3.1462 \(\int \frac{A+B x}{\sqrt{d+e x} (a-c x^2)^3} \, dx\)

Optimal. Leaf size=417 \[ -\frac{\left (3 A \left (-10 \sqrt{a} c d e+7 a \sqrt{c} e^2+4 c^{3/2} d^2\right )+a B e \left (2 \sqrt{c} d-5 \sqrt{a} e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right )^{5/2}}+\frac{\left (3 A \left (10 \sqrt{a} c d e+7 a \sqrt{c} e^2+4 c^{3/2} d^2\right )+a B e \left (5 \sqrt{a} e+2 \sqrt{c} d\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{3/4} \left (\sqrt{a} e+\sqrt{c} d\right )^{5/2}}-\frac{\sqrt{d+e x} \left (a e \left (-7 a A e^2+6 a B d e+A c d^2\right )-x \left (6 A c d \left (c d^2-2 a e^2\right )+a B e \left (5 a e^2+c d^2\right )\right )\right )}{16 a^2 \left (a-c x^2\right ) \left (c d^2-a e^2\right )^2}+\frac{\sqrt{d+e x} (x (A c d-a B e)+a (B d-A e))}{4 a \left (a-c x^2\right )^2 \left (c d^2-a e^2\right )} \]

[Out]

(Sqrt[d + e*x]*(a*(B*d - A*e) + (A*c*d - a*B*e)*x))/(4*a*(c*d^2 - a*e^2)*(a - c*x^2)^2) - (Sqrt[d + e*x]*(a*e*
(A*c*d^2 + 6*a*B*d*e - 7*a*A*e^2) - (6*A*c*d*(c*d^2 - 2*a*e^2) + a*B*e*(c*d^2 + 5*a*e^2))*x))/(16*a^2*(c*d^2 -
a*e^2)^2*(a - c*x^2)) - ((a*B*e*(2*Sqrt[c]*d - 5*Sqrt[a]*e) + 3*A*(4*c^(3/2)*d^2 - 10*Sqrt[a]*c*d*e + 7*a*Sqr
t[c]*e^2))*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(32*a^(5/2)*c^(3/4)*(Sqrt[c]*d - Sqrt
[a]*e)^(5/2)) + ((a*B*e*(2*Sqrt[c]*d + 5*Sqrt[a]*e) + 3*A*(4*c^(3/2)*d^2 + 10*Sqrt[a]*c*d*e + 7*a*Sqrt[c]*e^2)
)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(32*a^(5/2)*c^(3/4)*(Sqrt[c]*d + Sqrt[a]*e)^(5
/2))

________________________________________________________________________________________

Rubi [A]  time = 0.925666, antiderivative size = 417, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {823, 827, 1166, 208} \[ -\frac{\left (3 A \left (-10 \sqrt{a} c d e+7 a \sqrt{c} e^2+4 c^{3/2} d^2\right )+a B e \left (2 \sqrt{c} d-5 \sqrt{a} e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right )^{5/2}}+\frac{\left (3 A \left (10 \sqrt{a} c d e+7 a \sqrt{c} e^2+4 c^{3/2} d^2\right )+a B e \left (5 \sqrt{a} e+2 \sqrt{c} d\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{3/4} \left (\sqrt{a} e+\sqrt{c} d\right )^{5/2}}-\frac{\sqrt{d+e x} \left (a e \left (-7 a A e^2+6 a B d e+A c d^2\right )-x \left (6 A c d \left (c d^2-2 a e^2\right )+a B e \left (5 a e^2+c d^2\right )\right )\right )}{16 a^2 \left (a-c x^2\right ) \left (c d^2-a e^2\right )^2}+\frac{\sqrt{d+e x} (x (A c d-a B e)+a (B d-A e))}{4 a \left (a-c x^2\right )^2 \left (c d^2-a e^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(Sqrt[d + e*x]*(a - c*x^2)^3),x]

[Out]

(Sqrt[d + e*x]*(a*(B*d - A*e) + (A*c*d - a*B*e)*x))/(4*a*(c*d^2 - a*e^2)*(a - c*x^2)^2) - (Sqrt[d + e*x]*(a*e*
(A*c*d^2 + 6*a*B*d*e - 7*a*A*e^2) - (6*A*c*d*(c*d^2 - 2*a*e^2) + a*B*e*(c*d^2 + 5*a*e^2))*x))/(16*a^2*(c*d^2 -
a*e^2)^2*(a - c*x^2)) - ((a*B*e*(2*Sqrt[c]*d - 5*Sqrt[a]*e) + 3*A*(4*c^(3/2)*d^2 - 10*Sqrt[a]*c*d*e + 7*a*Sqr
t[c]*e^2))*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(32*a^(5/2)*c^(3/4)*(Sqrt[c]*d - Sqrt
[a]*e)^(5/2)) + ((a*B*e*(2*Sqrt[c]*d + 5*Sqrt[a]*e) + 3*A*(4*c^(3/2)*d^2 + 10*Sqrt[a]*c*d*e + 7*a*Sqrt[c]*e^2)
)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(32*a^(5/2)*c^(3/4)*(Sqrt[c]*d + Sqrt[a]*e)^(5
/2))

