3.1455 \(\int \frac{(A+B x) \sqrt{d+e x}}{(a-c x^2)^2} \, dx\)
Optimal. Leaf size=225 \[ -\frac{\left (-\sqrt{a} A \sqrt{c} e-a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} c^{5/4} \sqrt{\sqrt{c} d-\sqrt{a} e}}+\frac{\left (\sqrt{a} A \sqrt{c} e-a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} c^{5/4} \sqrt{\sqrt{a} e+\sqrt{c} d}}+\frac{\sqrt{d+e x} (a B+A c x)}{2 a c \left (a-c x^2\right )} \]
[Out]
((a*B + A*c*x)*Sqrt[d + e*x])/(2*a*c*(a - c*x^2)) - ((2*A*c*d - a*B*e - Sqrt[a]*A*Sqrt[c]*e)*ArcTanh[(c^(1/4)*
Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(4*a^(3/2)*c^(5/4)*Sqrt[Sqrt[c]*d - Sqrt[a]*e]) + ((2*A*c*d - a*B
*e + Sqrt[a]*A*Sqrt[c]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(5/4)*Sqr
t[Sqrt[c]*d + Sqrt[a]*e])
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Rubi [A] time = 0.397399, antiderivative size = 225, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used =
{821, 827, 1166, 208} \[ -\frac{\left (-\sqrt{a} A \sqrt{c} e-a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} c^{5/4} \sqrt{\sqrt{c} d-\sqrt{a} e}}+\frac{\left (\sqrt{a} A \sqrt{c} e-a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} c^{5/4} \sqrt{\sqrt{a} e+\sqrt{c} d}}+\frac{\sqrt{d+e x} (a B+A c x)}{2 a c \left (a-c x^2\right )} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*x)*Sqrt[d + e*x])/(a - c*x^2)^2,x]
[Out]
((a*B + A*c*x)*Sqrt[d + e*x])/(2*a*c*(a - c*x^2)) - ((2*A*c*d - a*B*e - Sqrt[a]*A*Sqrt[c]*e)*ArcTanh[(c^(1/4)*
Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(4*a^(3/2)*c^(5/4)*Sqrt[Sqrt[c]*d - Sqrt[a]*e]) + ((2*A*c*d - a*B
*e + Sqrt[a]*A*Sqrt[c]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(5/4)*Sqr
t[Sqrt[c]*d + Sqrt[a]*e])
Rule 821
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*
x^2)^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
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{(A+B x) \sqrt{d+e x}}{\left (a-c x^2\right )^2} \, dx &=\frac{(a B+A c x) \sqrt{d+e x}}{2 a c \left (a-c x^2\right )}-\frac{\int \frac{\frac{1}{2} (-2 A c d+a B e)-\frac{1}{2} A c e x}{\sqrt{d+e x} \left (a-c x^2\right )} \, dx}{2 a c}\\ &=\frac{(a B+A c x) \sqrt{d+e x}}{2 a c \left (a-c x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} A c d e+\frac{1}{2} e (-2 A c d+a B e)-\frac{1}{2} A c 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 c}\\ &=\frac{(a B+A c x) \sqrt{d+e x}}{2 a c \left (a-c x^2\right )}-\frac{\left (2 A c d-a B e-\sqrt{a} A \sqrt{c} e\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} \sqrt{c}}+\frac{\left (2 A c d-a B e+\sqrt{a} A \sqrt{c} e\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} \sqrt{c}}\\ &=\frac{(a B+A c x) \sqrt{d+e x}}{2 a c \left (a-c x^2\right )}-\frac{\left (2 A c d-a B e-\sqrt{a} A \sqrt{c} 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} c^{5/4} \sqrt{\sqrt{c} d-\sqrt{a} e}}+\frac{\left (2 A c d-a B e+\sqrt{a} A \sqrt{c} 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} c^{5/4} \sqrt{\sqrt{c} d+\sqrt{a} e}}\\ \end{align*}
