3.2.31 \(\int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx\) [131]

Optimal. Leaf size=310 \[ \frac {e^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}} \]

[Out]

1/2*e^(5/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a^2/d*2^(1/2)-1/2*e^(5/2)*arctan(1+2^(1/2)*(e*tan(d
*x+c))^(1/2)/e^(1/2))/a^2/d*2^(1/2)-1/4*e^(5/2)*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a^
2/d*2^(1/2)+1/4*e^(5/2)*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a^2/d*2^(1/2)-4*e^3/a^2/d/
(e*tan(d*x+c))^(1/2)+4*e^3*cos(d*x+c)/a^2/d/(e*tan(d*x+c))^(1/2)-4*e^2*cos(d*x+c)*(sin(c+1/4*Pi+d*x)^2)^(1/2)/
sin(c+1/4*Pi+d*x)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2))*(e*tan(d*x+c))^(1/2)/a^2/d/sin(2*d*x+2*c)^(1/2)

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Rubi [A]
time = 0.33, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 17, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3973, 3971, 3555, 3557, 335, 303, 1176, 631, 210, 1179, 642, 2688, 2695, 2652, 2719, 2687, 32} \begin {gather*} \frac {e^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a^2 d}+\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a^2 d}-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Tan[c + d*x])^(5/2)/(a + a*Sec[c + d*x])^2,x]

[Out]

(e^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a^2*d) - (e^(5/2)*ArcTan[1 + (Sqrt[2]*Sq
rt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a^2*d) - (e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*T
an[c + d*x]]])/(2*Sqrt[2]*a^2*d) + (e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]]
)/(2*Sqrt[2]*a^2*d) - (4*e^3)/(a^2*d*Sqrt[e*Tan[c + d*x]]) + (4*e^3*Cos[c + d*x])/(a^2*d*Sqrt[e*Tan[c + d*x]])
+ (4*e^2*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[e*Tan[c + d*x]])/(a^2*d*Sqrt[Sin[2*c + 2*d*x]])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2652

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a*Sin[e +
f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]), Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2688

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^2*(a*Sec[e
+ f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 1)/(b*f*(n + 1))), x] - Dist[a^2*((m - 2)/(b^2*(n + 1))), Int[(a*Sec[e
+ f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && (GtQ[m, 1] || (Eq
Q[m, 1] && EqQ[n, -3/2])) && IntegersQ[2*m, 2*n]

Rule 2695

Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[Sqrt[Cos[e + f*x]]*(Sqrt[b*
Tan[e + f*x]]/Sqrt[Sin[e + f*x]]), Int[Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, 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 3555

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3971

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3973

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx &=\frac {e^4 \int \frac {(-a+a \sec (c+d x))^2}{(e \tan (c+d x))^{3/2}} \, dx}{a^4}\\ &=\frac {e^4 \int \left (\frac {a^2}{(e \tan (c+d x))^{3/2}}-\frac {2 a^2 \sec (c+d x)}{(e \tan (c+d x))^{3/2}}+\frac {a^2 \sec ^2(c+d x)}{(e \tan (c+d x))^{3/2}}\right ) \, dx}{a^4}\\ &=\frac {e^4 \int \frac {1}{(e \tan (c+d x))^{3/2}} \, dx}{a^2}+\frac {e^4 \int \frac {\sec ^2(c+d x)}{(e \tan (c+d x))^{3/2}} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \frac {\sec (c+d x)}{(e \tan (c+d x))^{3/2}} \, dx}{a^2}\\ &=-\frac {2 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}-\frac {e^2 \int \sqrt {e \tan (c+d x)} \, dx}{a^2}+\frac {\left (4 e^2\right ) \int \cos (c+d x) \sqrt {e \tan (c+d x)} \, dx}{a^2}+\frac {e^4 \text {Subst}\left (\int \frac {1}{(e x)^{3/2}} \, dx,x,\tan (c+d x)\right )}{a^2 d}\\ &=-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}-\frac {e^3 \text {Subst}\left (\int \frac {\sqrt {x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a^2 d}+\frac {\left (4 e^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)} \, dx}{a^2 \sqrt {\sin (c+d x)}}\\ &=-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int \frac {x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d}+\frac {\left (4 e^2 \cos (c+d x) \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\sin (2 c+2 d x)} \, dx}{a^2 \sqrt {\sin (2 c+2 d x)}}\\ &=-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}+\frac {e^3 \text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d}-\frac {e^3 \text {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d}\\ &=-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}-\frac {e^{5/2} \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{5/2} \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {e^3 \text {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a^2 d}-\frac {e^3 \text {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a^2 d}\\ &=-\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}-\frac {e^{5/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}+\frac {e^{5/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}\\ &=\frac {e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.75, size = 812, normalized size = 2.62 \begin {gather*} \frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc ^2(c+d x) \left (\frac {32 \cos \left (\frac {c}{2}\right ) \cos (d x) \sec (2 c) \sin \left (\frac {c}{2}\right )}{d}+\frac {16 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}-\frac {16 \cos (c) \sec (2 c) \sin (d x)}{d}+\frac {16 \tan \left (\frac {c}{2}\right )}{d}\right ) (e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2}+\frac {e^{-2 i c} \left (-e^{4 i c} \sqrt {-1+e^{4 i (c+d x)}} \text {ArcTan}\left (\sqrt {-1+e^{4 i (c+d x)}}\right )+2 \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2}}{d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)}-\frac {e^{-2 i c} \left (\sqrt {-1+e^{4 i (c+d x)}} \text {ArcTan}\left (\sqrt {-1+e^{4 i (c+d x)}}\right )-2 e^{4 i c} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2}}{d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 e^{i (c-d x)} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (3-3 e^{4 i (c+d x)}+e^{4 i d x} \left (1+e^{4 i c}\right ) \sqrt {1-e^{4 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};e^{4 i (c+d x)}\right )\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2}}{3 d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Tan[c + d*x])^(5/2)/(a + a*Sec[c + d*x])^2,x]

