3.1.67 \(\int \frac {1}{(a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}} \, dx\) [67]

Optimal. Leaf size=501 \[ -\frac {b^{7/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{9/4} d e^{7/2}}+\frac {b^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{9/4} d e^{7/2}}+\frac {2 (b-a \cos (c+d x))}{5 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{5/2}}-\frac {2 \left (5 b^3+a \left (3 a^2-8 b^2\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^2 d e^3 \sqrt {e \sin (c+d x)}}+\frac {a b^3 \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right )^2 \left (b-\sqrt {-a^2+b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}+\frac {a b^3 \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right )^2 \left (b+\sqrt {-a^2+b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 a \left (3 a^2-8 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 \left (a^2-b^2\right )^2 d e^4 \sqrt {\sin (c+d x)}} \]

[Out]

-b^(7/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/(-a^2+b^2)^(9/4)/d/e^(7/2)+b^(7/2)*arct
anh(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/(-a^2+b^2)^(9/4)/d/e^(7/2)+2/5*(b-a*cos(d*x+c))/(a^
2-b^2)/d/e/(e*sin(d*x+c))^(5/2)-2/5*(5*b^3+a*(3*a^2-8*b^2)*cos(d*x+c))/(a^2-b^2)^2/d/e^3/(e*sin(d*x+c))^(1/2)-
a*b^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)/(a^2-b^2)^2/d/e^3/(b-(-a^2+b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-a*b^
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)/(a^2-b^2)^2/d/e^3/(b+(-a^2+b^2)^(1/2))/(e*sin(d*x+c))^(1/2)+2/5*a*(3
*a^2-8*b^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)/(a^2-b^2)^2/d/e^4/sin(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.87, antiderivative size = 501, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2775, 2945, 2946, 2721, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \begin {gather*} -\frac {b^{7/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{7/2} \left (b^2-a^2\right )^{9/4}}-\frac {2 a \left (3 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 d e^4 \left (a^2-b^2\right )^2 \sqrt {\sin (c+d x)}}+\frac {2 (b-a \cos (c+d x))}{5 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{5/2}}+\frac {b^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{7/2} \left (b^2-a^2\right )^{9/4}}-\frac {2 \left (a \left (3 a^2-8 b^2\right ) \cos (c+d x)+5 b^3\right )}{5 d e^3 \left (a^2-b^2\right )^2 \sqrt {e \sin (c+d x)}}+\frac {a b^3 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e^3 \left (a^2-b^2\right )^2 \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}+\frac {a b^3 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e^3 \left (a^2-b^2\right )^2 \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(7/2)),x]

[Out]

-((b^(7/2)*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(9/4)*d*e^(7/2))
) + (b^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(9/4)*d*e^(7/
2)) + (2*(b - a*Cos[c + d*x]))/(5*(a^2 - b^2)*d*e*(e*Sin[c + d*x])^(5/2)) - (2*(5*b^3 + a*(3*a^2 - 8*b^2)*Cos[
c + d*x]))/(5*(a^2 - b^2)^2*d*e^3*Sqrt[e*Sin[c + d*x]]) + (a*b^3*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c -
Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/((a^2 - b^2)^2*(b - Sqrt[-a^2 + b^2])*d*e^3*Sqrt[e*Sin[c + d*x]]) + (a*
b^3*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/((a^2 - b^2)^2*(b + Sq
rt[-a^2 + b^2])*d*e^3*Sqrt[e*Sin[c + d*x]]) - (2*a*(3*a^2 - 8*b^2)*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin
[c + d*x]])/(5*(a^2 - b^2)^2*d*e^4*Sqrt[Sin[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 2775

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*Cos
[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[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 2945

