\(\int \frac {(e \tan (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx\) [121]

Optimal result

Integrand size = 25, antiderivative size = 285 \[ \int \frac {(e \tan (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\frac {e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {2 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d} \]

[Out]

1/2*e^(5/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/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/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/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/d*2^(1/2)+2*e^2*cos(d*x+c)*(si
n(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/d/sin(2
*d*x+2*c)^(1/2)+2*e*cos(d*x+c)*(e*tan(d*x+c))^(3/2)/a/d

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3973, 3969, 3557, 335, 303, 1176, 631, 210, 1179, 642, 2693, 2695, 2652, 2719} \[ \int \frac {(e \tan (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\frac {e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{5/2} \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}-\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}+\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {2 e^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d} \]

[In]

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

[Out]

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

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 2693

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*(m + n - 1))), x] + Dist[a^2*((m - 2)/(m + n - 1)), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && 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 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 3969

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

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 16.52 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.37 \[ \int \frac {(e \tan (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\frac {4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \csc (c+d x) \left (\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\tan ^2(c+d x)\right )-\operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-\tan ^2(c+d x)\right )\right ) \left (1+\sqrt {\sec ^2(c+d x)}\right ) (e \tan (c+d x))^{5/2}}{3 a d (1+\sec (c+d x))^2} \]

[In]

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

[Out]

(4*Cos[(c + d*x)/2]^2*Csc[c + d*x]*(Hypergeometric2F1[1/2, 3/4, 7/4, -Tan[c + d*x]^2] - Hypergeometric2F1[3/4,
1, 7/4, -Tan[c + d*x]^2])*(1 + Sqrt[Sec[c + d*x]^2])*(e*Tan[c + d*x])^(5/2))/(3*a*d*(1 + Sec[c + d*x])^2)

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.64 (sec) , antiderivative size = 1100, normalized size of antiderivative = 3.86

method	result	size
default	\(\text {Expression too large to display}\)	\(1100\)

[In]

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

[Out]

-1/2/a/d*2^(1/2)*(-e/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(-cot(d*x+c)+csc(d*x+c)))^(5/2)*((1-cos(d*x+c))^2*csc(d
*x+c)^2-1)^3*(I*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(2-2*csc(d*x+c)+2*cot(d*x+c))^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1
/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*((1-cos(d*x+c))*(-cot(d*x+c)+csc(d*x+c)-
1)*(csc(d*x+c)-cot(d*x+c)+1)*csc(d*x+c))^(1/2)-I*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(2-2*csc(d*x+c)+2*cot(d*x+c))
^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*((1-cos
(d*x+c))*(-cot(d*x+c)+csc(d*x+c)-1)*(csc(d*x+c)-cot(d*x+c)+1)*csc(d*x+c))^(1/2)-4*(csc(d*x+c)-cot(d*x+c)+1)^(1
/2)*(2-2*csc(d*x+c)+2*cot(d*x+c))^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*EllipticE((csc(d*x+c)-cot(d*x+c)+1)^(1/2
),1/2*2^(1/2))*((1-cos(d*x+c))*(-cot(d*x+c)+csc(d*x+c)-1)*(csc(d*x+c)-cot(d*x+c)+1)*csc(d*x+c))^(1/2)+2*(csc(d
*x+c)-cot(d*x+c)+1)^(1/2)*(2-2*csc(d*x+c)+2*cot(d*x+c))^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*EllipticF((csc(d*x
+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*((1-cos(d*x+c))*(-cot(d*x+c)+csc(d*x+c)-1)*(csc(d*x+c)-cot(d*x+c)+1)*csc(
d*x+c))^(1/2)-(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(2-2*csc(d*x+c)+2*cot(d*x+c))^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2
)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*((1-cos(d*x+c))*(-cot(d*x+c)+csc(d*x+c)-1)
*(csc(d*x+c)-cot(d*x+c)+1)*csc(d*x+c))^(1/2)-(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(2-2*csc(d*x+c)+2*cot(d*x+c))^(1/
2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*((1-cos(d*x
+c))*(-cot(d*x+c)+csc(d*x+c)-1)*(csc(d*x+c)-cot(d*x+c)+1)*csc(d*x+c))^(1/2)-4*(1-cos(d*x+c))^2*((1-cos(d*x+c))
^3*csc(d*x+c)^3+cot(d*x+c)-csc(d*x+c))^(1/2)*csc(d*x+c)^2)/(1-cos(d*x+c))^2*sin(d*x+c)^2/((1-cos(d*x+c))*((1-c
os(d*x+c))^2*csc(d*x+c)^2-1)*csc(d*x+c))^(1/2)/((1-cos(d*x+c))*(-cot(d*x+c)+csc(d*x+c)-1)*(csc(d*x+c)-cot(d*x+
c)+1)*csc(d*x+c))^(1/2)/((1-cos(d*x+c))^3*csc(d*x+c)^3+cot(d*x+c)-csc(d*x+c))^(1/2)

Fricas [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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