3.1461 \(\int \frac{(A+B x) \sqrt{d+e x}}{(a-c x^2)^3} \, dx\)

Optimal. Leaf size=372 \[ \frac{\left (a B e \left (2 \sqrt{c} d-3 \sqrt{a} e\right )-A \left (-18 \sqrt{a} c d e+5 a \sqrt{c} e^2+12 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{5/4} \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2}}-\frac{\left (a B e \left (3 \sqrt{a} e+2 \sqrt{c} d\right )-A \left (18 \sqrt{a} c d e+5 a \sqrt{c} e^2+12 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{5/4} \left (\sqrt{a} e+\sqrt{c} d\right )^{3/2}}-\frac{\sqrt{d+e x} \left (a e (A c d-a B e)-c x \left (-5 a A e^2-a B d e+6 A c d^2\right )\right )}{16 a^2 c \left (a-c x^2\right ) \left (c d^2-a e^2\right )}+\frac{\sqrt{d+e x} (a B+A c x)}{4 a c \left (a-c x^2\right )^2} \]

[Out]

((a*B + A*c*x)*Sqrt[d + e*x])/(4*a*c*(a - c*x^2)^2) - (Sqrt[d + e*x]*(a*e*(A*c*d - a*B*e) - c*(6*A*c*d^2 - a*B
*d*e - 5*a*A*e^2)*x))/(16*a^2*c*(c*d^2 - a*e^2)*(a - c*x^2)) + ((a*B*e*(2*Sqrt[c]*d - 3*Sqrt[a]*e) - A*(12*c^(
3/2)*d^2 - 18*Sqrt[a]*c*d*e + 5*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^(5/4)*(Sqrt[c]*d - Sqrt[a]*e)^(3/2)) - ((a*B*e*(2*Sqrt[c]*d + 3*Sqrt[a]*e) - A*(12*c^(3/2)*d^2 +
18*Sqrt[a]*c*d*e + 5*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^(5/4)*(Sqrt[c]*d + Sqrt[a]*e)^(3/2))

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Rubi [A]  time = 0.760527, antiderivative size = 372, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {821, 823, 827, 1166, 208} \[ \frac{\left (a B e \left (2 \sqrt{c} d-3 \sqrt{a} e\right )-A \left (-18 \sqrt{a} c d e+5 a \sqrt{c} e^2+12 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{5/4} \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2}}-\frac{\left (a B e \left (3 \sqrt{a} e+2 \sqrt{c} d\right )-A \left (18 \sqrt{a} c d e+5 a \sqrt{c} e^2+12 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{5/4} \left (\sqrt{a} e+\sqrt{c} d\right )^{3/2}}-\frac{\sqrt{d+e x} \left (a e (A c d-a B e)-c x \left (-5 a A e^2-a B d e+6 A c d^2\right )\right )}{16 a^2 c \left (a-c x^2\right ) \left (c d^2-a e^2\right )}+\frac{\sqrt{d+e x} (a B+A c x)}{4 a c \left (a-c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*B + A*c*x)*Sqrt[d + e*x])/(4*a*c*(a - c*x^2)^2) - (Sqrt[d + e*x]*(a*e*(A*c*d - a*B*e) - c*(6*A*c*d^2 - a*B
*d*e - 5*a*A*e^2)*x))/(16*a^2*c*(c*d^2 - a*e^2)*(a - c*x^2)) + ((a*B*e*(2*Sqrt[c]*d - 3*Sqrt[a]*e) - A*(12*c^(
3/2)*d^2 - 18*Sqrt[a]*c*d*e + 5*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^(5/4)*(Sqrt[c]*d - Sqrt[a]*e)^(3/2)) - ((a*B*e*(2*Sqrt[c]*d + 3*Sqrt[a]*e) - A*(12*c^(3/2)*d^2 +
18*Sqrt[a]*c*d*e + 5*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^(5/4)*(Sqrt[c]*d + Sqrt[a]*e)^(3/2))