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (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 )^3} \, dx &=\frac{\sqrt{d+e x} (a (B d-A e)+(A c d-a B e) x)}{4 a \left (c d^2-a e^2\right ) \left (a-c x^2\right )^2}-\frac{\int \frac{-\frac{1}{2} c \left (6 A c d^2+a B d e-7 a A e^2\right )-\frac{5}{2} c e (A c d-a B e) x}{\sqrt{d+e x} \left (a-c x^2\right )^2} \, dx}{4 a c \left (c d^2-a e^2\right )}\\ &=\frac{\sqrt{d+e x} (a (B d-A e)+(A c d-a B e) x)}{4 a \left (c d^2-a e^2\right ) \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a e \left (A c d^2+6 a B d e-7 a A e^2\right )-\left (6 A c d \left (c d^2-2 a e^2\right )+a B e \left (c d^2+5 a e^2\right )\right ) x\right )}{16 a^2 \left (c d^2-a e^2\right )^2 \left (a-c x^2\right )}+\frac{\int \frac{\frac{1}{4} c^2 \left (2 a B d e \left (c d^2-4 a e^2\right )+3 A \left (4 c^2 d^4-9 a c d^2 e^2+7 a^2 e^4\right )\right )+\frac{1}{4} c^2 e \left (6 A c d \left (c d^2-2 a e^2\right )+a B e \left (c d^2+5 a e^2\right )\right ) x}{\sqrt{d+e x} \left (a-c x^2\right )} \, dx}{8 a^2 c^2 \left (c d^2-a e^2\right )^2}\\ &=\frac{\sqrt{d+e x} (a (B d-A e)+(A c d-a B e) x)}{4 a \left (c d^2-a e^2\right ) \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a e \left (A c d^2+6 a B d e-7 a A e^2\right )-\left (6 A c d \left (c d^2-2 a e^2\right )+a B e \left (c d^2+5 a e^2\right )\right ) x\right )}{16 a^2 \left (c d^2-a e^2\right )^2 \left (a-c x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{4} c^2 d e \left (6 A c d \left (c d^2-2 a e^2\right )+a B e \left (c d^2+5 a e^2\right )\right )+\frac{1}{4} c^2 e \left (2 a B d e \left (c d^2-4 a e^2\right )+3 A \left (4 c^2 d^4-9 a c d^2 e^2+7 a^2 e^4\right )\right )+\frac{1}{4} c^2 e \left (6 A c d \left (c d^2-2 a e^2\right )+a B e \left (c d^2+5 a e^2\right )\right ) x^2}{-c d^2+a e^2+2 c d x^2-c x^4} \, dx,x,\sqrt{d+e x}\right )}{4 a^2 c^2 \left (c d^2-a e^2\right )^2}\\ &=\frac{\sqrt{d+e x} (a (B d-A e)+(A c d-a B e) x)}{4 a \left (c d^2-a e^2\right ) \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a e \left (A c d^2+6 a B d e-7 a A e^2\right )-\left (6 A c d \left (c d^2-2 a e^2\right )+a B e \left (c d^2+5 a e^2\right )\right ) x\right )}{16 a^2 \left (c d^2-a e^2\right )^2 \left (a-c x^2\right )}-\frac{\left (a B e \left (2 \sqrt{c} d-5 \sqrt{a} e\right )+3 A \left (4 c^{3/2} d^2-10 \sqrt{a} c d e+7 a \sqrt{c} e^2\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 )}{32 a^{5/2} \left (\sqrt{c} d-\sqrt{a} e\right )^2}+\frac{\left (a B e \left (2 \sqrt{c} d+5 \sqrt{a} e\right )+3 A \left (4 c^{3/2} d^2+10 \sqrt{a} c d e+7 a \sqrt{c} e^2\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 )}{32 a^{5/2} \left (\sqrt{c} d+\sqrt{a} e\right )^2}\\ &=\frac{\sqrt{d+e x} (a (B d-A e)+(A c d-a B e) x)}{4 a \left (c d^2-a e^2\right ) \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a e \left (A c d^2+6 a B d e-7 a A e^2\right )-\left (6 A c d \left (c d^2-2 a e^2\right )+a B e \left (c d^2+5 a e^2\right )\right ) x\right )}{16 a^2 \left (c d^2-a e^2\right )^2 \left (a-c x^2\right )}-\frac{\left (a B e \left (2 \sqrt{c} d-5 \sqrt{a} e\right )+3 A \left (4 c^{3/2} d^2-10 \sqrt{a} c d e+7 a \sqrt{c} e^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right )^{5/2}}+\frac{\left (a B e \left (2 \sqrt{c} d+5 \sqrt{a} e\right )+3 A \left (4 c^{3/2} d^2+10 \sqrt{a} c d e+7 a \sqrt{c} e^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d+\sqrt{a} e}}\right )}{32 a^{5/2} c^{3/4} \left (\sqrt{c} d+\sqrt{a} e\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.14865, size = 536, normalized size = 1.29 \[ \frac{\frac{c^2 \sqrt{d+e x} \left (a^2 e^2 (7 A e-6 B d+5 B e x)+a c d e (B d x-A (d+12 e x))+6 A c^2 d^3 x\right )}{2 \left (a-c x^2\right )}+\frac{c^{7/4} \left (3 A \left (7 a^2 e^4-5 a c d^2 e^2+2 c^2 d^4\right )+a B d e \left (c d^2-13 a e^2\right )\right ) \left (\sqrt{\sqrt{c} d-\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )-\sqrt{\sqrt{a} e+\sqrt{c} d} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )\right )}{4 \sqrt{a} \sqrt{\sqrt{c} d-\sqrt{a} e} \sqrt{\sqrt{a} e+\sqrt{c} d}}+\frac{2 a c^2 \sqrt{d+e x} \left (c d^2-a e^2\right ) (-a A e+a B (d-e x)+A c d x)}{\left (a-c x^2\right )^2}-\frac{c^{5/4} \left (6 A c d \left (c d^2-2 a e^2\right )+a B e \left (5 a e^2+c d^2\right )\right ) \left (\sqrt{\sqrt{c} d-\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-\sqrt{\sqrt{a} e+\sqrt{c} d} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )\right )}{4 \sqrt{a}}}{8 a^2 c^2 \left (c d^2-a e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(Sqrt[d + e*x]*(a - c*x^2)^3),x]

