\(\int \frac {a+a \sec (c+d x)}{(e \tan (c+d x))^{5/2}} \, dx\) [109]

Optimal result

Integrand size = 23, antiderivative size = 282 \[ \int \frac {a+a \sec (c+d x)}{(e \tan (c+d x))^{5/2}} \, dx=\frac {a \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}-\frac {a \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}+\frac {a \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {a \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {2 (a+a \sec (c+d x))}{3 d e (e \tan (c+d x))^{3/2}}-\frac {a \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 d e^2 \sqrt {e \tan (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3967, 3969, 3557, 335, 217, 1179, 642, 1176, 631, 210, 2694, 2653, 2720} \[ \int \frac {a+a \sec (c+d x)}{(e \tan (c+d x))^{5/2}} \, dx=\frac {a \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}-\frac {a \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{5/2}}+\frac {a \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {a \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{3 d e^2 \sqrt {e \tan (c+d x)}}-\frac {2 (a \sec (c+d x)+a)}{3 d e (e \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

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

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 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{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 2653

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2694

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 3967

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-(e*Cot[c
+ d*x])^(m + 1))*((a + b*Csc[c + d*x])/(d*e*(m + 1))), x] - Dist[1/(e^2*(m + 1)), Int[(e*Cot[c + d*x])^(m + 2)
*(a*(m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[m, -1]

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]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.64 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.71 \[ \int \frac {a+a \sec (c+d x)}{(e \tan (c+d x))^{5/2}} \, dx=-\frac {a \csc (c+d x) \left (\sqrt {\sec ^2(c+d x)} \left (2 \cot \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {3}{2} (c+d x)\right ) \csc \left (\frac {1}{2} (c+d x)\right )-3 \arcsin (\cos (c+d x)-\sin (c+d x)) \sqrt {\sin (2 (c+d x))}+3 \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right ) \sqrt {\sin (2 (c+d x))}\right )-4 \sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ),-1\right ) \sqrt {\tan (c+d x)}\right ) \sqrt {e \tan (c+d x)}}{6 d e^3 \sqrt {\sec ^2(c+d x)}} \]

[In]

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

[Out]

-1/6*(a*Csc[c + d*x]*(Sqrt[Sec[c + d*x]^2]*(2*Cot[(c + d*x)/2] + 2*Cos[(3*(c + d*x))/2]*Csc[(c + d*x)/2] - 3*A
rcSin[Cos[c + d*x] - Sin[c + d*x]]*Sqrt[Sin[2*(c + d*x)]] + 3*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c
+ d*x)]]]*Sqrt[Sin[2*(c + d*x)]]) - 4*(-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]], -1]*Sqrt[
Tan[c + d*x]])*Sqrt[e*Tan[c + d*x]])/(d*e^3*Sqrt[Sec[c + d*x]^2])

Maple [A] (verified)

Time = 4.78 (sec) , antiderivative size = 458, normalized size of antiderivative = 1.62

method	result	size
parts	\(\frac {2 a e \left (-\frac {1}{3 e^{2} \left (e \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \tan \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \tan \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e^{4}}\right )}{d}+\frac {a \sqrt {2}\, \left (1-\cos \left (d x +c \right )\right )^{2} \left (2 \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {2-2 \csc \left (d x +c \right )+2 \cot \left (d x +c \right )}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-\left (1-\cos \left (d x +c \right )\right )^{4} \csc \left (d x +c \right )^{4}+1\right ) \csc \left (d x +c \right )^{2}}{6 d \sqrt {\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sqrt {\left (1-\cos \left (d x +c \right )\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right ) \csc \left (d x +c \right )}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2} \left (-\frac {e \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )^{\frac {5}{2}}}\)	\(458\)
default	\(\frac {a \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \left (3 i \sin \left (d x +c \right ) \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 i \sin \left (d x +c \right ) \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 \sin \left (d x +c \right ) \operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}-3 \sin \left (d x +c \right ) \operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}+4 \sin \left (d x +c \right ) \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}-2 \sqrt {2}\, \cos \left (d x +c \right )\right ) \sec \left (d x +c \right ) \csc \left (d x +c \right )}{6 e^{2} d \sqrt {e \tan \left (d x +c \right )}}\)	\(494\)

[In]

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

[Out]

2*a/d*e*(-1/3/e^2/(e*tan(d*x+c))^(3/2)-1/8/e^4*(e^2)^(1/4)*2^(1/2)*(ln((e*tan(d*x+c)+(e^2)^(1/4)*(e*tan(d*x+c)
)^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*tan(d*x+c)-(e^2)^(1/4)*(e*tan(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+2*arctan(2^(
1/2)/(e^2)^(1/4)*(e*tan(d*x+c))^(1/2)+1)-2*arctan(-2^(1/2)/(e^2)^(1/4)*(e*tan(d*x+c))^(1/2)+1)))+1/6*a/d*2^(1/
2)*(1-cos(d*x+c))^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))*(-cot(d*x+c)+csc(d*x+c))-(1-cos(d*x+c))^4*cs
c(d*x+c)^4+1)/((1-cos(d*x+c))^3*csc(d*x+c)^3+cot(d*x+c)-csc(d*x+c))^(1/2)/((1-cos(d*x+c))*((1-cos(d*x+c))^2*cs
c(d*x+c)^2-1)*csc(d*x+c))^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^2/(-e/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(-co
t(d*x+c)+csc(d*x+c)))^(5/2)*csc(d*x+c)^2

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+a \sec (c+d x)}{(e \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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