\(\int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^3} \, dx\) [82]
Optimal result
Integrand size = 25, antiderivative size = 520 \[
\int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^3} \, dx=-\frac {3 \left (a^2-2 b^2\right ) e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{5/2} \left (-a^2+b^2\right )^{5/4} d}+\frac {3 \left (a^2-2 b^2\right ) e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{5/2} \left (-a^2+b^2\right )^{5/4} d}-\frac {3 a \left (a^2-2 b^2\right ) e^3 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b^3 \left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-2 b^2\right ) e^3 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b^3 \left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{4 b^2 \left (a^2-b^2\right ) d \sqrt {\sin (c+d x)}}+\frac {e (e \sin (c+d x))^{3/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {3 a e (e \sin (c+d x))^{3/2}}{4 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}
\]
[Out]
-3/8*(a^2-2*b^2)*e^(5/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(5/2)/(-a^2+b^2)^(5/4
)/d+3/8*(a^2-2*b^2)*e^(5/2)*arctanh(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(5/2)/(-a^2+b^2)^
(5/4)/d+1/2*e*(e*sin(d*x+c))^(3/2)/b/d/(a+b*cos(d*x+c))^2-3/4*a*e*(e*sin(d*x+c))^(3/2)/b/(a^2-b^2)/d/(a+b*cos(
d*x+c))+3/8*a*(a^2-2*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*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/b^3/(a^2-b^2)/d/(b-(-a^2+b^2)^(1/2))/(e*
sin(d*x+c))^(1/2)+3/8*a*(a^2-2*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)*Elliptic
Pi(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^3/(a^2-b^2)/d/(b+(-a^2+b^2)^
(1/2))/(e*sin(d*x+c))^(1/2)-3/4*a*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)*EllipticE(
cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/b^2/(a^2-b^2)/d/sin(d*x+c)^(1/2)
Rubi [A] (verified)
Time = 1.56 (sec) , antiderivative size = 520, 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 = {2772, 2943, 2946, 2721,
2719, 2780, 2886, 2884, 335, 304, 211, 214} \[
\int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^3} \, dx=-\frac {3 e^{5/2} \left (a^2-2 b^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{5/2} d \left (b^2-a^2\right )^{5/4}}+\frac {3 e^{5/2} \left (a^2-2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{5/2} d \left (b^2-a^2\right )^{5/4}}+\frac {3 a e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{4 b^2 d \left (a^2-b^2\right ) \sqrt {\sin (c+d x)}}-\frac {3 a e (e \sin (c+d x))^{3/2}}{4 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {3 a e^3 \left (a^2-2 b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{8 b^3 d \left (a^2-b^2\right ) \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {3 a e^3 \left (a^2-2 b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{8 b^3 d \left (a^2-b^2\right ) \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}+\frac {e (e \sin (c+d x))^{3/2}}{2 b d (a+b \cos (c+d x))^2}
\]
[In]
Int[(e*Sin[c + d*x])^(5/2)/(a + b*Cos[c + d*x])^3,x]
[Out]
(-3*(a^2 - 2*b^2)*e^(5/2)*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(8*b^(5/2)*(-a^
2 + b^2)^(5/4)*d) + (3*(a^2 - 2*b^2)*e^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e
])])/(8*b^(5/2)*(-a^2 + b^2)^(5/4)*d) - (3*a*(a^2 - 2*b^2)*e^3*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - P
i/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b^3*(a^2 - b^2)*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) - (3*a*
(a^2 - 2*b^2)*e^3*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b^3*(
a^2 - b^2)*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (3*a*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*S
in[c + d*x]])/(4*b^2*(a^2 - b^2)*d*Sqrt[Sin[c + d*x]]) + (e*(e*Sin[c + d*x])^(3/2))/(2*b*d*(a + b*Cos[c + d*x]
)^2) - (3*a*e*(e*Sin[c + d*x])^(3/2))/(4*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))
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 2772
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 + 1))), x] + Dist[g^2*((p - 1)/(b*(m + 1))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && 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 2943
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m + 1)/(f*g*(
a^2 - b^2)*(m + 1))), x] + Dist[1/((a^2 - b^2)*(m + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*S
imp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p},
x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && 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 {e (e \sin (c+d x))^{3/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {\left (3 e^2\right ) \int \frac {\cos (c+d x) \sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^2} \, dx}{4 b} \\ & = \frac {e (e \sin (c+d x))^{3/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {3 a e (e \sin (c+d x))^{3/2}}{4 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (3 e^2\right ) \int \frac {\left (b+\frac {1}{2} a \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{4 b \left (a^2-b^2\right )} \\ & = \frac {e (e \sin (c+d x))^{3/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {3 a e (e \sin (c+d x))^{3/2}}{4 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (3 a e^2\right ) \int \sqrt {e \sin (c+d x)} \, dx}{8 b^2 \left (a^2-b^2\right )}-\frac {\left (3 \left (a^2-2 b^2\right ) e^2\right ) \int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{8 b^2 \left (a^2-b^2\right )} \\ & = \frac {e (e \sin (c+d x))^{3/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {3 a e (e \sin (c+d x))^{3/2}}{4 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (3 a \left (a^2-2 b^2\right ) e^3\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{16 b^3 \left (a^2-b^2\right )}-\frac {\left (3 a \left (a^2-2 b^2\right ) e^3\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{16 b^3 \left (a^2-b^2\right )}+\frac {\left (3 \left (a^2-2 b^2\right ) e^3\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 )}{8 b \left (a^2-b^2\right ) d}+\frac {\left (3 a e^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{8 b^2 \left (a^2-b^2\right ) \sqrt {\sin (c+d x)}} \\ & = \frac {3 a 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)}}{4 b^2 \left (a^2-b^2\right ) d \sqrt {\sin (c+d x)}}+\frac {e (e \sin (c+d x))^{3/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {3 a e (e \sin (c+d x))^{3/2}}{4 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (3 \left (a^2-2 b^2\right ) e^3\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 )}{4 b \left (a^2-b^2\right ) d}+\frac {\left (3 a \left (a^2-2 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}-b \sin (c+d x)\right )} \, dx}{16 b^3 \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}-\frac {\left (3 a \left (a^2-2 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}+b \sin (c+d x)\right )} \, dx}{16 b^3 \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}} \\ & = -\frac {3 a \left (a^2-2 b^2\right ) e^3 \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)}}{8 b^3 \left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-2 b^2\right ) e^3 \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)}}{8 b^3 \left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a 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)}}{4 b^2 \left (a^2-b^2\right ) d \sqrt {\sin (c+d x)}}+\frac {e (e \sin (c+d x))^{3/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {3 a e (e \sin (c+d x))^{3/2}}{4 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (3 \left (a^2-2 b^2\right ) e^3\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 )}{8 b^2 \left (a^2-b^2\right ) d}+\frac {\left (3 \left (a^2-2 b^2\right ) e^3\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 )}{8 b^2 \left (a^2-b^2\right ) d} \\ & = -\frac {3 \left (a^2-2 b^2\right ) e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{5/2} \left (-a^2+b^2\right )^{5/4} d}+\frac {3 \left (a^2-2 b^2\right ) e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{5/2} \left (-a^2+b^2\right )^{5/4} d}-\frac {3 a \left (a^2-2 b^2\right ) e^3 \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)}}{8 b^3 \left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {3 a \left (a^2-2 b^2\right ) e^3 \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)}}{8 b^3 \left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a 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)}}{4 b^2 \left (a^2-b^2\right ) d \sqrt {\sin (c+d x)}}+\frac {e (e \sin (c+d x))^{3/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {3 a e (e \sin (c+d x))^{3/2}}{4 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \\
\end{align*}
Mathematica [C] (warning: unable to verify)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 14.24 (sec) , antiderivative size = 831, normalized size of antiderivative = 1.60
\[
\int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^3} \, dx=\frac {\csc ^2(c+d x) (e \sin (c+d x))^{5/2} \left (\frac {\sin (c+d x)}{2 b (a+b \cos (c+d x))^2}+\frac {3 a \sin (c+d x)}{4 b \left (-a^2+b^2\right ) (a+b \cos (c+d x))}\right )}{d}+\frac {3 (e \sin (c+d x))^{5/2} \left (\frac {a \cos ^2(c+d x) \left (3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )\right )+8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right ) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right )}{12 b^{3/2} \left (-a^2+b^2\right ) (a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {4 b \cos (c+d x) \left (\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}+\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}\right ) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right )}{(a+b \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{8 (a-b) b (a+b) d \sin ^{\frac {5}{2}}(c+d x)}
\]
[In]
Integrate[(e*Sin[c + d*x])^(5/2)/(a + b*Cos[c + d*x])^3,x]
[Out]
(Csc[c + d*x]^2*(e*Sin[c + d*x])^(5/2)*(Sin[c + d*x]/(2*b*(a + b*Cos[c + d*x])^2) + (3*a*Sin[c + d*x])/(4*b*(-
a^2 + b^2)*(a + b*Cos[c + d*x]))))/d + (3*(e*Sin[c + d*x])^(5/2)*((a*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]*Sqr
t[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)*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))*(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))
+ (4*b*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*A
rcTan[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, Si
n[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])))/(8*(a - b)*b*(a + b)*d*Sin[c + d*x]^(5/2))
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(2611\) vs. \(2(544)=1088\).
