### 3.628 $$\int \frac{1}{\sqrt{d+e x} (a-c x^2)^2} \, dx$$

Optimal. Leaf size=222 $-\frac{\left (2 \sqrt{c} d-3 \sqrt{a} e\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} \sqrt [4]{c} \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2}}+\frac{\left (3 \sqrt{a} e+2 \sqrt{c} d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} \sqrt [4]{c} \left (\sqrt{a} e+\sqrt{c} d\right )^{3/2}}-\frac{\sqrt{d+e x} (a e-c d x)}{2 a \left (a-c x^2\right ) \left (c d^2-a e^2\right )}$

[Out]

-((a*e - c*d*x)*Sqrt[d + e*x])/(2*a*(c*d^2 - a*e^2)*(a - c*x^2)) - ((2*Sqrt[c]*d - 3*Sqrt[a]*e)*ArcTanh[(c^(1/
4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(4*a^(3/2)*c^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)^(3/2)) + ((2*Sqrt[c
]*d + 3*Sqrt[a]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(1/4)*(Sqrt[c]*d
+ Sqrt[a]*e)^(3/2))

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Rubi [A]  time = 0.338558, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {741, 827, 1166, 208} $-\frac{\left (2 \sqrt{c} d-3 \sqrt{a} e\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} \sqrt [4]{c} \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2}}+\frac{\left (3 \sqrt{a} e+2 \sqrt{c} d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} \sqrt [4]{c} \left (\sqrt{a} e+\sqrt{c} d\right )^{3/2}}-\frac{\sqrt{d+e x} (a e-c d x)}{2 a \left (a-c x^2\right ) \left (c d^2-a e^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(Sqrt[d + e*x]*(a - c*x^2)^2),x]

[Out]

-((a*e - c*d*x)*Sqrt[d + e*x])/(2*a*(c*d^2 - a*e^2)*(a - c*x^2)) - ((2*Sqrt[c]*d - 3*Sqrt[a]*e)*ArcTanh[(c^(1/
4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(4*a^(3/2)*c^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)^(3/2)) + ((2*Sqrt[c
]*d + 3*Sqrt[a]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(1/4)*(Sqrt[c]*d
+ Sqrt[a]*e)^(3/2))

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{d+e x} \left (a-c x^2\right )^2} \, dx &=-\frac{(a e-c d x) \sqrt{d+e x}}{2 a \left (c d^2-a e^2\right ) \left (a-c x^2\right )}+\frac{\int \frac{\frac{1}{2} \left (2 c d^2-3 a e^2\right )+\frac{1}{2} c d e x}{\sqrt{d+e x} \left (a-c x^2\right )} \, dx}{2 a \left (c d^2-a e^2\right )}\\ &=-\frac{(a e-c d x) \sqrt{d+e x}}{2 a \left (c d^2-a e^2\right ) \left (a-c x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2} c d^2 e+\frac{1}{2} e \left (2 c d^2-3 a e^2\right )+\frac{1}{2} c d e x^2}{-c d^2+a e^2+2 c d x^2-c x^4} \, dx,x,\sqrt{d+e x}\right )}{a \left (c d^2-a e^2\right )}\\ &=-\frac{(a e-c d x) \sqrt{d+e x}}{2 a \left (c d^2-a e^2\right ) \left (a-c x^2\right )}-\frac{\left (\sqrt{c} \left (2 \sqrt{c} d-3 \sqrt{a} e\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c d-\sqrt{a} \sqrt{c} e-c x^2} \, dx,x,\sqrt{d+e x}\right )}{4 a^{3/2} \left (\sqrt{c} d-\sqrt{a} e\right )}+\frac{\left (\sqrt{c} \left (2 \sqrt{c} d+3 \sqrt{a} e\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c d+\sqrt{a} \sqrt{c} e-c x^2} \, dx,x,\sqrt{d+e x}\right )}{4 a^{3/2} \left (\sqrt{c} d+\sqrt{a} e\right )}\\ &=-\frac{(a e-c d x) \sqrt{d+e x}}{2 a \left (c d^2-a e^2\right ) \left (a-c x^2\right )}-\frac{\left (2 \sqrt{c} d-3 \sqrt{a} e\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} \sqrt [4]{c} \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2}}+\frac{\left (2 \sqrt{c} d+3 \sqrt{a} e\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d+\sqrt{a} e}}\right )}{4 a^{3/2} \sqrt [4]{c} \left (\sqrt{c} d+\sqrt{a} e\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.491688, size = 248, normalized size = 1.12 $\frac{\frac{\frac{\left (c x^2-a\right ) \left (\sqrt{a} \sqrt{c} d e-3 a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{\sqrt{a} \sqrt [4]{c} \sqrt{\sqrt{a} e+\sqrt{c} d}}+2 \sqrt{d+e x} (a e-c d x)}{a-c x^2}+\frac{\left (-\sqrt{a} \sqrt{c} d e-3 a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{\sqrt{a} \sqrt [4]{c} \sqrt{\sqrt{c} d-\sqrt{a} e}}}{4 a \left (a e^2-c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/(Sqrt[d + e*x]*(a - c*x^2)^2),x]