Mathematica [A] time = 0.518935, size = 375, normalized size = 1.67 \[ \frac{-\frac{\sqrt [4]{c} \left (a A e^2+2 a B d e-3 A c d^2\right ) \left (\sqrt{\sqrt{c} d-\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-\sqrt{\sqrt{a} e+\sqrt{c} d} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )\right )}{2 \sqrt{a}}+\frac{c (d+e x)^{3/2} (-a A e+a B (d-e x)+A c d x)}{c x^2-a}+\frac{(a B e-A c d) \left (2 \sqrt{a} \sqrt [4]{c} e \sqrt{d+e x}+\left (\sqrt{c} d-\sqrt{a} e\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-\left (\sqrt{a} e+\sqrt{c} d\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )\right )}{2 \sqrt{a} \sqrt [4]{c}}}{2 a c \left (a e^2-c d^2\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*x)*Sqrt[d + e*x])/(a - c*x^2)^2,x]
[Out]
((c*(d + e*x)^(3/2)*(-(a*A*e) + A*c*d*x + a*B*(d - e*x)))/(-a + c*x^2) - (c^(1/4)*(-3*A*c*d^2 + 2*a*B*d*e + a*
A*e^2)*(Sqrt[Sqrt[c]*d - Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]] - Sqrt[Sqrt[c
]*d + Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]]))/(2*Sqrt[a]) + ((-(A*c*d) + a*B
*e)*(2*Sqrt[a]*c^(1/4)*e*Sqrt[d + e*x] + (Sqrt[c]*d - Sqrt[a]*e)^(3/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sq
rt[c]*d - Sqrt[a]*e]] - (Sqrt[c]*d + Sqrt[a]*e)^(3/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]
*e]]))/(2*Sqrt[a]*c^(1/4)))/(2*a*c*(-(c*d^2) + a*e^2))
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Maple [B] time = 0.031, size = 432, normalized size = 1.9 \begin{align*} -{\frac{Ae}{ \left ( 2\,c{e}^{2}{x}^{2}-2\,a{e}^{2} \right ) a} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{Aed}{ \left ( 2\,c{e}^{2}{x}^{2}-2\,a{e}^{2} \right ) a}\sqrt{ex+d}}-{\frac{{e}^{2}B}{ \left ( 2\,c{e}^{2}{x}^{2}-2\,a{e}^{2} \right ) c}\sqrt{ex+d}}+{\frac{Acde}{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}}}}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}}-{\frac{{e}^{2}B}{4}{\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}}}}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{\frac{Ae}{4\,a}{\it Artanh} \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{\frac{Acde}{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}}}}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}}-{\frac{{e}^{2}B}{4}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ac{e}^{2}}}}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}}-{\frac{Ae}{4\,a}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\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((B*x+A)*(e*x+d)^(1/2)/(-c*x^2+a)^2,x)
[Out]
-1/2*e/(c*e^2*x^2-a*e^2)*A/a*(e*x+d)^(3/2)+1/2*e/(c*e^2*x^2-a*e^2)/a*(e*x+d)^(1/2)*A*d-1/2*e^2/(c*e^2*x^2-a*e^
2)/c*(e*x+d)^(1/2)*B+1/2*e/a/(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))*A*c*d-1/4*e^2/(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))*B+1/4*e/a/((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))*A+1/2*e/a/(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))*A*c*d-1/4*e^2/(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))*B-1/4*e/a/((-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))*A