[Out]

(Cos[c/2 + (d*x)/2]^4*Csc[c + d*x]^2*((32*Cos[c/2]*Cos[d*x]*Sec[2*c]*Sin[c/2])/d + (16*Sec[c/2]*Sec[c/2 + (d*x
)/2]*Sin[(d*x)/2])/d - (16*Cos[c]*Sec[2*c]*Sin[d*x])/d + (16*Tan[c/2])/d)*(e*Tan[c + d*x])^(5/2))/(a + a*Sec[c
+ d*x])^2 + ((-(E^((4*I)*c)*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]]) + 2*Sqrt[-
1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(
c + d*x)))]])*Cos[c/2 + (d*x)/2]^4*Sec[2*c]*Sec[c + d*x]^2*(e*Tan[c + d*x])^(5/2))/(d*E^((2*I)*c)*Sqrt[((-I)*(
-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Tan[c +
d*x]^(5/2)) - ((Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]] - 2*E^((4*I)*c)*Sqrt[-1
+ E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c
+ d*x)))]])*Cos[c/2 + (d*x)/2]^4*Sec[2*c]*Sec[c + d*x]^2*(e*Tan[c + d*x])^(5/2))/(d*E^((2*I)*c)*Sqrt[((-I)*(-
1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Tan[c +
d*x]^(5/2)) - (8*E^(I*(c - d*x))*Cos[c/2 + (d*x)/2]^4*(3 - 3*E^((4*I)*(c + d*x)) + E^((4*I)*d*x)*(1 + E^((4*I)
*c))*Sqrt[1 - E^((4*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x
]^2*(e*Tan[c + d*x])^(5/2))/(3*d*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*
I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Tan[c + d*x]^(5/2))

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Maple [C] Result contains complex when optimal does not.
time = 0.24, size = 367, normalized size = 1.18

method	result	size
default	\(\frac {\left (1+\cos \left (d x +c \right )\right )^{2} \left (i \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-4 \EllipticF \left (\sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )+8 \EllipticE \left (\sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (\frac {e \sin \left (d x +c \right )}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}}{2 a^{2} d \sin \left (d x +c \right )^{5}}\)	\(367\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/2/a^2/d*(1+cos(d*x+c))^2*(I*EllipticPi((-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))
-I*EllipticPi((-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-4*EllipticF((-(-1+cos(d*x+
c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+8*EllipticE((-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2
^(1/2))-EllipticPi((-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-EllipticPi((-(-1+cos(
d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2)))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c
)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2)*(-1+cos(d*x+c))*(e*sin(d*x+c)/c
os(d*x+c))^(5/2)*cos(d*x+c)^2/sin(d*x+c)^5*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

e^(5/2)*integrate(tan(d*x + c)^(5/2)/(a*sec(d*x + c) + a)^2, x)

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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((e*tan(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))**(5/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Integral((e*tan(c + d*x))**(5/2)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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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((e*tan(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*tan(d*x + c))^(5/2)/(a*sec(d*x + c) + a)^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 )}^2\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(c + d*x))^(5/2)/(a + a/cos(c + d*x))^2,x)

[Out]

int((cos(c + d*x)^2*(e*tan(c + d*x))^(5/2))/(a^2*(cos(c + d*x) + 1)^2), x)

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