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c -
b*d)*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -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*} \int \frac {1}{(a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}} \, dx &=\frac {2 (b-a \cos (c+d x))}{5 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{5/2}}-\frac {2 \int \frac {-\frac {3 a^2}{2}+\frac {5 b^2}{2}-\frac {3}{2} a b \cos (c+d x)}{(a+b \cos (c+d x)) (e \sin (c+d x))^{3/2}} \, dx}{5 \left (a^2-b^2\right ) e^2}\\ &=\frac {2 (b-a \cos (c+d x))}{5 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{5/2}}-\frac {2 \left (5 b^3+a \left (3 a^2-8 b^2\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^2 d e^3 \sqrt {e \sin (c+d x)}}+\frac {4 \int \frac {\left (\frac {1}{4} \left (-3 a^4+8 a^2 b^2+5 b^4\right )-\frac {1}{4} a b \left (3 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 \left (a^2-b^2\right )^2 e^4}\\ &=\frac {2 (b-a \cos (c+d x))}{5 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{5/2}}-\frac {2 \left (5 b^3+a \left (3 a^2-8 b^2\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^2 d e^3 \sqrt {e \sin (c+d x)}}+\frac {b^4 \int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2 e^4}-\frac {\left (a \left (3 a^2-8 b^2\right )\right ) \int \sqrt {e \sin (c+d x)} \, dx}{5 \left (a^2-b^2\right )^2 e^4}\\ &=\frac {2 (b-a \cos (c+d x))}{5 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{5/2}}-\frac {2 \left (5 b^3+a \left (3 a^2-8 b^2\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^2 d e^3 \sqrt {e \sin (c+d x)}}-\frac {\left (a b^3\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 \left (a^2-b^2\right )^2 e^3}+\frac {\left (a b^3\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 \left (a^2-b^2\right )^2 e^3}-\frac {b^5 \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 )}{\left (a^2-b^2\right )^2 d e^3}-\frac {\left (a \left (3 a^2-8 b^2\right ) \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 \left (a^2-b^2\right )^2 e^4 \sqrt {\sin (c+d x)}}\\ &=\frac {2 (b-a \cos (c+d x))}{5 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{5/2}}-\frac {2 \left (5 b^3+a \left (3 a^2-8 b^2\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^2 d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 a \left (3 a^2-8 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 \left (a^2-b^2\right )^2 d e^4 \sqrt {\sin (c+d x)}}-\frac {\left (2 b^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 )}{\left (a^2-b^2\right )^2 d e^3}-\frac {\left (a b^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}{2 \left (a^2-b^2\right )^2 e^3 \sqrt {e \sin (c+d x)}}+\frac {\left (a b^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}{2 \left (a^2-b^2\right )^2 e^3 \sqrt {e \sin (c+d x)}}\\ &=\frac {2 (b-a \cos (c+d x))}{5 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{5/2}}-\frac {2 \left (5 b^3+a \left (3 a^2-8 b^2\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^2 d e^3 \sqrt {e \sin (c+d x)}}+\frac {a b^3 \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right )^2 \left (b-\sqrt {-a^2+b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}+\frac {a b^3 \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right )^2 \left (b+\sqrt {-a^2+b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 a \left (3 a^2-8 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 \left (a^2-b^2\right )^2 d e^4 \sqrt {\sin (c+d x)}}+\frac {b^4 \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (a^2-b^2\right )^2 d e^3}-\frac {b^4 \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (a^2-b^2\right )^2 d e^3}\\ &=-\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{9/4} d e^{7/2}}+\frac {b^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{9/4} d e^{7/2}}+\frac {2 (b-a \cos (c+d x))}{5 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{5/2}}-\frac {2 \left (5 b^3+a \left (3 a^2-8 b^2\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^2 d e^3 \sqrt {e \sin (c+d x)}}+\frac {a b^3 \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right )^2 \left (b-\sqrt {-a^2+b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}+\frac {a b^3 \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right )^2 \left (b+\sqrt {-a^2+b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 a \left (3 a^2-8 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 \left (a^2-b^2\right )^2 d e^4 \sqrt {\sin (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 26.57, size = 881, normalized size = 1.76 \begin {gather*} \frac {\left (-\frac {2 \left (5 b^3+3 a^3 \cos (c+d x)-8 a b^2 \cos (c+d x)\right ) \csc (c+d x)}{5 \left (a^2-b^2\right )^2}-\frac {2 (-b+a \cos (c+d x)) \csc ^3(c+d x)}{5 \left (a^2-b^2\right )}\right ) \sin ^4(c+d x)}{d (e \sin (c+d x))^{7/2}}-\frac {\sin ^{\frac {7}{2}}(c+d x) \left (\frac {\left (3 a^3 b-8 a b^3\right ) \cos ^2(c+d x) \left (3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \text {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} F_1\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 (3 a^4-8 a^2 b^2-5 b^4\right ) \cos (c+d x) \left (\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \text {ArcTan}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \text {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 F_1\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 (a-b)^2 (a+b)^2 d (e \sin (c+d x))^{7/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(7/2)),x]