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 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{(a B+A c x) \sqrt{d+e x}}{4 a c \left (a-c x^2\right )^2}-\frac{\int \frac{\frac{1}{2} (-6 A c d+a B e)-\frac{5}{2} A c e x}{\sqrt{d+e x} \left (a-c x^2\right )^2} \, dx}{4 a c}\\ &=\frac{(a B+A c x) \sqrt{d+e x}}{4 a c \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a e (A c d-a B e)-c \left (6 A c d^2-a B d e-5 a A e^2\right ) x\right )}{16 a^2 c \left (c d^2-a e^2\right ) \left (a-c x^2\right )}+\frac{\int \frac{\frac{1}{4} c \left (A c d \left (12 c d^2-13 a e^2\right )-a B e \left (2 c d^2-3 a e^2\right )\right )+\frac{1}{4} c^2 e \left (6 A c d^2-a B d e-5 a A e^2\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 )}\\ &=\frac{(a B+A c x) \sqrt{d+e x}}{4 a c \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a e (A c d-a B e)-c \left (6 A c d^2-a B d e-5 a A e^2\right ) x\right )}{16 a^2 c \left (c d^2-a e^2\right ) \left (a-c x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{4} c^2 d e \left (6 A c d^2-a B d e-5 a A e^2\right )+\frac{1}{4} c e \left (A c d \left (12 c d^2-13 a e^2\right )-a B e \left (2 c d^2-3 a e^2\right )\right )+\frac{1}{4} c^2 e \left (6 A c d^2-a B d e-5 a A e^2\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 )}\\ &=\frac{(a B+A c x) \sqrt{d+e x}}{4 a c \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a e (A c d-a B e)-c \left (6 A c d^2-a B d e-5 a A e^2\right ) x\right )}{16 a^2 c \left (c d^2-a e^2\right ) \left (a-c x^2\right )}+\frac{\left (a B e \left (2 \sqrt{c} d-3 \sqrt{a} e\right )-A \left (12 c^{3/2} d^2-18 \sqrt{a} c d e+5 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} \sqrt{c} \left (\sqrt{c} d-\sqrt{a} e\right )}-\frac{\left (a B e \left (2 \sqrt{c} d+3 \sqrt{a} e\right )-A \left (12 c^{3/2} d^2+18 \sqrt{a} c d e+5 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} \sqrt{c} \left (\sqrt{c} d+\sqrt{a} e\right )}\\ &=\frac{(a B+A c x) \sqrt{d+e x}}{4 a c \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a e (A c d-a B e)-c \left (6 A c d^2-a B d e-5 a A e^2\right ) x\right )}{16 a^2 c \left (c d^2-a e^2\right ) \left (a-c x^2\right )}+\frac{\left (a B e \left (2 \sqrt{c} d-3 \sqrt{a} e\right )-A \left (12 c^{3/2} d^2-18 \sqrt{a} c d e+5 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^{5/4} \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2}}-\frac{\left (a B e \left (2 \sqrt{c} d+3 \sqrt{a} e\right )-A \left (12 c^{3/2} d^2+18 \sqrt{a} c d e+5 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^{5/4} \left (\sqrt{c} d+\sqrt{a} e\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.14241, size = 522, normalized size = 1.4 \[ \frac{\frac{c^2 (d+e x)^{3/2} \left (a^2 e^2 (5 A e-2 B d+3 B e x)-a c d e (3 A d+8 A e x+B d x)+6 A c^2 d^3 x\right )}{2 \left (a-c x^2\right )}-\frac{c^{5/4} \left (A \left (5 a^2 e^4-27 a c d^2 e^2+18 c^2 d^4\right )+a B d e \left (7 a e^2-3 c d^2\right )\right ) \left (\sqrt{\sqrt{c} d-\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-\sqrt{\sqrt{a} e+\sqrt{c} d} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )\right )}{4 \sqrt{a}}+\frac{2 a c^2 (d+e x)^{3/2} \left (c d^2-a e^2\right ) (-a A e+a B (d-e x)+A c d x)}{\left (a-c x^2\right )^2}+\frac{c^{3/4} \left (2 A c d \left (3 c d^2-4 a e^2\right )+a B e \left (3 a e^2-c d^2\right )\right ) \left (2 \sqrt{a} \sqrt [4]{c} e \sqrt{d+e x}+\left (\sqrt{c} d-\sqrt{a} e\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-\left (\sqrt{a} e+\sqrt{c} d\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )\right )}{4 \sqrt{a}}}{8 a^2 c^2 \left (c d^2-a e^2\right )^2} \]

