3.3.40 \(\int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{7/2}} \, dx\) [240]

Optimal. Leaf size=511 \[ \frac {a^{5/2} b \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\left (a^2-b^2\right )^{9/4} d e^{7/2}}-\frac {a^{5/2} b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\left (a^2-b^2\right )^{9/4} d e^{7/2}}+\frac {2 (b-a \cos (c+d x))}{5 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{5/2}}+\frac {2 \left (5 a^2 b-a \left (3 a^2+2 b^2\right ) \cos (c+d x)\right )}{5 \left (a^2-b^2\right )^2 d e^3 \sqrt {e \sin (c+d x)}}-\frac {a^2 b^2 \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right )^2 \left (a-\sqrt {a^2-b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}-\frac {a^2 b^2 \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right )^2 \left (a+\sqrt {a^2-b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 a \left (3 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 \left (a^2-b^2\right )^2 d e^4 \sqrt {\sin (c+d x)}} \]

[Out]

a^(5/2)*b*arctan(a^(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)-a^(5/2)*b*arc
tanh(a^(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*a^2*b-a*(3*a^2+2*b^2)*cos(d*x+c))/(a^2-b^2)^2/d/e^3/(e*sin(d*x+c))^(1/2)
+a^2*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)*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*
a/(a-(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/(a^2-b^2)^2/d/e^3/(a-(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)+a^2
*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)*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*a/(a
+(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/(a^2-b^2)^2/d/e^3/(a+(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)+2/5*a*(
3*a^2+2*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 = 0.99, antiderivative size = 511, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3957, 2945, 2946, 2721, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \begin {gather*} -\frac {2 a \left (3 a^2+2 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 \left (5 a^2 b-a \left (3 a^2+2 b^2\right ) \cos (c+d x)\right )}{5 d e^3 \left (a^2-b^2\right )^2 \sqrt {e \sin (c+d x)}}-\frac {a^2 b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^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 (a-\sqrt {a^2-b^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {a^2 b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^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 {a^2-b^2}+a\right ) \sqrt {e \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 {a^{5/2} b \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{7/2} \left (a^2-b^2\right )^{9/4}}-\frac {a^{5/2} b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{7/2} \left (a^2-b^2\right )^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{7/2}} \, dx &=-\int \frac {\cos (c+d x)}{(-b-a \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 {a b-\frac {3}{2} a^2 \cos (c+d x)}{(-b-a \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 a^2 b-a \left (3 a^2+2 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}{2} a b \left (4 a^2+b^2\right )+\frac {1}{4} a^2 \left (3 a^2+2 b^2\right ) \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{-b-a \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 a^2 b-a \left (3 a^2+2 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^3 b\right ) \int \frac {\sqrt {e \sin (c+d x)}}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2 e^4}-\frac {\left (a \left (3 a^2+2 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 a^2 b-a \left (3 a^2+2 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^2 b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )} \, dx}{2 \left (a^2-b^2\right )^2 e^3}-\frac {\left (a^2 b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}+a \sin (c+d x)\right )} \, dx}{2 \left (a^2-b^2\right )^2 e^3}+\frac {\left (a^4 b\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (-a^2+b^2\right ) e^2+a^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+2 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 a^2 b-a \left (3 a^2+2 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+2 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 a^4 b\right ) \text {Subst}\left (\int \frac {x^2}{\left (-a^2+b^2\right ) e^2+a^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^2 b^2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {a^2-b^2}-a \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^2 b^2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {a^2-b^2}+a \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 a^2 b-a \left (3 a^2+2 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^2 b^2 \Pi \left (\frac {2 a}{a-\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 (a-\sqrt {a^2-b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}-\frac {a^2 b^2 \Pi \left (\frac {2 a}{a+\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 (a+\sqrt {a^2-b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 a \left (3 a^2+2 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 (a^3 b\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a^2-b^2} e-a x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (a^2-b^2\right )^2 d e^3}+\frac {\left (a^3 b\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a^2-b^2} e+a x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (a^2-b^2\right )^2 d e^3}\\ &=\frac {a^{5/2} b \tan ^{-1}\left (\frac {\sqrt {a} \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 {a^{5/2} b \tanh ^{-1}\left (\frac {\sqrt {a} \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 a^2 b-a \left (3 a^2+2 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^2 b^2 \Pi \left (\frac {2 a}{a-\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 (a-\sqrt {a^2-b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}-\frac {a^2 b^2 \Pi \left (\frac {2 a}{a+\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 (a+\sqrt {a^2-b^2}\right ) d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 a \left (3 a^2+2 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 = 15.33, size = 797, normalized size = 1.56 \begin {gather*} \frac {(b+a \cos (c+d x)) \sin ^3(c+d x) \left (-\frac {2 \left (\left (a^2-b^2\right ) (-b+a \cos (c+d x)) \csc ^2(c+d x) \sec (c+d x)+a \left (3 a^2+2 b^2-5 a b \sec (c+d x)\right )\right )}{\left (a^2-b^2\right )^2}-\frac {\left (b+a \sqrt {\cos ^2(c+d x)}\right ) \sec ^2(c+d x) \sqrt {\sin (c+d x)} \left (\left (3 a^3+2 a b^2\right ) \cos (c+d x) \left (3 \sqrt {2} b \left (-a^2+b^2\right )^{3/4} \left (2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )\right )+8 a^{5/2} F_1\left (\frac {3}{4};-\frac {1}{2},1;\frac {7}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right )+(2+2 i) a b \left (4 a^2+b^2\right ) \sqrt {\cos ^2(c+d x)} \left (3 \left (a^2-b^2\right )^{3/4} \left (2 \text {ArcTan}\left (1-\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \text {ArcTan}\left (1+\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )\right )-(4-4 i) \sqrt {a} b F_1\left (\frac {3}{4};\frac {1}{2},1;\frac {7}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right )\right )}{12 \sqrt {a} (a-b)^2 (a+b)^2 \left (a^2-b^2\right ) (b+a \cos (c+d x))}\right )}{5 d (a+b \sec (c+d x)) (e \sin (c+d x))^{7/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])*Sin[c + d*x]^3*((-2*((a^2 - b^2)*(-b + a*Cos[c + d*x])*Csc[c + d*x]^2*Sec[c + d*x] + a*(
3*a^2 + 2*b^2 - 5*a*b*Sec[c + d*x])))/(a^2 - b^2)^2 - ((b + a*Sqrt[Cos[c + d*x]^2])*Sec[c + d*x]^2*Sqrt[Sin[c
+ d*x]]*((3*a^3 + 2*a*b^2)*Cos[c + d*x]*(3*Sqrt[2]*b*(-a^2 + b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[a]*Sqrt[Si
n[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log
[Sqrt[-a^2 + b^2] - Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + a*Sin[c + d*x]] + Log[Sqrt[-a^2 +
b^2] + Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + a*Sin[c + d*x]]) + 8*a^(5/2)*AppellF1[3/4, -1/2
, 1, 7/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Sin[c + d*x]^(3/2)) + (2 + 2*I)*a*b*(4*a^2 + b^2)*
Sqrt[Cos[c + d*x]^2]*(3*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)
] - 2*ArcTan[1 + ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - (1 + I)*Sqrt[
a]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*a*Sin[c + d*x]] + Log[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[a]*(a^2 - b^2
)^(1/4)*Sqrt[Sin[c + d*x]] + I*a*Sin[c + d*x]]) - (4 - 4*I)*Sqrt[a]*b*AppellF1[3/4, 1/2, 1, 7/4, Sin[c + d*x]^
2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Sin[c + d*x]^(3/2))))/(12*Sqrt[a]*(a - b)^2*(a + b)^2*(a^2 - b^2)*(b + a*
Cos[c + d*x]))))/(5*d*(a + b*Sec[c + d*x])*(e*Sin[c + d*x])^(7/2))