[Out]

(((-2*(5*b^3 + 3*a^3*Cos[c + d*x] - 8*a*b^2*Cos[c + d*x])*Csc[c + d*x])/(5*(a^2 - b^2)^2) - (2*(-b + a*Cos[c +
d*x])*Csc[c + d*x]^3)/(5*(a^2 - b^2)))*Sin[c + d*x]^4)/(d*(e*Sin[c + d*x])^(7/2)) - (Sin[c + d*x]^(7/2)*(((3*
a^3*b - 8*a*b^3)*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)*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)) + (2*(3*a^4 - 8*a^2*b^2 - 5*b^4)*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*ArcTan[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])))/(5*(a - b)^2*(a + b)^2*d*(e*Sin[c + d*x])^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1079\) vs. \(2(535)=1070\).
time = 0.24, size = 1080, normalized size = 2.16

method	result	size
default	\(\text {Expression too large to display}\)	\(1080\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(d*x+c))/(e*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

(-1/4*b^3/e^3/(a-b)^2/(a+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)))-1/2*b^3/e^3/(a-b)^2/(a+b)^2/(e^2*(a^2-b^2)/b^2)^(1/4)*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)-1/2*b^3/e^3/(a-b)^2/(a+b)^2/(e^2*(a^2-b^2)/b^2)^(1/4)
*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)+2/5*b/e/(a+b)/(a-b)/(e*sin(d*x+c))^(
5/2)-2*b^3/e^3/(a-b)^2/(a+b)^2/(e*sin(d*x+c))^(1/2)+1/10*a/e^3*(5*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)
*sin(d*x+c)^(7/2)*EllipticPi((-sin(d*x+c)+1)^(1/2),-b/((-a^2+b^2)^(1/2)-b),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*b^3+5
*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticPi((-sin(d*x+c)+1)^(1/2),-b/((-a^2+b^2)
^(1/2)-b),1/2*2^(1/2))*b^4-5*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticPi((-sin(d*
x+c)+1)^(1/2),b/(b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*b^3+5*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2
)^(1/2)*sin(d*x+c)^(7/2)*EllipticPi((-sin(d*x+c)+1)^(1/2),b/(b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*b^4-12*(-sin(d*x
+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^4+32*(-sin
(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2*b^2+
6*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a
^4-16*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2
))*a^2*b^2-12*a^4*cos(d*x+c)^4*sin(d*x+c)+32*a^2*b^2*cos(d*x+c)^4*sin(d*x+c)+16*a^4*cos(d*x+c)^2*sin(d*x+c)-36
*a^2*b^2*cos(d*x+c)^2*sin(d*x+c))/sin(d*x+c)^3/(a^2-b^2)^2/(b+(-a^2+b^2)^(1/2))/((-a^2+b^2)^(1/2)-b)/cos(d*x+c
)/(e*sin(d*x+c))^(1/2))/d

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(sin(d*x + c))/(b*cos(d*x + c)^5*e^(7/2) + a*cos(d*x + c)^4*e^(7/2) - 2*b*cos(d*x + c)^3*e^(7/2)
- 2*a*cos(d*x + c)^2*e^(7/2) + b*cos(d*x + c)*e^(7/2) + a*e^(7/2)), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))/(e*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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