 (g \cos (e+f x))^{3/2}}{5 a d f g (d \sin (e+f x))^{5/2}}+\frac {2 b (g \cos (e+f x))^{3/2}}{3 a^2 d^2 f g (d \sin (e+f x))^{3/2}}-\frac {4 (g \cos (e+f x))^{3/2}}{5 a d^3 f g \sqrt {d \sin (e+f x)}}-\frac {2 b^2 (g \cos (e+f x))^{3/2}}{a^3 d^3 f g \sqrt {d \sin (e+f x)}}-\frac {\left (2 b^2\right ) \int \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)} \, dx}{a^3 d^4}-\frac {\left (b^3 \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}{a^3 d^3 \sqrt {d \sin (e+f x)}}-\frac {\left (4 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}\right ) \int \sqrt {\sin (2 e+2 f x)} \, dx}{5 a d^4 \sqrt {\sin (2 e+2 f x)}}\\ &=-\frac {2 (g \cos (e+f x))^{3/2}}{5 a d f g (d \sin (e+f x))^{5/2}}+\frac {2 b (g \cos (e+f x))^{3/2}}{3 a^2 d^2 f g (d \sin (e+f x))^{3/2}}-\frac {4 (g \cos (e+f x))^{3/2}}{5 a d^3 f g \sqrt {d \sin (e+f x)}}-\frac {2 b^2 (g \cos (e+f x))^{3/2}}{a^3 d^3 f g \sqrt {d \sin (e+f x)}}-\frac {4 \sqrt {g \cos (e+f x)} E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {d \sin (e+f x)}}{5 a d^4 f \sqrt {\sin (2 e+2 f x)}}+\frac {\left (4 \sqrt {2} b^3 g \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 )}{a^3 d^3 f \sqrt {d \sin (e+f x)}}-\frac {\left (2 b^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}\right ) \int \sqrt {\sin (2 e+2 f x)} \, dx}{a^3 d^4 \sqrt {\sin (2 e+2 f x)}}\\ &=-\frac {2 (g \cos (e+f x))^{3/2}}{5 a d f g (d \sin (e+f x))^{5/2}}+\frac {2 b (g \cos (e+f x))^{3/2}}{3 a^2 d^2 f g (d \sin (e+f x))^{3/2}}-\frac {4 (g \cos (e+f x))^{3/2}}{5 a d^3 f g \sqrt {d \sin (e+f x)}}-\frac {2 b^2 (g \cos (e+f x))^{3/2}}{a^3 d^3 f g \sqrt {d \sin (e+f x)}}-\frac {4 \sqrt {g \cos (e+f x)} E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {d \sin (e+f x)}}{5 a d^4 f \sqrt {\sin (2 e+2 f x)}}-\frac {2 b^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)}}{a^3 d^4 f \sqrt {\sin (2 e+2 f x)}}+\frac {\left (2 \sqrt {2} b^3 g \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 )}{a^3 \sqrt {-a+b} d^3 f \sqrt {d \sin (e+f x)}}-\frac {\left (2 \sqrt {2} b^3 g \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 )}{a^3 \sqrt {-a+b} d^3 f \sqrt {d \sin (e+f x)}}\\ &=-\frac {2 (g \cos (e+f x))^{3/2}}{5 a d f g (d \sin (e+f x))^{5/2}}+\frac {2 b (g \cos (e+f x))^{3/2}}{3 a^2 d^2 f g (d \sin (e+f x))^{3/2}}-\frac {4 (g \cos (e+f x))^{3/2}}{5 a d^3 f g \sqrt {d \sin (e+f x)}}-\frac {2 b^2 (g \cos (e+f x))^{3/2}}{a^3 d^3 f g \sqrt {d \sin (e+f x)}}-\frac {2 \sqrt {2} b^3 \sqrt {g} \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)}}{a^3 \sqrt {-a+b} \sqrt {a+b} d^3 f \sqrt {d \sin (e+f x)}}+\frac {2 \sqrt {2} b^3 \sqrt {g} \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)}}{a^3 \sqrt {-a+b} \sqrt {a+b} d^3 f \sqrt {d \sin (e+f x)}}-\frac {4 \sqrt {g \cos (e+f x)} E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {d \sin (e+f x)}}{5 a d^4 f \sqrt {\sin (2 e+2 f x)}}-\frac {2 b^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)}}{a^3 d^4 f \sqrt {\sin (2 e+2 f x)}}\\ \end {align*}

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Mathematica [C]  time = 23.61, size = 1726, normalized size = 3.36 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[g*Cos[e + f*x]]/((d*Sin[e + f*x])^(7/2)*(a + b*Sin[e + f*x])),x]

[Out]

(Sqrt[g*Cos[e + f*x]]*((-2*(2*a^2*Cos[e + f*x] + 5*b^2*Cos[e + f*x])*Csc[e + f*x])/(5*a^3) + (2*b*Cot[e + f*x]
*Csc[e + f*x])/(3*a^2) - (2*Cot[e + f*x]*Csc[e + f*x]^2)/(5*a))*Sin[e + f*x]^4)/(f*(d*Sin[e + f*x])^(7/2)) - (
Sqrt[g*Cos[e + f*x]]*Sin[e + f*x]^(7/2)*((-2*(4*a^3 + 10*a*b^2)*(-(b*AppellF1[3/4, -1/4, 1, 7/4, Cos[e + f*x]^
2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]) + a*AppellF1[3/4, 1/4, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a
^2 + b^2)])*Cos[e + f*x]^(3/2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Sin[e + f*x]^(3/2))/(3*(a^2 - b^2)*(1 - Cos[e
+ f*x]^2)^(3/4)*(a + b*Sin[e + f*x])) + ((2*a^2*b + 10*b^3)*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]])/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]]))/(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)) + ((-2*a^2*b - 5*b^3)*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[T
an[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]*Sqrt[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]*Ta
n[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])))/(5*a^
3*f*Sqrt[Cos[e + f*x]]*(d*Sin[e + f*x])^(7/2))

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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((g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(7/2)/(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 \[ \int \frac {\sqrt {g \cos \left (f x + e\right )}}{{\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(7/2)/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate(sqrt(g*cos(f*x + e))/((b*sin(f*x + e) + a)*(d*sin(f*x + e))^(7/2)), x)

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maple [B]  time = 0.75, size = 6208, normalized size = 12.10 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(7/2)/(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 \[ \int \frac {\sqrt {g \cos \left (f x + e\right )}}{{\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(7/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate(sqrt(g*cos(f*x + e))/((b*sin(f*x + e) + a)*(d*sin(f*x + e))^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {g\,\cos \left (e+f\,x\right )}}{{\left (d\,\sin \left (e+f\,x\right )\right )}^{7/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(e + f*x))^(1/2)/((d*sin(e + f*x))^(7/2)*(a + b*sin(e + f*x))),x)

[Out]

int((g*cos(e + f*x))^(1/2)/((d*sin(e + f*x))^(7/2)*(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))**(1/2)/(d*sin(f*x+e))**(7/2)/(a+b*sin(f*x+e)),x)

[Out]

Timed out

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