\(\int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx\) [59]

Optimal result

Integrand size = 25, antiderivative size = 461 \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=-\frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}+\frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}+\frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 a \left (5 a^2-8 b^2\right ) e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 b^4 d \sqrt {\sin (c+d x)}}+\frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d} \]

[Out]

-(-a^2+b^2)^(7/4)*e^(9/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(9/2)/d+(-a^2+b^2)^(
7/4)*e^(9/2)*arctanh(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(9/2)/d+2/15*e^3*(5*a^2-5*b^2-3*
a*b*cos(d*x+c))*(e*sin(d*x+c))^(3/2)/b^3/d-2/7*e*(e*sin(d*x+c))^(7/2)/b/d-a*(a^2-b^2)^2*e^5*(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*b/(b-(-a^2+b^2)^(1/2)),2^(1
/2))*sin(d*x+c)^(1/2)/b^5/d/(b-(-a^2+b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-a*(a^2-b^2)^2*e^5*(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*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))
*sin(d*x+c)^(1/2)/b^5/d/(b+(-a^2+b^2)^(1/2))/(e*sin(d*x+c))^(1/2)+2/5*a*(5*a^2-8*b^2)*e^4*(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)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/b^
4/d/sin(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 1.53 (sec) , antiderivative size = 461, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2774, 2944, 2946, 2721, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=-\frac {e^{9/2} \left (b^2-a^2\right )^{7/4} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} d}+\frac {e^{9/2} \left (b^2-a^2\right )^{7/4} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} d}+\frac {a e^5 \left (a^2-b^2\right )^2 \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b^5 d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}+\frac {a e^5 \left (a^2-b^2\right )^2 \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b^5 d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}-\frac {2 a e^4 \left (5 a^2-8 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 b^4 d \sqrt {\sin (c+d x)}}+\frac {2 e^3 (e \sin (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right )}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d} \]

[In]

Int[(e*Sin[c + d*x])^(9/2)/(a + b*Cos[c + d*x]),x]

[Out]

-(((-a^2 + b^2)^(7/4)*e^(9/2)*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(b^(9/2)*d)
) + ((-a^2 + b^2)^(7/4)*e^(9/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(b^(9/2)
*d) + (a*(a^2 - b^2)^2*e^5*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])
/(b^5*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (a*(a^2 - b^2)^2*e^5*EllipticPi[(2*b)/(b + Sqrt[-a^2 +
b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(b^5*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) - (2*a*(
5*a^2 - 8*b^2)*e^4*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*b^4*d*Sqrt[Sin[c + d*x]]) + (2*e^
3*(5*(a^2 - b^2) - 3*a*b*Cos[c + d*x])*(e*Sin[c + d*x])^(3/2))/(15*b^3*d) - (2*e*(e*Sin[c + d*x])^(7/2))/(7*b*
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 2774

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

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac {e^2 \int \frac {(-b-a \cos (c+d x)) (e \sin (c+d x))^{5/2}}{a+b \cos (c+d x)} \, dx}{b} \\ & = \frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac {\left (2 e^4\right ) \int \frac {\left (\frac {1}{2} b \left (2 a^2-5 b^2\right )+\frac {1}{2} a \left (5 a^2-8 b^2\right ) \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{5 b^3} \\ & = \frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac {\left (a \left (5 a^2-8 b^2\right ) e^4\right ) \int \sqrt {e \sin (c+d x)} \, dx}{5 b^4}+\frac {\left (\left (a^2-b^2\right )^2 e^4\right ) \int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{b^4} \\ & = \frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac {\left (a \left (a^2-b^2\right )^2 e^5\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 b^5}+\frac {\left (a \left (a^2-b^2\right )^2 e^5\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 b^5}-\frac {\left (\left (a^2-b^2\right )^2 e^5\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \sin (c+d x)\right )}{b^3 d}-\frac {\left (a \left (5 a^2-8 b^2\right ) e^4 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 b^4 \sqrt {\sin (c+d x)}} \\ & = -\frac {2 a \left (5 a^2-8 b^2\right ) e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 b^4 d \sqrt {\sin (c+d x)}}+\frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac {\left (2 \left (a^2-b^2\right )^2 e^5\right ) \text {Subst}\left (\int \frac {x^2}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{b^3 d}-\frac {\left (a \left (a^2-b^2\right )^2 e^5 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 b^5 \sqrt {e \sin (c+d x)}}+\frac {\left (a \left (a^2-b^2\right )^2 e^5 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 b^5 \sqrt {e \sin (c+d x)}} \\ & = \frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 a \left (5 a^2-8 b^2\right ) e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 b^4 d \sqrt {\sin (c+d x)}}+\frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d}+\frac {\left (\left (a^2-b^2\right )^2 e^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{b^4 d}-\frac {\left (\left (a^2-b^2\right )^2 e^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{b^4 d} \\ & = -\frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}+\frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}+\frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 a \left (5 a^2-8 b^2\right ) e^4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 b^4 d \sqrt {\sin (c+d x)}}+\frac {2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac {2 e (e \sin (c+d x))^{7/2}}{7 b d} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 36.31 (sec) , antiderivative size = 834, normalized size of antiderivative = 1.81 \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=-\frac {(e \sin (c+d x))^{9/2} \left (\frac {\left (5 a^3-8 a b^2\right ) \cos ^2(c+d x) \left (3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )\right )+8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right ) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right )}{12 b^{3/2} \left (-a^2+b^2\right ) (a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {2 \left (2 a^2 b-5 b^3\right ) \cos (c+d x) \left (\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}+\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {b^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 (a+b \sqrt {1-\sin ^2(c+d x)}\right )}{(a+b \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{5 b^3 d \sin ^{\frac {9}{2}}(c+d x)}+\frac {\csc ^4(c+d x) (e \sin (c+d x))^{9/2} \left (-\frac {\left (-28 a^2+37 b^2\right ) \sin (c+d x)}{42 b^3}-\frac {a \sin (2 (c+d x))}{5 b^2}+\frac {\sin (3 (c+d x))}{14 b}\right )}{d} \]