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )} \sqrt{e x + d}}{{\left (c x^{2} - a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(e*x+d)^(1/2)/(-c*x^2+a)^2,x, algorithm="maxima")
[Out]
integrate((B*x + A)*sqrt(e*x + d)/(c*x^2 - a)^2, x)
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Fricas [B] time = 18.4197, size = 6109, normalized size = 27.15 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(e*x+d)^(1/2)/(-c*x^2+a)^2,x, algorithm="fricas")
[Out]
1/8*((a*c^2*x^2 - a^2*c)*sqrt((4*A^2*c^2*d^3 - 4*A*B*a*c*d^2*e + 2*A*B*a^2*e^3 + (B^2*a^2 - 3*A^2*a*c)*d*e^2 +
(a^3*c^3*d^2 - a^4*c^2*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^4 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^5 + (B^4*a^2 + 2*A^2*
B^2*a*c + A^4*c^2)*e^6)/(a^3*c^7*d^4 - 2*a^4*c^6*d^2*e^2 + a^5*c^5*e^4)))/(a^3*c^3*d^2 - a^4*c^2*e^2))*log((8*
A^3*B*c^3*d^3*e^2 - 4*(3*A^2*B^2*a*c^2 + A^4*c^3)*d^2*e^3 + 2*(3*A*B^3*a^2*c + A^3*B*a*c^2)*d*e^4 - (B^4*a^3 -
A^4*a*c^2)*e^5)*sqrt(e*x + d) + (2*A^2*B*a^2*c^3*d^2*e^3 - (3*A*B^2*a^3*c^2 + A^3*a^2*c^3)*d*e^4 + (B^3*a^4*c
+ A^2*B*a^3*c^2)*e^5 + (2*A*a^3*c^6*d^4 - B*a^4*c^5*d^3*e - 3*A*a^4*c^5*d^2*e^2 + B*a^5*c^4*d*e^3 + A*a^5*c^4
*e^4)*sqrt((4*A^2*B^2*c^2*d^2*e^4 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^5 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^6)
/(a^3*c^7*d^4 - 2*a^4*c^6*d^2*e^2 + a^5*c^5*e^4)))*sqrt((4*A^2*c^2*d^3 - 4*A*B*a*c*d^2*e + 2*A*B*a^2*e^3 + (B^
2*a^2 - 3*A^2*a*c)*d*e^2 + (a^3*c^3*d^2 - a^4*c^2*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^4 - 4*(A*B^3*a*c + A^3*B*c^2)
*d*e^5 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^6)/(a^3*c^7*d^4 - 2*a^4*c^6*d^2*e^2 + a^5*c^5*e^4)))/(a^3*c^3*d
^2 - a^4*c^2*e^2))) - (a*c^2*x^2 - a^2*c)*sqrt((4*A^2*c^2*d^3 - 4*A*B*a*c*d^2*e + 2*A*B*a^2*e^3 + (B^2*a^2 - 3
*A^2*a*c)*d*e^2 + (a^3*c^3*d^2 - a^4*c^2*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^4 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^5 +
(B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^6)/(a^3*c^7*d^4 - 2*a^4*c^6*d^2*e^2 + a^5*c^5*e^4)))/(a^3*c^3*d^2 - a^4*
c^2*e^2))*log((8*A^3*B*c^3*d^3*e^2 - 4*(3*A^2*B^2*a*c^2 + A^4*c^3)*d^2*e^3 + 2*(3*A*B^3*a^2*c + A^3*B*a*c^2)*d
*e^4 - (B^4*a^3 - A^4*a*c^2)*e^5)*sqrt(e*x + d) - (2*A^2*B*a^2*c^3*d^2*e^3 - (3*A*B^2*a^3*c^2 + A^3*a^2*c^3)*d
*e^4 + (B^3*a^4*c + A^2*B*a^3*c^2)*e^5 + (2*A*a^3*c^6*d^4 - B*a^4*c^5*d^3*e - 3*A*a^4*c^5*d^2*e^2 + B*a^5*c^4*
d*e^3 + A*a^5*c^4*e^4)*sqrt((4*A^2*B^2*c^2*d^2*e^4 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^5 + (B^4*a^2 + 2*A^2*B^2*a*
c + A^4*c^2)*e^6)/(a^3*c^7*d^4 - 2*a^4*c^6*d^2*e^2 + a^5*c^5*e^4)))*sqrt((4*A^2*c^2*d^3 - 4*A*B*a*c*d^2*e + 2*
A*B*a^2*e^3 + (B^2*a^2 - 3*A^2*a*c)*d*e^2 + (a^3*c^3*d^2 - a^4*c^2*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^4 - 4*(A*B^3
*a*c + A^3*B*c^2)*d*e^5 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^6)/(a^3*c^7*d^4 - 2*a^4*c^6*d^2*e^2 + a^5*c^5*