Time = 98.23 (sec) , antiderivative size = 2612, normalized size of antiderivative =
5.02
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method | result | size |
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default |
\(\text {Expression too large to display}\) |
\(2612\) |
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[In]
int((e*sin(d*x+c))^(5/2)/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)
[Out]
(e^3*b*(-1/4*(e*sin(d*x+c))^(3/2)*e^2*(-5*a^2*b^2*cos(d*x+c)^2+2*b^4*cos(d*x+c)^2+a^4+2*a^2*b^2)/b^2/(a^2-b^2)
/(-b^2*cos(d*x+c)^2*e^2+a^2*e^2)^2+3/32*(a^2-2*b^2)/b^4/(a^2-b^2)/(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^3*a*(3/b^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))-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)))-(7*a^
2-3*b^2)/b^2*(1/2*b^2/e/a^2/(a^2-b^2)*sin(d*x+c)*(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-b^2*cos(d*x+c)^2+a^2)-1/2
/a^2/(a^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)*
EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))+1/4/a^2/(a^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)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-3/8/(a^2-b^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))+1/4/a^2/(a^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)*Ellipti
cPi((1-sin(d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-3/8/(a^2-b^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))+1/4/a^2/(a^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)))+4*a^2*(a^2-b^2)/b^2*(1/4*b^2/e/a^2/(a^2-b^2)*sin(d*x+c)*(cos(d*x+c)^2*e*si
n(d*x+c))^(1/2)/(-b^2*cos(d*x+c)^2+a^2)^2+1/16*b^2*(11*a^2-6*b^2)/a^4/(a^2-b^2)^2/e*sin(d*x+c)*(cos(d*x+c)^2*e
*sin(d*x+c))^(1/2)/(-b^2*cos(d*x+c)^2+a^2)-11/16/a^2/(a^2-b^2)^2*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*s
in(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))+3/8/a^4/(a^2-b^2
)^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)*EllipticE((
1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2+11/32/a^2/(a^2-b^2)^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)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-3/16/a^4/(a^2-b^2)^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)*EllipticF((1-sin
(d*x+c))^(1/2),1/2*2^(1/2))*b^2-21/64/(a^2-b^2)^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))+7/16/a^2/(a^2-b^2)^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))-3/16/a^4/(a^2-b^2)^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))-21/64/(a^2-b^2)^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))+7/16/a
^2/(a^2-b^2)^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))-3/16/a^4/(a^2-b^2
)^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. \[
\int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((e*sin(d*x+c))^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
[Out]
Timed out
Sympy [F(-1)]
Timed out. \[
\int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((e*sin(d*x+c))**(5/2)/(a+b*cos(d*x+c))**3,x)
[Out]
Timed out
Maxima [F]
\[
\int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x }
\]
[In]
integrate((e*sin(d*x+c))^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
[Out]
integrate((e*sin(d*x + c))^(5/2)/(b*cos(d*x + c) + a)^3, x)
Giac [F]
\[
\int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x }
\]
[In]
integrate((e*sin(d*x+c))^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((e*sin(d*x + c))^(5/2)/(b*cos(d*x + c) + a)^3, x)
Mupad [F(-1)]
Timed out. \[
\int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^3} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x
\]
[In]
int((e*sin(c + d*x))^(5/2)/(a + b*cos(c + d*x))^3,x)
[Out]
int((e*sin(c + d*x))^(5/2)/(a + b*cos(c + d*x))^3, x)