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)*(d + e*x)^(3/2)*(-(a*A*e) + A*c*d*x + a*B*(d - e*x)))/(a - c*x^2)^2 + (c^2*(d + e*x)
^(3/2)*(6*A*c^2*d^3*x - a*c*d*e*(3*A*d + B*d*x + 8*A*e*x) + a^2*e^2*(-2*B*d + 5*A*e + 3*B*e*x)))/(2*(a - c*x^2
)) - (c^(5/4)*(a*B*d*e*(-3*c*d^2 + 7*a*e^2) + A*(18*c^2*d^4 - 27*a*c*d^2*e^2 + 5*a^2*e^4))*(Sqrt[Sqrt[c]*d - S
qrt[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]) + (c^(3/4)*(2*A*c*d*(3*c*d^2 - 4*a*e^2) + a*
B*e*(-(c*d^2) + 3*a*e^2))*(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[Sqrt[c]*d - Sqrt[a]*e]] - (Sqrt[c]*d + Sqrt[a]*e)^(3/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sq
rt[Sqrt[c]*d + Sqrt[a]*e]]))/(4*Sqrt[a]))/(8*a^2*c^2*(c*d^2 - a*e^2)^2)

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Maple [B]  time = 0.041, size = 1733, normalized size = 4.7 \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]

1/16*e^2/a/(a*e^2-c*d^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*c*d^2+13/32*e^3/a/(a*e^2-c*d^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))*A*c*d+1/16*e^2/a/(a*e^2-c*d^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*c*d^2+13/32*e^3/a/(a*e^2-c*d^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))*A*c*
d-3/8*e/a^2/(a*e^2-c*d^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))*A*d^3*c^2-3/8*e/a^2/(a*e^2-c*d^2)/(a*c*e^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan
h((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*A*d^3*c^2+5/32*e^3/a/(a*e^2-c*d^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-5/32*e^3/a/(a*e^2-c*d^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-1/16*e^4/(c*e^2*x^2-a*e^2)^2/(a*e^2-
c*d^2)*(e*x+d)^(3/2)*B*d-3/32*e^4/(a*e^2-c*d^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-3/32*e^4/(a*e^2-c*d^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-1/2*e^3/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(1/2)*
A*d-1/16*e^2/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(1/2)*B*d^2+3/16*e^4/(c*e^2*x^2-a*e^2)^2/c*(e*x+d)^(1/2)*B+9/16*e^5
/(c*e^2*x^2-a*e^2)^2/(a*e^2-c*d^2)*(e*x+d)^(3/2)*A+1/16*e^4/(c*e^2*x^2-a*e^2)^2/(a*e^2-c*d^2)*(e*x+d)^(5/2)*B-
3/16*e/a^2/(a*e^2-c*d^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^2-9/8*e/(c*e^2*x^2-a*e^2)^2/a^2/(a*e^2-c*d^2)*(e*x+d)^(5/2)*A*d^3*c^2-1/16*e^2/(c*e^2*x^2-a*e^2)^2*c
/a/(a*e^2-c*d^2)*(e*x+d)^(7/2)*B*d+7/8*e^3/(c*e^2*x^2-a*e^2)^2/a/(a*e^2-c*d^2)*(e*x+d)^(5/2)*A*c*d+3/16*e/a^2/
(a*e^2-c*d^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^
2+3/16*e^2/(c*e^2*x^2-a*e^2)^2/a/(a*e^2-c*d^2)*(e*x+d)^(5/2)*B*c*d^2-23/16*e^3/(c*e^2*x^2-a*e^2)^2/a/(a*e^2-c*
d^2)*(e*x+d)^(3/2)*A*c*d^2+9/8*e/(c*e^2*x^2-a*e^2)^2/a^2/(a*e^2-c*d^2)*(e*x+d)^(3/2)*A*c^2*d^4+1/32*e^2/a/(a*e
^2-c*d^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*d-1/32*e^
2/a/(a*e^2-c*d^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-5/16*e^3/(c*e^2*x^2-a*e^2)^2*c/a/(a*e^2-c*d^2)*(e*x+d)^(7/2)*A+3/8*e/(c*e^2*x^2-a*e^2)^2*c/a^2*(e*x+d)^(1/2)
*A*d^3-3/16*e^2/(c*e^2*x^2-a*e^2)^2/a/(a*e^2-c*d^2)*(e*x+d)^(3/2)*B*c*d^3+3/8*e/(c*e^2*x^2-a*e^2)^2*c^2/a^2/(a
*e^2-c*d^2)*(e*x+d)^(7/2)*A*d^2

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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 )}^{3}}\,{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)*sqrt(e*x + d)/(c*x^2 - a)^3, 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