3.108 \(\int \frac{a+a \sec (c+d x)}{(e \tan (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=305 \[ \frac{a \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{3/2}}-\frac{a \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d e^{3/2}}-\frac{a \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{3/2}}+\frac{a \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{3/2}}+\frac{2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e^3}-\frac{2 a \cos (c+d x) E\left (\left .c+d x-\frac{\pi }{4}\right |2\right ) \sqrt{e \tan (c+d x)}}{d e^2 \sqrt{\sin (2 c+2 d x)}}-\frac{2 (a \sec (c+d x)+a)}{d e \sqrt{e \tan (c+d x)}} \]
[Out]
(a*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(3/2)) - (a*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan
[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(3/2)) - (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x
]]])/(2*Sqrt[2]*d*e^(3/2)) + (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]
*d*e^(3/2)) - (2*(a + a*Sec[c + d*x]))/(d*e*Sqrt[e*Tan[c + d*x]]) - (2*a*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x
, 2]*Sqrt[e*Tan[c + d*x]])/(d*e^2*Sqrt[Sin[2*c + 2*d*x]]) + (2*a*Cos[c + d*x]*(e*Tan[c + d*x])^(3/2))/(d*e^3)
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Rubi [A] time = 0.303329, antiderivative size = 305, normalized size of antiderivative = 1.,
number of steps used = 17, number of rules used = 14, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.609, Rules used
= {3882, 3884, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2613, 2615, 2572, 2639}
\[ \frac{a \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{3/2}}-\frac{a \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d e^{3/2}}-\frac{a \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{3/2}}+\frac{a \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{3/2}}+\frac{2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e^3}-\frac{2 a \cos (c+d x) E\left (\left .c+d x-\frac{\pi }{4}\right |2\right ) \sqrt{e \tan (c+d x)}}{d e^2 \sqrt{\sin (2 c+2 d x)}}-\frac{2 (a \sec (c+d x)+a)}{d e \sqrt{e \tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[c + d*x])/(e*Tan[c + d*x])^(3/2),x]
[Out]
(a*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(3/2)) - (a*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan
[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(3/2)) - (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x
]]])/(2*Sqrt[2]*d*e^(3/2)) + (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]
*d*e^(3/2)) - (2*(a + a*Sec[c + d*x]))/(d*e*Sqrt[e*Tan[c + d*x]]) - (2*a*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x
, 2]*Sqrt[e*Tan[c + d*x]])/(d*e^2*Sqrt[Sin[2*c + 2*d*x]]) + (2*a*Cos[c + d*x]*(e*Tan[c + d*x])^(3/2))/(d*e^3)
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{a+a \sec (c+d x)}{(e \tan (c+d x))^{3/2}} \, dx &=-\frac{2 (a+a \sec (c+d x))}{d e \sqrt{e \tan (c+d x)}}+\frac{2 \int \left (-\frac{a}{2}+\frac{1}{2} a \sec (c+d x)\right ) \sqrt{e \tan (c+d x)} \, dx}{e^2}\\ &=-\frac{2 (a+a \sec (c+d x))}{d e \sqrt{e \tan (c+d x)}}-\frac{a \int \sqrt{e \tan (c+d x)} \, dx}{e^2}+\frac{a \int \sec (c+d x) \sqrt{e \tan (c+d x)} \, dx}{e^2}\\ &=-\frac{2 (a+a \sec (c+d x))}{d e \sqrt{e \tan (c+d x)}}+\frac{2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e^3}-\frac{(2 a) \int \cos (c+d x) \sqrt{e \tan (c+d x)} \, dx}{e^2}-\frac{a \operatorname{Subst}\left (\int \frac{\sqrt{x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{d e}\\ &=-\frac{2 (a+a \sec (c+d x))}{d e \sqrt{e \tan (c+d x)}}+\frac{2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e^3}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d e}-\frac{\left (2 a \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \sqrt{\sin (c+d x)} \, dx}{e^2 \sqrt{\sin (c+d x)}}\\ &=-\frac{2 (a+a \sec (c+d x))}{d e \sqrt{e \tan (c+d x)}}+\frac{2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e^3}+\frac{a \operatorname{Subst}\left (\int \frac{e-x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d e}-\frac{a \operatorname{Subst}\left (\int \frac{e+x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d e}-\frac{\left (2 a \cos (c+d x) \sqrt{e \tan (c+d x)}\right ) \int \sqrt{\sin (2 c+2 d x)} \, dx}{e^2 \sqrt{\sin (2 c+2 d x)}}\\ &=-\frac{2 (a+a \sec (c+d x))}{d e \sqrt{e \tan (c+d x)}}-\frac{2 a \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{d e^2 \sqrt{\sin (2 c+2 d x)}}+\frac{2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e^3}-\frac{a \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} d e^{3/2}}-\frac{a \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} d e^{3/2}}-\frac{a \operatorname{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}-\frac{a \operatorname{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}\\ &=-\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^{3/2}}+\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^{3/2}}-\frac{2 (a+a \sec (c+d x))}{d e \sqrt{e \tan (c+d x)}}-\frac{2 a \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{d e^2 \sqrt{\sin (2 c+2 d x)}}+\frac{2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e^3}-\frac{a \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} d e^{3/2}}+\frac{a \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} d e^{3/2}}\\ &=\frac{a \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{3/2}}-\frac{a \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{3/2}}-\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^{3/2}}+\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^{3/2}}-\frac{2 (a+a \sec (c+d x))}{d e \sqrt{e \tan (c+d x)}}-\frac{2 a \cos (c+d x) E\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sqrt{e \tan (c+d x)}}{d e^2 \sqrt{\sin (2 c+2 d x)}}+\frac{2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e^3}\\ \end{align*}
Mathematica [C] time = 2.60705, size = 196, normalized size = 0.64 \[ -\frac{a (\cos (c+d x)+1) \csc (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{e \tan (c+d x)} \left (8 \tan ^2(c+d x) \text{Hypergeometric2F1}\left (\frac{3}{4},\frac{3}{2},\frac{7}{4},-\tan ^2(c+d x)\right )+3 \sqrt{\sec ^2(c+d x)} \left (4 \cos (c+d x)+2 \cos (2 (c+d x))-\sqrt{\sin (2 (c+d x))} \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-\sqrt{\sin (2 (c+d x))} \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )+2\right )\right )}{12 d e^2 \sqrt{\sec ^2(c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sec[c + d*x])/(e*Tan[c + d*x])^(3/2),x]
[Out]
-(a*(1 + Cos[c + d*x])*Csc[c + d*x]*Sec[(c + d*x)/2]^2*Sqrt[e*Tan[c + d*x]]*(3*Sqrt[Sec[c + d*x]^2]*(2 + 4*Cos
[c + d*x] + 2*Cos[2*(c + d*x)] - ArcSin[Cos[c + d*x] - Sin[c + d*x]]*Sqrt[Sin[2*(c + d*x)]] - Log[Cos[c + d*x]
+ Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]]) + 8*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[c
+ d*x]^2]*Tan[c + d*x]^2))/(12*d*e^2*Sqrt[Sec[c + d*x]^2])
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Maple [C] time = 0.247, size = 1414, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x)
[Out]
-1/2*a/d*2^(1/2)*(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)-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)+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))-(-(-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)-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))-(-(-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)-4*(-(-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))*cos(d*x+c)+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))/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)-(-(-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))-(-(-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)*Ellipti
cPi((-(-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*(-(-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))+2*(-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+co
s(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))+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
)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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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((a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \frac{1}{\left (e \tan{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx + \int \frac{\sec{\left (c + d x \right )}}{\left (e \tan{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))/(e*tan(d*x+c))**(3/2),x)
[Out]
a*(Integral((e*tan(c + d*x))**(-3/2), x) + Integral(sec(c + d*x)/(e*tan(c + d*x))**(3/2), x))
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a \sec \left (d x + c\right ) + a}{\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((a+a*sec(d*x+c))/(e*tan(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate((a*sec(d*x + c) + a)/(e*tan(d*x + c))^(3/2), x)