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Maple [A]
time = 0.35, size = 924, normalized size = 1.81

method	result	size
default	\(\text {Expression too large to display}\)	\(924\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2/5/e*b/(a-b)/(a+b)/(e*sin(d*x+c))^(5/2)+2/e^3*b/(a+b)^2/(a-b)^2*a^2/(e*sin(d*x+c))^(1/2)+1/e^3*b*a^2/(a+b)^2
/(a-b)^2/(e^2*(a^2-b^2)/a^2)^(1/4)*arctan((e*sin(d*x+c))^(1/2)/(e^2*(a^2-b^2)/a^2)^(1/4))-1/2/e^3*b*a^2/(a+b)^
2/(a-b)^2/(e^2*(a^2-b^2)/a^2)^(1/4)*ln(((e*sin(d*x+c))^(1/2)+(e^2*(a^2-b^2)/a^2)^(1/4))/((e*sin(d*x+c))^(1/2)-
(e^2*(a^2-b^2)/a^2)^(1/4)))-1/10/e^3*b^2*(12*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*Ell
ipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2+8*(-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))*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^2-4*(-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))*b^2+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),-a/((a^2-b^2)^(1/2)-a),1/2*2^(1/2))*(a^2-b^2)^(1/2)*a+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),-a/((a^2-b^2)^(1/2)-a),1/2*
2^(1/2))*a^2-5*(-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),
a/(a+(a^2-b^2)^(1/2)),1/2*2^(1/2))*(a^2-b^2)^(1/2)*a+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),a/(a+(a^2-b^2)^(1/2)),1/2*2^(1/2))*a^2+12*a^2*cos(d*x+c)^4*sin(d*x+c)+
8*b^2*cos(d*x+c)^4*sin(d*x+c)-16*a^2*cos(d*x+c)^2*sin(d*x+c)-4*b^2*cos(d*x+c)^2*sin(d*x+c))*a/sin(d*x+c)^3/(a+
(a^2-b^2)^(1/2))/((a^2-b^2)^(1/2)-a)/(a-b)^2/(a+b)^2/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*sec(d*x+c))/(e*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [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*sec(d*x+c))/(e*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

Timed out

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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*sec(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*sec(d*x+c))/(e*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(1/((b*sec(d*x + c) + a)*(e*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 {\cos \left (c+d\,x\right )}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}\,\left (b+a\,\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(cos(c + d*x)/((e*sin(c + d*x))^(7/2)*(b + a*cos(c + d*x))), x)

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