[In]

Integrate[(e*Sin[c + d*x])^(9/2)/(a + b*Cos[c + d*x]),x]

[Out]

-1/5*((e*Sin[c + d*x])^(9/2)*(((5*a^3 - 8*a*b^2)*Cos[c + d*x]^2*(3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - (
Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^
2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]]
+ Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]]) + 8*b^(5/2)*A
ppellF1[3/4, -1/2, 1, 7/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(3/2))*(a + b*Sqrt[
1 - Sin[c + d*x]^2]))/(12*b^(3/2)*(-a^2 + b^2)*(a + b*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (2*(2*a^2*b - 5*b^
3)*Cos[c + d*x]*(((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTa
n[1 + ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2
+ b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]] + Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4
)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)) + (a*AppellF1[3/4, 1/2, 1, 7/4, Sin[c
+ d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(3/2))/(3*(a^2 - b^2)))*(a + b*Sqrt[1 - Sin[c + d*x]
^2]))/((a + b*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2])))/(b^3*d*Sin[c + d*x]^(9/2)) + (Csc[c + d*x]^4*(e*Sin[c
+ d*x])^(9/2)*(-1/42*((-28*a^2 + 37*b^2)*Sin[c + d*x])/b^3 - (a*Sin[2*(c + d*x)])/(5*b^2) + Sin[3*(c + d*x)]/(
14*b)))/d

Maple [A] (verified)

Time = 5.37 (sec) , antiderivative size = 851, normalized size of antiderivative = 1.85

method	result	size
default	\(\text {Expression too large to display}\)	\(851\)

[In]

int((e*sin(d*x+c))^(9/2)/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)

[Out]

(-2*e*b*(-1/21/b^4*(e*sin(d*x+c))^(3/2)*e^2*(3*b^2*cos(d*x+c)^2+7*a^2-10*b^2)+1/8*e^4*(a^4-2*a^2*b^2+b^4)/b^6/
(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((e*sin(d*x+c)-(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^
2*(a^2-b^2)/b^2)^(1/2))/(e*sin(d*x+c)+(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^
2)^(1/2)))+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)+1)+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/
b^2)^(1/4)*(e*sin(d*x+c))^(1/2)-1)))+(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*e^5*a*(1/5/b^4/(cos(d*x+c)^2*e*sin(d*x+
c))^(1/2)*(10*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*
2^(1/2))*a^2-16*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),1/
2*2^(1/2))*b^2-5*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1
/2*2^(1/2))*a^2+8*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),
1/2*2^(1/2))*b^2+2*b^2*cos(d*x+c)^4-2*b^2*cos(d*x+c)^2)+(a^4-2*a^2*b^2+b^4)/b^4*(-1/2/b^2*(1-sin(d*x+c))^(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)/b)*EllipticPi((
1-sin(d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-1/2/b^2*(1-sin(d*x+c))^(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)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1
+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))))/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

Fricas [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((e*sin(d*x+c))^(9/2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((e*sin(d*x+c))**(9/2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {9}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \]

[In]

integrate((e*sin(d*x+c))^(9/2)/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate((e*sin(d*x + c))^(9/2)/(b*cos(d*x + c) + a), x)

Giac [F]

\[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {9}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \]

[In]

integrate((e*sin(d*x+c))^(9/2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*sin(d*x + c))^(9/2)/(b*cos(d*x + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{9/2}}{a+b\,\cos \left (c+d\,x\right )} \,d x \]

[In]

int((e*sin(c + d*x))^(9/2)/(a + b*cos(c + d*x)),x)

[Out]

int((e*sin(c + d*x))^(9/2)/(a + b*cos(c + d*x)), x)