\(\int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx\) [81]

Optimal result

Integrand size = 25, antiderivative size = 512 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}+\frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {15 a e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{8 b^4 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 512, 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, 2942, 2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\frac {5 e^{7/2} \left (3 a^2-2 b^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}+\frac {5 e^{7/2} \left (3 a^2-2 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}+\frac {5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{8 b^4 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{8 b^4 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}}-\frac {15 a e^4 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 \sqrt {e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2} \]

[In]

Int[(e*Sin[c + d*x])^(7/2)/(a + b*Cos[c + d*x])^3,x]

[Out]

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

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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 2781

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[
-a^2 + b^2, 2]}, Dist[-a/(2*q), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Dist[b*(g/f), Sub
st[Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e
+ f*x]]*(q - b*Cos[e + f*x])), 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 2942

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 + 1)*Sin[e + f*x])/(b^2*f*(m + 1)*(m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(m + 1)*(m + p + 1
))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[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 {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {\left (5 e^2\right ) \int \frac {\cos (c+d x) (e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2} \, dx}{4 b} \\ & = \frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}+\frac {\left (5 e^4\right ) \int \frac {-b-\frac {3}{2} a \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{4 b^3} \\ & = \frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {\left (15 a e^4\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{8 b^4}+\frac {\left (5 \left (3 a^2-2 b^2\right ) e^4\right ) \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{8 b^4} \\ & = \frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {\left (5 a \left (3 a^2-2 b^2\right ) e^4\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^4 \sqrt {-a^2+b^2}}-\frac {\left (5 a \left (3 a^2-2 b^2\right ) e^4\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^4 \sqrt {-a^2+b^2}}-\frac {\left (5 \left (3 a^2-2 b^2\right ) e^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \sin (c+d x)\right )}{8 b^3 d}-\frac {\left (15 a e^4 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{8 b^4 \sqrt {e \sin (c+d x)}} \\ & = -\frac {15 a e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac {\left (5 \left (3 a^2-2 b^2\right ) e^5\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{4 b^3 d}-\frac {\left (5 a \left (3 a^2-2 b^2\right ) e^4 \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^4 \sqrt {-a^2+b^2} \sqrt {e \sin (c+d x)}}-\frac {\left (5 a \left (3 a^2-2 b^2\right ) e^4 \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^4 \sqrt {-a^2+b^2} \sqrt {e \sin (c+d x)}} \\ & = -\frac {15 a e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \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^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}-\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \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^4 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}+\frac {\left (5 \left (3 a^2-2 b^2\right ) e^4\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^3 \sqrt {-a^2+b^2} d}+\frac {\left (5 \left (3 a^2-2 b^2\right ) e^4\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^3 \sqrt {-a^2+b^2} d} \\ & = \frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}+\frac {5 \left (3 a^2-2 b^2\right ) e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {15 a e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{4 b^4 d \sqrt {e \sin (c+d x)}}+\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \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^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}-\frac {5 a \left (3 a^2-2 b^2\right ) e^4 \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^4 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 e^3 (3 a+2 b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac {e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 10.94 (sec) , antiderivative size = 946, normalized size of antiderivative = 1.85 \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\frac {\csc ^3(c+d x) (e \sin (c+d x))^{7/2} \left (7 a^2+2 b^2+9 a b \cos (c+d x)+\frac {(a+b \cos (c+d x)) (-6 b-7 a \cos (c+d x)+4 b \cos (2 (c+d x))) \left (8 (a+b)-5 (3 a+2 b) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}+(3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{\frac {1}{9} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sin (c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \left (-45 (3 a+2 b) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+18 (3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+\frac {9 (3 a+2 b) \left (2 (a-b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}+9 (3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )-\frac {5 (3 a-2 b) \left (2 (a-b) \operatorname {AppellF1}\left (\frac {9}{4},\frac {1}{2},2,\frac {13}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) \operatorname {AppellF1}\left (\frac {9}{4},\frac {3}{2},1,\frac {13}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+\cos (c+d x) \left (8 (a+b)-5 (3 a+2 b) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}+(3 a-2 b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}\right )}{4 b^3 d (a+b \cos (c+d x))^2} \]

[In]

Integrate[(e*Sin[c + d*x])^(7/2)/(a + b*Cos[c + d*x])^3,x]

[Out]

