3.125 \(\int \frac{1}{(a+a \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=359 \[ \frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d e^{3/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} a d e^{3/2}}-\frac{\log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d e^{3/2}}+\frac{\log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d e^{3/2}}-\frac{6 \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 a d e^3}+\frac{6 \cos (c+d x) E\left (\left .c+d x-\frac{\pi }{4}\right |2\right ) \sqrt{e \tan (c+d x)}}{5 a d e^2 \sqrt{\sin (2 c+2 d x)}}-\frac{2 (5-3 \sec (c+d x))}{5 a d e \sqrt{e \tan (c+d x)}}+\frac{2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}} \]

[Out]

ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]]/(Sqrt[2]*a*d*e^(3/2)) - ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c +
d*x]])/Sqrt[e]]/(Sqrt[2]*a*d*e^(3/2)) - Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x]]]/(2*
Sqrt[2]*a*d*e^(3/2)) + Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]]/(2*Sqrt[2]*a*d*e^(3/
2)) + (2*e*(1 - Sec[c + d*x]))/(5*a*d*(e*Tan[c + d*x])^(5/2)) - (2*(5 - 3*Sec[c + d*x]))/(5*a*d*e*Sqrt[e*Tan[c
+ d*x]]) + (6*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[e*Tan[c + d*x]])/(5*a*d*e^2*Sqrt[Sin[2*c + 2*d*x
]]) - (6*Cos[c + d*x]*(e*Tan[c + d*x])^(3/2))/(5*a*d*e^3)

________________________________________________________________________________________

Rubi [A]  time = 0.452068, antiderivative size = 359, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 15, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3888, 3882, 3884, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2613, 2615, 2572, 2639} \[ \frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d e^{3/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} a d e^{3/2}}-\frac{\log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d e^{3/2}}+\frac{\log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d e^{3/2}}-\frac{6 \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 a d e^3}+\frac{6 \cos (c+d x) E\left (\left .c+d x-\frac{\pi }{4}\right |2\right ) \sqrt{e \tan (c+d x)}}{5 a d e^2 \sqrt{\sin (2 c+2 d x)}}-\frac{2 (5-3 \sec (c+d x))}{5 a d e \sqrt{e \tan (c+d x)}}+\frac{2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[c + d*x])*(e*Tan[c + d*x])^(3/2)),x]

[Out]

ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]]/(Sqrt[2]*a*d*e^(3/2)) - ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c +
d*x]])/Sqrt[e]]/(Sqrt[2]*a*d*e^(3/2)) - Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x]]]/(2*
Sqrt[2]*a*d*e^(3/2)) + Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]]/(2*Sqrt[2]*a*d*e^(3/
2)) + (2*e*(1 - Sec[c + d*x]))/(5*a*d*(e*Tan[c + d*x])^(5/2)) - (2*(5 - 3*Sec[c + d*x]))/(5*a*d*e*Sqrt[e*Tan[c
+ d*x]]) + (6*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[e*Tan[c + d*x]])/(5*a*d*e^2*Sqrt[Sin[2*c + 2*d*x
]]) - (6*Cos[c + d*x]*(e*Tan[c + d*x])^(3/2))/(5*a*d*e^3)

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]

Rule 3882

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 3884

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 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 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 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 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 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 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 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 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 2613

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sec (c+d x)) (e \tan (c+d x))^{3/2}} \, dx &=\frac{e^2 \int \frac{-a+a \sec (c+d x)}{(e \tan (c+d x))^{7/2}} \, dx}{a^2}\\ &=\frac{2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}}+\frac{2 \int \frac{\frac{5 a}{2}-\frac{3}{2} a \sec (c+d x)}{(e \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}}-\frac{2 (5-3 \sec (c+d x))}{5 a d e \sqrt{e \tan (c+d x)}}+\frac{4 \int \left (-\frac{5 a}{4}-\frac{3}{4} a \sec (c+d x)\right ) \sqrt{e \tan (c+d x)} \, dx}{5 a^2 e^2}\\ &=\frac{2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}}-\frac{2 (5-3 \sec (c+d x))}{5 a d e \sqrt{e \tan (c+d x)}}-\frac{3 \int \sec (c+d x) \sqrt{e \tan (c+d x)} \, dx}{5 a e^2}-\frac{\int \sqrt{e \tan (c+d x)} \, dx}{a e^2}\\ &=\frac{2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}}-\frac{2 (5-3 \sec (c+d x))}{5 a d e \sqrt{e \tan (c+d x)}}-\frac{6 \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 a d e^3}+\frac{6 \int \cos (c+d x) \sqrt{e \tan (c+d x)} \, dx}{5 a e^2}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a d e}\\ &=\frac{2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}}-\frac{2 (5-3 \sec (c+d x))}{5 a d e \sqrt{e \tan (c+d x)}}-\frac{6 \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 a d e^3}-\frac{2 \operatorname{Subst}\left (\int \frac{x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a d e}+\frac{\left (6 \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \sqrt{\sin (c+d x)} \, dx}{5 a e^2 \sqrt{\sin (c+d x)}}\\ &=\frac{2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}}-\frac{2 (5-3 \sec (c+d x))}{5 a d e \sqrt{e \tan (c+d x)}}-\frac{6 \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 a d e^3}+\frac{\operatorname{Subst}\left (\int \frac{e-x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a d e}-\frac{\operatorname{Subst}\left (\int \frac{e+x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a d e}+\frac{\left (6 \cos (c+d x) \sqrt{e \tan (c+d x)}\right ) \int \sqrt{\sin (2 c+2 d x)} \, dx}{5 a e^2 \sqrt{\sin (2 c+2 d x)}}\\ &=\frac{2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}}-\frac{2 (5-3 \sec (c+d x))}{5 a d e \sqrt{e \tan (c+d x)}}+\frac{6 \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{5 a d e^2 \sqrt{\sin (2 c+2 d x)}}-\frac{6 \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 a d e^3}-\frac{\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 d e^{3/2}}-\frac{\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 d e^{3/2}}-\frac{\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 d e}-\frac{\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 d e}\\ &=-\frac{\log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a d e^{3/2}}+\frac{\log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a d e^{3/2}}+\frac{2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}}-\frac{2 (5-3 \sec (c+d x))}{5 a d e \sqrt{e \tan (c+d x)}}+\frac{6 \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{5 a d e^2 \sqrt{\sin (2 c+2 d x)}}-\frac{6 \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 a d e^3}-\frac{\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 d e^{3/2}}+\frac{\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 d e^{3/2}}\\ &=\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d e^{3/2}}-\frac{\tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d e^{3/2}}-\frac{\log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a d e^{3/2}}+\frac{\log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a d e^{3/2}}+\frac{2 e (1-\sec (c+d x))}{5 a d (e \tan (c+d x))^{5/2}}-\frac{2 (5-3 \sec (c+d x))}{5 a d e \sqrt{e \tan (c+d x)}}+\frac{6 \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{5 a d e^2 \sqrt{\sin (2 c+2 d x)}}-\frac{6 \cos (c+d x) (e \tan (c+d x))^{3/2}}{5 a d e^3}\\ \end{align*}