[Out]

((2*a*c^2*(c*d^2 - a*e^2)*Sqrt[d + e*x]*(-(a*A*e) + A*c*d*x + a*B*(d - e*x)))/(a - c*x^2)^2 + (c^2*Sqrt[d + e*
x]*(6*A*c^2*d^3*x + a^2*e^2*(-6*B*d + 7*A*e + 5*B*e*x) + a*c*d*e*(B*d*x - A*(d + 12*e*x))))/(2*(a - c*x^2)) +
(c^(7/4)*(a*B*d*e*(c*d^2 - 13*a*e^2) + 3*A*(2*c^2*d^4 - 5*a*c*d^2*e^2 + 7*a^2*e^4))*(-(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]]))/(4*Sqrt[a]*Sqrt[Sqrt[c]*d - Sqrt[a]*e]*Sqrt[Sqrt[c]*d + Sqrt
[a]*e]) - (c^(5/4)*(6*A*c*d*(c*d^2 - 2*a*e^2) + a*B*e*(c*d^2 + 5*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]]))/(4*Sqrt[a]))/(8*a^2*c^2*(c*d^2 - a*e^2)^2)

________________________________________________________________________________________

Maple [B]  time = 0.115, size = 1778, normalized size = 4.3 \begin{align*} \text{result too large to display} \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)^3,x)

[Out]

