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

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

[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 = 0.47, 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 = {3888, 3886, 3474, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2608, 2615, 2572, 2639, 2607, 32} \[ \frac {e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a^2 d}-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}-\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a^2 d}+\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a^2 d}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}} \]

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 204

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

Rule 297

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 329

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 617

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 628

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 1162

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 1165

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 2572

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 2607

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 2608

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 2615

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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3474

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 3476

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 3886

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 3888

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 \operatorname {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 \operatorname {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 ) \operatorname {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 \operatorname {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 \operatorname {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} \operatorname {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} \operatorname {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 \operatorname {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 \operatorname {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} \operatorname {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} \operatorname {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]  time = 6.71, size = 812, normalized size = 2.62 \[ \frac {\csc ^2(c+d x) \left (\frac {32 \cos \left (\frac {c}{2}\right ) \cos (d x) \sec (2 c) \sin \left (\frac {c}{2}\right )}{d}+\frac {16 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}-\frac {16 \cos (c) \sec (2 c) \sin (d x)}{d}+\frac {16 \tan \left (\frac {c}{2}\right )}{d}\right ) (e \tan (c+d x))^{5/2} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{(\sec (c+d x) a+a)^2}+\frac {e^{-2 i c} \left (2 \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )-e^{4 i c} \sqrt {-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt {-1+e^{4 i (c+d x)}}\right )\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a)^2 \tan ^{\frac {5}{2}}(c+d x)}-\frac {e^{-2 i c} \left (\sqrt {-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt {-1+e^{4 i (c+d x)}}\right )-2 e^{4 i c} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a)^2 \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 e^{i (c-d x)} \left (e^{4 i d x} \left (1+e^{4 i c}\right ) \sqrt {1-e^{4 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};e^{4 i (c+d x)}\right )-3 e^{4 i (c+d x)}+3\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a)^2 \tan ^{\frac {5}{2}}(c+d x)} \]

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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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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maple [C]  time = 1.72, size = 360, normalized size = 1.16 \[ \frac {\left (1+\cos \left (d x +c \right )\right )^{2} \left (i \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-4 \EllipticF \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )+8 \EllipticE \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (\frac {e \sin \left (d x +c \right )}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}}{2 a^{2} d \sin \left (d x +c \right )^{5}} \]

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)

[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)/cos(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 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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]

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 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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