e^4)))/(a^3*c^3*d^2 - a^4*c^2*e^2))) + (a*c^2*x^2 - a^2*c)*sqrt((4*A^2*c^2*d^3 - 4*A*B*a*c*d^2*e + 2*A*B*a^2*e
^3 + (B^2*a^2 - 3*A^2*a*c)*d*e^2 - (a^3*c^3*d^2 - a^4*c^2*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^4 - 4*(A*B^3*a*c + A^
3*B*c^2)*d*e^5 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^6)/(a^3*c^7*d^4 - 2*a^4*c^6*d^2*e^2 + a^5*c^5*e^4)))/(a
^3*c^3*d^2 - a^4*c^2*e^2))*log((8*A^3*B*c^3*d^3*e^2 - 4*(3*A^2*B^2*a*c^2 + A^4*c^3)*d^2*e^3 + 2*(3*A*B^3*a^2*c
+ A^3*B*a*c^2)*d*e^4 - (B^4*a^3 - A^4*a*c^2)*e^5)*sqrt(e*x + d) + (2*A^2*B*a^2*c^3*d^2*e^3 - (3*A*B^2*a^3*c^2
+ A^3*a^2*c^3)*d*e^4 + (B^3*a^4*c + A^2*B*a^3*c^2)*e^5 - (2*A*a^3*c^6*d^4 - B*a^4*c^5*d^3*e - 3*A*a^4*c^5*d^2
*e^2 + B*a^5*c^4*d*e^3 + A*a^5*c^4*e^4)*sqrt((4*A^2*B^2*c^2*d^2*e^4 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^5 + (B^4*a
^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^6)/(a^3*c^7*d^4 - 2*a^4*c^6*d^2*e^2 + a^5*c^5*e^4)))*sqrt((4*A^2*c^2*d^3 - 4*A
*B*a*c*d^2*e + 2*A*B*a^2*e^3 + (B^2*a^2 - 3*A^2*a*c)*d*e^2 - (a^3*c^3*d^2 - a^4*c^2*e^2)*sqrt((4*A^2*B^2*c^2*d
^2*e^4 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^5 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^6)/(a^3*c^7*d^4 - 2*a^4*c^6*d
^2*e^2 + a^5*c^5*e^4)))/(a^3*c^3*d^2 - a^4*c^2*e^2))) - (a*c^2*x^2 - a^2*c)*sqrt((4*A^2*c^2*d^3 - 4*A*B*a*c*d^
2*e + 2*A*B*a^2*e^3 + (B^2*a^2 - 3*A^2*a*c)*d*e^2 - (a^3*c^3*d^2 - a^4*c^2*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^4 -
4*(A*B^3*a*c + A^3*B*c^2)*d*e^5 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^6)/(a^3*c^7*d^4 - 2*a^4*c^6*d^2*e^2 +
a^5*c^5*e^4)))/(a^3*c^3*d^2 - a^4*c^2*e^2))*log((8*A^3*B*c^3*d^3*e^2 - 4*(3*A^2*B^2*a*c^2 + A^4*c^3)*d^2*e^3 +
2*(3*A*B^3*a^2*c + A^3*B*a*c^2)*d*e^4 - (B^4*a^3 - A^4*a*c^2)*e^5)*sqrt(e*x + d) - (2*A^2*B*a^2*c^3*d^2*e^3 -
(3*A*B^2*a^3*c^2 + A^3*a^2*c^3)*d*e^4 + (B^3*a^4*c + A^2*B*a^3*c^2)*e^5 - (2*A*a^3*c^6*d^4 - B*a^4*c^5*d^3*e
- 3*A*a^4*c^5*d^2*e^2 + B*a^5*c^4*d*e^3 + A*a^5*c^4*e^4)*sqrt((4*A^2*B^2*c^2*d^2*e^4 - 4*(A*B^3*a*c + A^3*B*c^
2)*d*e^5 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^6)/(a^3*c^7*d^4 - 2*a^4*c^6*d^2*e^2 + a^5*c^5*e^4)))*sqrt((4*
A^2*c^2*d^3 - 4*A*B*a*c*d^2*e + 2*A*B*a^2*e^3 + (B^2*a^2 - 3*A^2*a*c)*d*e^2 - (a^3*c^3*d^2 - a^4*c^2*e^2)*sqrt
((4*A^2*B^2*c^2*d^2*e^4 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^5 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^6)/(a^3*c^7*
d^4 - 2*a^4*c^6*d^2*e^2 + a^5*c^5*e^4)))/(a^3*c^3*d^2 - a^4*c^2*e^2))) - 4*(A*c*x + B*a)*sqrt(e*x + d))/(a*c^2
*x^2 - a^2*c)
________________________________________________________________________________________
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((B*x+A)*(e*x+d)**(1/2)/(-c*x**2+a)**2,x)
[Out]
Timed out
________________________________________________________________________________________
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((B*x+A)*(e*x+d)^(1/2)/(-c*x^2+a)^2,x, algorithm="giac")
[Out]
Timed out