[Out]

(((2*c*d^2 - Sqrt[a]*Sqrt[c]*d*e - 3*a*e^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(Sqr
t[a]*c^(1/4)*Sqrt[Sqrt[c]*d - Sqrt[a]*e]) + (2*(a*e - c*d*x)*Sqrt[d + e*x] + ((2*c*d^2 + Sqrt[a]*Sqrt[c]*d*e -
3*a*e^2)*(-a + c*x^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(Sqrt[a]*c^(1/4)*Sqrt[Sqr
t[c]*d + Sqrt[a]*e]))/(a - c*x^2))/(4*a*(-(c*d^2) + a*e^2))

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Maple [B]  time = 0.227, size = 375, normalized size = 1.7 \begin{align*} -{\frac{e}{4\,a}\sqrt{ex+d} \left ( cd-\sqrt{ac{e}^{2}} \right ) ^{-1} \left ( ex+{\frac{1}{c}\sqrt{ac{e}^{2}}} \right ) ^{-1}}-{\frac{{c}^{2}ed}{2\,a}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ac{e}^{2}}}} \left ( -cd+\sqrt{ac{e}^{2}} \right ) ^{-1}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{\frac{3\,ce}{4\,a}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ) \left ( -cd+\sqrt{ac{e}^{2}} \right ) ^{-1}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}}-{\frac{e}{4\,a}\sqrt{ex+d} \left ( cd+\sqrt{ac{e}^{2}} \right ) ^{-1} \left ( ex-{\frac{1}{c}\sqrt{ac{e}^{2}}} \right ) ^{-1}}+{\frac{{c}^{2}ed}{2\,a}{\it Artanh} \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ac{e}^{2}}}} \left ( cd+\sqrt{ac{e}^{2}} \right ) ^{-1}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{\frac{3\,ce}{4\,a}{\it Artanh} \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ) \left ( cd+\sqrt{ac{e}^{2}} \right ) ^{-1}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-c*x^2+a)^2/(e*x+d)^(1/2),x)

[Out]

-1/4*e/a/(c*d-(a*c*e^2)^(1/2))*(e*x+d)^(1/2)/(e*x+(a*c*e^2)^(1/2)/c)-1/2*e*c^2/a/(a*c*e^2)^(1/2)/(-c*d+(a*c*e^
2)^(1/2))/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*d+3/4*e*c/
a/(-c*d+(a*c*e^2)^(1/2))/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1
/2))-1/4*e/a/(c*d+(a*c*e^2)^(1/2))*(e*x+d)^(1/2)/(e*x-(a*c*e^2)^(1/2)/c)+1/2*e*c^2/a/(a*c*e^2)^(1/2)/(c*d+(a*c
*e^2)^(1/2))/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*d+3/4*e*
c/a/(c*d+(a*c*e^2)^(1/2))/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1
/2))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c*x^2 - a)^2*sqrt(e*x + d)), x)

