3.1460 \(\int \frac{(A+B x) (d+e x)^{3/2}}{(a-c x^2)^3} \, dx\)

Optimal. Leaf size=350 \[ \frac{3 \left (a B e \left (2 \sqrt{c} d-\sqrt{a} e\right )-A \left (-2 \sqrt{a} c d e-a \sqrt{c} e^2+4 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{7/4} \sqrt{\sqrt{c} d-\sqrt{a} e}}-\frac{3 \left (a B e \left (\sqrt{a} e+2 \sqrt{c} d\right )-A \left (2 \sqrt{a} c d e-a \sqrt{c} e^2+4 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{7/4} \sqrt{\sqrt{a} e+\sqrt{c} d}}-\frac{\sqrt{d+e x} (a A e-3 x (2 A c d-a B e))}{16 a^2 c \left (a-c x^2\right )}+\frac{\sqrt{d+e x} (x (a B e+A c d)+a (A e+B d))}{4 a c \left (a-c x^2\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.801277, antiderivative size = 350, 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 = {819, 823, 827, 1166, 208} \[ \frac{3 \left (a B e \left (2 \sqrt{c} d-\sqrt{a} e\right )-A \left (-2 \sqrt{a} c d e-a \sqrt{c} e^2+4 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{7/4} \sqrt{\sqrt{c} d-\sqrt{a} e}}-\frac{3 \left (a B e \left (\sqrt{a} e+2 \sqrt{c} d\right )-A \left (2 \sqrt{a} c d e-a \sqrt{c} e^2+4 c^{3/2} d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{7/4} \sqrt{\sqrt{a} e+\sqrt{c} d}}-\frac{\sqrt{d+e x} (a A e-3 x (2 A c d-a B e))}{16 a^2 c \left (a-c x^2\right )}+\frac{\sqrt{d+e x} (x (a B e+A c d)+a (A e+B d))}{4 a c \left (a-c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^(3/2))/(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*(a - c*x^2)^2) - (Sqrt[d + e*x]*(a*A*e - 3*(2*A*c*d
- a*B*e)*x))/(16*a^2*c*(a - c*x^2)) + (3*(a*B*e*(2*Sqrt[c]*d - Sqrt[a]*e) - A*(4*c^(3/2)*d^2 - 2*Sqrt[a]*c*d*
e - 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^(7/4)*Sqrt[Sqr
t[c]*d - Sqrt[a]*e]) - (3*(a*B*e*(2*Sqrt[c]*d + Sqrt[a]*e) - A*(4*c^(3/2)*d^2 + 2*Sqrt[a]*c*d*e - 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^(7/4)*Sqrt[Sqrt[c]*d + Sqrt[a]
*e])

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*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, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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) (d+e x)^{3/2}}{\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 c \left (a-c x^2\right )^2}-\frac{\int \frac{\frac{1}{2} \left (-6 A c d^2+3 a B d e+a A e^2\right )-\frac{1}{2} e (5 A c d-3 a B e) x}{\sqrt{d+e x} \left (a-c x^2\right )^2} \, dx}{4 a c}\\ &=\frac{\sqrt{d+e x} (a (B d+A e)+(A c d+a B e) x)}{4 a c \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} (a A e-3 (2 A c d-a B e) x)}{16 a^2 c \left (a-c x^2\right )}+\frac{\int \frac{\frac{3}{4} c \left (c d^2-a e^2\right ) \left (4 A c d^2-2 a B d e-a A e^2\right )+\frac{3}{4} c e (2 A c d-a B e) \left (c d^2-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{\sqrt{d+e x} (a (B d+A e)+(A c d+a B e) x)}{4 a c \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} (a A e-3 (2 A c d-a B e) x)}{16 a^2 c \left (a-c x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{3}{4} c d e (2 A c d-a B e) \left (c d^2-a e^2\right )+\frac{3}{4} c e \left (c d^2-a e^2\right ) \left (4 A c d^2-2 a B d e-a A e^2\right )+\frac{3}{4} c e (2 A c d-a B e) \left (c d^2-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{\sqrt{d+e x} (a (B d+A e)+(A c d+a B e) x)}{4 a c \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} (a A e-3 (2 A c d-a B e) x)}{16 a^2 c \left (a-c x^2\right )}+\frac{\left (3 \left (a B e \left (2 \sqrt{c} d-\sqrt{a} e\right )-A \left (4 c^{3/2} d^2-2 \sqrt{a} c d e-a \sqrt{c} e^2\right )\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} c}-\frac{\left (3 \left (a B e \left (2 \sqrt{c} d+\sqrt{a} e\right )-A \left (4 c^{3/2} d^2+2 \sqrt{a} c d e-a \sqrt{c} e^2\right )\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} c}\\ &=\frac{\sqrt{d+e x} (a (B d+A e)+(A c d+a B e) x)}{4 a c \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} (a A e-3 (2 A c d-a B e) x)}{16 a^2 c \left (a-c x^2\right )}+\frac{3 \left (a B e \left (2 \sqrt{c} d-\sqrt{a} e\right )-A \left (4 c^{3/2} d^2-2 \sqrt{a} c d e-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^{7/4} \sqrt{\sqrt{c} d-\sqrt{a} e}}-\frac{3 \left (a B e \left (2 \sqrt{c} d+\sqrt{a} e\right )-A \left (4 c^{3/2} d^2+2 \sqrt{a} c d e-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^{7/4} \sqrt{\sqrt{c} d+\sqrt{a} e}}\\ \end{align*}