5/32*e^2/a/(a*e^2+c*d^2+2*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))*B+5/32*e^2/a/(-a*e^2-c*d^2+2*d*(a*c*e^2)^(1/2))/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arcta
n((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*B-11/32*e/c*(a*c*e^2)^(1/2)/a^2/(e*x+(a*c*e^2)^(1/2)/c)^2/
(c*d-(a*c*e^2)^(1/2))*(e*x+d)^(1/2)*A-21/32*e^3*c/(a*c*e^2)^(1/2)/a/(-a*e^2-c*d^2+2*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))*A-3/8*e*c^2/(a*c*e^2)^(1/2)
/a^2/(-a*e^2-c*d^2+2*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))*A*d^2-1/16*e^2*c/(a*c*e^2)^(1/2)/a/(-a*e^2-c*d^2+2*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))*B*d+3/8*e*c^2/(a*c*e^2)^(1/2)/a^2/(a*e^2
+c*d^2+2*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))*A*d^2+1/16*e^2*c/(a*c*e^2)^(1/2)/a/(a*e^2+c*d^2+2*d*(a*c*e^2)^(1/2))/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*ar
ctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*B*d-5/32*e^2/c/a/(e*x+(a*c*e^2)^(1/2)/c)^2/(a*e^2+c*d^2
-2*d*(a*c*e^2)^(1/2))*(e*x+d)^(3/2)*B+7/32*e^2/c/a/(e*x+(a*c*e^2)^(1/2)/c)^2/(c*d-(a*c*e^2)^(1/2))*(e*x+d)^(1/
2)*B-5/32*e^2/c/a/(e*x-(a*c*e^2)^(1/2)/c)^2/(a*e^2+c*d^2+2*d*(a*c*e^2)^(1/2))*(e*x+d)^(3/2)*B+7/32*e^2/c/a/(e*
x-(a*c*e^2)^(1/2)/c)^2/(c*d+(a*c*e^2)^(1/2))*(e*x+d)^(1/2)*B+3/16*e/a^2/(e*x+(a*c*e^2)^(1/2)/c)^2/(c*d-(a*c*e^
2)^(1/2))*(e*x+d)^(1/2)*A*d-3/16*e/a^2/(e*x-(a*c*e^2)^(1/2)/c)^2/(a*e^2+c*d^2+2*d*(a*c*e^2)^(1/2))*(e*x+d)^(3/
2)*A*d-3/16*e/a^2/(e*x+(a*c*e^2)^(1/2)/c)^2/(a*e^2+c*d^2-2*d*(a*c*e^2)^(1/2))*(e*x+d)^(3/2)*A*d+3/16*e/a^2/(e*
x-(a*c*e^2)^(1/2)/c)^2/(c*d+(a*c*e^2)^(1/2))*(e*x+d)^(1/2)*A*d-1/16/c*(a*c*e^2)^(1/2)/a^2/(e*x-(a*c*e^2)^(1/2)
/c)^2/(a*e^2+c*d^2+2*d*(a*c*e^2)^(1/2))*(e*x+d)^(3/2)*B*d+1/16/c*(a*c*e^2)^(1/2)/a^2/(e*x+(a*c*e^2)^(1/2)/c)^2
/(a*e^2+c*d^2-2*d*(a*c*e^2)^(1/2))*(e*x+d)^(3/2)*B*d-1/16/c*(a*c*e^2)^(1/2)/a^2/(e*x+(a*c*e^2)^(1/2)/c)^2/(c*d
-(a*c*e^2)^(1/2))*(e*x+d)^(1/2)*B*d+1/16/c*(a*c*e^2)^(1/2)/a^2/(e*x-(a*c*e^2)^(1/2)/c)^2/(c*d+(a*c*e^2)^(1/2))
*(e*x+d)^(1/2)*B*d+9/32*e/c*(a*c*e^2)^(1/2)/a^2/(e*x+(a*c*e^2)^(1/2)/c)^2/(a*e^2+c*d^2-2*d*(a*c*e^2)^(1/2))*(e
*x+d)^(3/2)*A-9/32*e/c*(a*c*e^2)^(1/2)/a^2/(e*x-(a*c*e^2)^(1/2)/c)^2/(a*e^2+c*d^2+2*d*(a*c*e^2)^(1/2))*(e*x+d)
^(3/2)*A+11/32*e/c*(a*c*e^2)^(1/2)/a^2/(e*x-(a*c*e^2)^(1/2)/c)^2/(c*d+(a*c*e^2)^(1/2))*(e*x+d)^(1/2)*A+21/32*e
^3*c/(a*c*e^2)^(1/2)/a/(a*e^2+c*d^2+2*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))*A+15/16*e*c/a^2/(-a*e^2-c*d^2+2*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))*A*d+15/16*e*c/a^2/(a*e^2+c*d^2+2*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))*A*d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{B x + A}{{\left (c x^{2} - a\right )}^{3} \sqrt{e x + d}}\,{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)^3,x, algorithm="maxima")

[Out]

-integrate((B*x + A)/((c*x^2 - a)^3*sqrt(e*x + d)), x)

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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((B*x+A)/(e*x+d)^(1/2)/(-c*x^2+a)^3,x, algorithm="fricas")

[Out]

Timed out

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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((B*x+A)/(e*x+d)**(1/2)/(-c*x**2+a)**3,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((B*x+A)/(e*x+d)^(1/2)/(-c*x^2+a)^3,x, algorithm="giac")

[Out]

Timed out