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Fricas [B]  time = 3.06926, size = 6498, normalized size = 29.27 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a^2*c*d^2 - a^3*e^2 - (a*c^2*d^2 - a^2*c*e^2)*x^2)*sqrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 + (a^
3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6)*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10
)/(a^3*c^7*d^12 - 6*a^4*c^6*d^10*e^2 + 15*a^5*c^5*d^8*e^4 - 20*a^6*c^4*d^6*e^6 + 15*a^7*c^3*d^4*e^8 - 6*a^8*c^
2*d^2*e^10 + a^9*c*e^12)))/(a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6))*log((20*c^2*d^4*e^3
- 81*a*c*d^2*e^5 + 81*a^2*e^7)*sqrt(e*x + d) + (5*a^2*c^2*d^4*e^4 - 24*a^3*c*d^2*e^6 + 27*a^4*e^8 + 2*(a^3*c^5
*d^9 - 5*a^4*c^4*d^7*e^2 + 9*a^5*c^3*d^5*e^4 - 7*a^6*c^2*d^3*e^6 + 2*a^7*c*d*e^8)*sqrt((25*c^2*d^4*e^6 - 90*a*
c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7*d^12 - 6*a^4*c^6*d^10*e^2 + 15*a^5*c^5*d^8*e^4 - 20*a^6*c^4*d^6*e^6 + 15*a^7
*c^3*d^4*e^8 - 6*a^8*c^2*d^2*e^10 + a^9*c*e^12)))*sqrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 + (a^3*c^3*d
^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6)*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*
c^7*d^12 - 6*a^4*c^6*d^10*e^2 + 15*a^5*c^5*d^8*e^4 - 20*a^6*c^4*d^6*e^6 + 15*a^7*c^3*d^4*e^8 - 6*a^8*c^2*d^2*e
^10 + a^9*c*e^12)))/(a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6))) - (a^2*c*d^2 - a^3*e^2 - (
a*c^2*d^2 - a^2*c*e^2)*x^2)*sqrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 + (a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2
+ 3*a^5*c*d^2*e^4 - a^6*e^6)*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7*d^12 - 6*a^4*c^6*d
^10*e^2 + 15*a^5*c^5*d^8*e^4 - 20*a^6*c^4*d^6*e^6 + 15*a^7*c^3*d^4*e^8 - 6*a^8*c^2*d^2*e^10 + a^9*c*e^12)))/(a
^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6))*log((20*c^2*d^4*e^3 - 81*a*c*d^2*e^5 + 81*a^2*e^7
)*sqrt(e*x + d) - (5*a^2*c^2*d^4*e^4 - 24*a^3*c*d^2*e^6 + 27*a^4*e^8 + 2*(a^3*c^5*d^9 - 5*a^4*c^4*d^7*e^2 + 9*
a^5*c^3*d^5*e^4 - 7*a^6*c^2*d^3*e^6 + 2*a^7*c*d*e^8)*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3
*c^7*d^12 - 6*a^4*c^6*d^10*e^2 + 15*a^5*c^5*d^8*e^4 - 20*a^6*c^4*d^6*e^6 + 15*a^7*c^3*d^4*e^8 - 6*a^8*c^2*d^2*
e^10 + a^9*c*e^12)))*sqrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 + (a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^
5*c*d^2*e^4 - a^6*e^6)*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7*d^12 - 6*a^4*c^6*d^10*e^2
+ 15*a^5*c^5*d^8*e^4 - 20*a^6*c^4*d^6*e^6 + 15*a^7*c^3*d^4*e^8 - 6*a^8*c^2*d^2*e^10 + a^9*c*e^12)))/(a^3*c^3*
d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6))) + (a^2*c*d^2 - a^3*e^2 - (a*c^2*d^2 - a^2*c*e^2)*x^2)*s
qrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 - (a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6)