Mathematica [C]  time = 12.8607, size = 180, normalized size = 0.5 \[ -\frac{4 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \csc (c+d x) \left (\sqrt{\sec ^2(c+d x)}+1\right ) \sqrt{e \tan (c+d x)} \left (-5 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-\tan ^2(c+d x)\right )+5 \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},-\tan ^2(c+d x)\right )+3 \cot ^4(c+d x) \text{Hypergeometric2F1}\left (-\frac{5}{4},-\frac{1}{2},-\frac{1}{4},-\tan ^2(c+d x)\right )-15 \cot ^2(c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{2},-\frac{1}{4},\frac{3}{4},-\tan ^2(c+d x)\right )-3 \cot ^4(c+d x)+15 \cot ^2(c+d x)\right )}{15 a d e^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + a*Sec[c + d*x])*(e*Tan[c + d*x])^(3/2)),x]

[Out]

(-4*Csc[c + d*x]*(15*Cot[c + d*x]^2 - 3*Cot[c + d*x]^4 + 3*Cot[c + d*x]^4*Hypergeometric2F1[-5/4, -1/2, -1/4,
-Tan[c + d*x]^2] - 15*Cot[c + d*x]^2*Hypergeometric2F1[-1/2, -1/4, 3/4, -Tan[c + d*x]^2] - 5*Hypergeometric2F1
[1/2, 3/4, 7/4, -Tan[c + d*x]^2] + 5*Hypergeometric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2])*(1 + Sqrt[Sec[c + d*x]^2
])*Sin[(c + d*x)/2]^2*Sqrt[e*Tan[c + d*x]])/(15*a*d*e^2)

________________________________________________________________________________________

Maple [C]  time = 0.245, size = 2113, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x)

[Out]

-1/10/a/d*2^(1/2)*(-1+cos(d*x+c))*(5*I*((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))/sin(d*x+c))^(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))*cos(d*x+c)^2+10*I*((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))/sin(d*x+c))^(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))*cos(d*x+c)+5*((-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)*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))*cos(d*x+c)^2-10*I*((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))/sin(d*x+c))^(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))*cos(d*x+c)+5*((-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)*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))*cos(d*x+c)^2-5*I*((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))/sin(d*x+c))^(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))*cos(d*x+c)^2-12*((-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)*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(
1/2),1/2*2^(1/2))*cos(d*x+c)^2+6*((-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)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^
(1/2))*cos(d*x+c)^2+10*cos(d*x+c)*((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))/sin(d*x+c))^(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))-5*I*((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))/sin(d*x+c))^(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))+10*cos(d*x+c)*((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))/sin(d*x+c))^(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))+5*I*((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))/sin(d*x+c))^(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))-24*c
os(d*x+c)*((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))/sin(d*x+c))^(1/2)*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+12*cos(d*x+c)*((
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))/sin(
d*x+c))^(1/2)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+5*((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))/sin(d*x+c))^(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))+5*((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))/sin(d*x+c))^(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))-12*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+c
os(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticE(((1-cos(d*x+c)+sin(d*x+c)
)/sin(d*x+c))^(1/2),1/2*2^(1/2))+6*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/si
n(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*
2^(1/2))-6*cos(d*x+c)^2*2^(1/2)-4*cos(d*x+c)*2^(1/2))/sin(d*x+c)/cos(d*x+c)^2/(e*sin(d*x+c)/cos(d*x+c))^(3/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((a*sec(d*x + c) + a)*(e*tan(d*x + c))^(3/2)), x)

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\left (e \tan{\left (c + d x \right )}\right )^{\frac{3}{2}} \sec{\left (c + d x \right )} + \left (e \tan{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))/(e*tan(d*x+c))**(3/2),x)

[Out]

Integral(1/((e*tan(c + d*x))**(3/2)*sec(c + d*x) + (e*tan(c + d*x))**(3/2)), x)/a

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)*(e*tan(d*x + c))^(3/2)), x)