(Csc[c + d*x]^3*(e*Sin[c + d*x])^(7/2)*(7*a^2 + 2*b^2 + 9*a*b*Cos[c + d*x] + ((a + b*Cos[c + d*x])*(-6*b - 7*a
*Cos[c + d*x] + 4*b*Cos[2*(c + d*x)])*(8*(a + b) - 5*(3*a + 2*b)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[(c + d*x)/2]^
2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sqrt[Sec[(c + d*x)/2]^2] + (3*a - 2*b)*AppellF1[5/4, 1/2, 1, 9/4, -T
an[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sqrt[Sec[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2))/((Sqrt
[Sec[(c + d*x)/2]^2]*Sin[c + d*x]*Tan[(c + d*x)/2]*(-45*(3*a + 2*b)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[(c + d*x)/
2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)] + 18*(3*a - 2*b)*AppellF1[5/4, 1/2, 1, 9/4, -Tan[(c + d*x)/2]^2,
((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sec[(c + d*x)/2]^2 + (9*(3*a + 2*b)*(2*(a - b)*AppellF1[5/4, 1/2, 2, 9/
4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*AppellF1[5/4, 3/2, 1, 9/4, -Tan[(c +
d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + d*x)/2]^2)/(a + b) + 9*(3*a - 2*b)*AppellF1[5/4, 1
/2, 1, 9/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Tan[(c + d*x)/2]^2 - (5*(3*a - 2*b)*(2
*(a - b)*AppellF1[9/4, 1/2, 2, 13/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*App
ellF1[9/4, 3/2, 1, 13/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + d*x)/2]^2*Tan[(
c + d*x)/2]^2)/(a + b)))/9 + Cos[c + d*x]*(8*(a + b) - 5*(3*a + 2*b)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[(c + d*x)
/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sqrt[Sec[(c + d*x)/2]^2] + (3*a - 2*b)*AppellF1[5/4, 1/2, 1, 9/4
, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sqrt[Sec[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2))))/
(4*b^3*d*(a + b*Cos[c + d*x])^2)

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2588\) vs. \(2(536)=1072\).

Time = 98.09 (sec) , antiderivative size = 2589, normalized size of antiderivative = 5.06

method	result	size
default	\(\text {Expression too large to display}\)	\(2589\)

[In]

int((e*sin(d*x+c))^(7/2)/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)

[Out]

(2*e^3*b*((e*sin(d*x+c))^(1/2)/b^4-e^2/b^4*(-1/8*(e*sin(d*x+c))^(1/2)*e^2*(-11*a^2*b^2*cos(d*x+c)^2+2*b^4*cos(
d*x+c)^2+7*a^4+2*a^2*b^2)/(-b^2*cos(d*x+c)^2*e^2+a^2*e^2)^2+5/64*(3*a^2-2*b^2)*(e^2*(a^2-b^2)/b^2)^(1/4)/(a^2*
e^2-b^2*e^2)*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^4*a*(-3/b^4*(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))+(-10*a^2
+6*b^2)/b^4*(-1/2/(-a^2+b^2)^(1/2)/b*(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/(-a^2+b^2)^(1/2)/b*(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/b
^4*(11*a^4-14*a^2*b^2+3*b^4)*(1/2*b^2/e/a^2/(a^2-b^2)*(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-b^2*cos(d*x+c)^2+a^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))-5/8/(a^2-b^2)/(-a^2+b^2)^(1/2)/b*(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)/(-a^2+b^2)^(1/2)*b*(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-s
in(d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))+5/8/(a^2-b^2)/(-a^2+b^2)^(1/2)/b*(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)/(-a^2+b^2)^(1/2)*b*(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)*EllipticP
i((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^4-2*a^2*b^2+b^4)/b^4*(1/4*b^2/e/a^2/(a^
2-b^2)*(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-b^2*cos(d*x+c)^2+a^2)^2+1/16*b^2*(13*a^2-6*b^2)/a^4/(a^2-b^2)^2/e*(
cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-b^2*cos(d*x+c)^2+a^2)+13/32/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^2-45/64/(a^2-b^2)^2/(-a^2+b^2)^(1/2)/b*(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))+9/16/a^2/(a^2-b^2)^2/(-a^2+b^2)^(1/2)*b*(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)*Ellip
ticPi((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/(-a^2+b^2)^(1/2)*b^3*(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))+45/64/(a^2-b^2)^2/(-a^2+b^2)^(1/2)/
b*(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))-9/16/a^2/(a^2-b^2)^2/(-a^2+b^2
)^(1/2)*b*(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/(
-a^2+b^2)^(1/2)*b^3*(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. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((e*sin(d*x+c))^(7/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((e*sin(d*x+c))**(7/2)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate((e*sin(d*x+c))^(7/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((e*sin(d*x + c))^(7/2)/(b*cos(d*x + c) + a)^3, x)

Giac [F]

\[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate((e*sin(d*x+c))^(7/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*sin(d*x + c))^(7/2)/(b*cos(d*x + c) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]

[In]

int((e*sin(c + d*x))^(7/2)/(a + b*cos(c + d*x))^3,x)

[Out]

int((e*sin(c + d*x))^(7/2)/(a + b*cos(c + d*x))^3, x)