*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7*d^12 - 6*a^4*c^6*d^10*e^2 + 15*a^5*c^5*d^8*e^4
- 20*a^6*c^4*d^6*e^6 + 15*a^7*c^3*d^4*e^8 - 6*a^8*c^2*d^2*e^10 + a^9*c*e^12)))/(a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^
2 + 3*a^5*c*d^2*e^4 - a^6*e^6))*log((20*c^2*d^4*e^3 - 81*a*c*d^2*e^5 + 81*a^2*e^7)*sqrt(e*x + d) + (5*a^2*c^2*
d^4*e^4 - 24*a^3*c*d^2*e^6 + 27*a^4*e^8 - 2*(a^3*c^5*d^9 - 5*a^4*c^4*d^7*e^2 + 9*a^5*c^3*d^5*e^4 - 7*a^6*c^2*d
^3*e^6 + 2*a^7*c*d*e^8)*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7*d^12 - 6*a^4*c^6*d^10*e^
2 + 15*a^5*c^5*d^8*e^4 - 20*a^6*c^4*d^6*e^6 + 15*a^7*c^3*d^4*e^8 - 6*a^8*c^2*d^2*e^10 + a^9*c*e^12)))*sqrt((4*
c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 - (a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6)*sqrt((
25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7*d^12 - 6*a^4*c^6*d^10*e^2 + 15*a^5*c^5*d^8*e^4 - 20*a^
6*c^4*d^6*e^6 + 15*a^7*c^3*d^4*e^8 - 6*a^8*c^2*d^2*e^10 + a^9*c*e^12)))/(a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a
^5*c*d^2*e^4 - a^6*e^6))) - (a^2*c*d^2 - a^3*e^2 - (a*c^2*d^2 - a^2*c*e^2)*x^2)*sqrt((4*c^2*d^5 - 15*a*c*d^3*e
^2 + 15*a^2*d*e^4 - (a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6)*sqrt((25*c^2*d^4*e^6 - 90*a*
c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7*d^12 - 6*a^4*c^6*d^10*e^2 + 15*a^5*c^5*d^8*e^4 - 20*a^6*c^4*d^6*e^6 + 15*a^7
*c^3*d^4*e^8 - 6*a^8*c^2*d^2*e^10 + a^9*c*e^12)))/(a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6
))*log((20*c^2*d^4*e^3 - 81*a*c*d^2*e^5 + 81*a^2*e^7)*sqrt(e*x + d) - (5*a^2*c^2*d^4*e^4 - 24*a^3*c*d^2*e^6 +
27*a^4*e^8 - 2*(a^3*c^5*d^9 - 5*a^4*c^4*d^7*e^2 + 9*a^5*c^3*d^5*e^4 - 7*a^6*c^2*d^3*e^6 + 2*a^7*c*d*e^8)*sqrt(
(25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7*d^12 - 6*a^4*c^6*d^10*e^2 + 15*a^5*c^5*d^8*e^4 - 20*a
^6*c^4*d^6*e^6 + 15*a^7*c^3*d^4*e^8 - 6*a^8*c^2*d^2*e^10 + a^9*c*e^12)))*sqrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15
*a^2*d*e^4 - (a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6)*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e
^8 + 81*a^2*e^10)/(a^3*c^7*d^12 - 6*a^4*c^6*d^10*e^2 + 15*a^5*c^5*d^8*e^4 - 20*a^6*c^4*d^6*e^6 + 15*a^7*c^3*d^
4*e^8 - 6*a^8*c^2*d^2*e^10 + a^9*c*e^12)))/(a^3*c^3*d^6 - 3*a^4*c^2*d^4*e^2 + 3*a^5*c*d^2*e^4 - a^6*e^6))) + 4
*(c*d*x - a*e)*sqrt(e*x + d))/(a^2*c*d^2 - a^3*e^2 - (a*c^2*d^2 - a^2*c*e^2)*x^2)

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Sympy [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/(-c*x**2+a)**2/(e*x+d)**(1/2),x)

[Out]

Timed out

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Giac [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/(-c*x^2+a)^2/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

Timed out