Mathematica [A]  time = 1.5855, size = 550, normalized size = 1.57 \[ \frac{\frac{c^2 (d+e x)^{5/2} \left (a^2 e^2 (3 A e+2 B d+B e x)-a c d e (5 A d+4 A e x+3 B d x)+6 A c^2 d^3 x\right )}{2 \left (a-c x^2\right )}+\frac{3 c^{3/4} \left (A \left (a^2 e^4+5 a c d^2 e^2-10 c^2 d^4\right )+a B d e \left (5 c d^2-a e^2\right )\right ) \left (2 \sqrt{a} \sqrt [4]{c} e \sqrt{d+e x}+\left (\sqrt{c} d-\sqrt{a} e\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-\left (\sqrt{a} e+\sqrt{c} d\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )\right )}{4 \sqrt{a}}+\frac{2 a c^2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (-a A e+a B (d-e x)+A c d x)}{\left (a-c x^2\right )^2}+\frac{\sqrt [4]{c} \left (2 A c d \left (3 c d^2-2 a e^2\right )+a B e \left (a e^2-3 c d^2\right )\right ) \left (2 \sqrt{a} c^{3/4} e \sqrt{d+e x} (7 d+e x)+3 \left (\sqrt{c} d-\sqrt{a} e\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-3 \left (\sqrt{a} e+\sqrt{c} d\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )\right )}{4 \sqrt{a}}}{8 a^2 c^2 \left (c d^2-a e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*c^2*(c*d^2 - a*e^2)*(d + e*x)^(5/2)*(-(a*A*e) + A*c*d*x + a*B*(d - e*x)))/(a - c*x^2)^2 + (c^2*(d + e*x)
^(5/2)*(6*A*c^2*d^3*x - a*c*d*e*(5*A*d + 3*B*d*x + 4*A*e*x) + a^2*e^2*(2*B*d + 3*A*e + B*e*x)))/(2*(a - c*x^2)
) + (3*c^(3/4)*(a*B*d*e*(5*c*d^2 - a*e^2) + A*(-10*c^2*d^4 + 5*a*c*d^2*e^2 + a^2*e^4))*(2*Sqrt[a]*c^(1/4)*e*Sq
rt[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]] - (Sq
rt[c]*d + Sqrt[a]*e)^(3/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]]))/(4*Sqrt[a]) + (c^(1/
4)*(2*A*c*d*(3*c*d^2 - 2*a*e^2) + a*B*e*(-3*c*d^2 + a*e^2))*(2*Sqrt[a]*c^(3/4)*e*Sqrt[d + e*x]*(7*d + e*x) + 3
*(Sqrt[c]*d - Sqrt[a]*e)^(5/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]] - 3*(Sqrt[c]*d + S
qrt[a]*e)^(5/2)*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)

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Maple [B]  time = 0.042, size = 1060, normalized size = 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)^(3/2)/(-c*x^2+a)^3,x)

[Out]

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

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

[Out]

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

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Fricas [B]  time = 10.2314, size = 8124, normalized size = 23.21 \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)^(3/2)/(-c*x^2+a)^3,x, algorithm="fricas")

[Out]

-1/64*(3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)*sqrt((16*A^2*c^3*d^5 - 16*A*B*a*c^2*d^4*e + 16*A*B*a^2*c*d^2*e^
3 - 2*A*B*a^3*e^5 + 4*(B^2*a^2*c - 5*A^2*a*c^2)*d^3*e^2 - (3*B^2*a^3 - 5*A^2*a^2*c)*d*e^4 + (a^5*c^4*d^2 - a^6
*c^3*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^8 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^9 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*
e^10)/(a^5*c^9*d^4 - 2*a^6*c^8*d^2*e^2 + a^7*c^7*e^4)))/(a^5*c^4*d^2 - a^6*c^3*e^2))*log(-27*(32*A^3*B*c^4*d^5
*e^4 - 16*(3*A^2*B^2*a*c^3 + A^4*c^4)*d^4*e^5 + 8*(3*A*B^3*a^2*c^2 - A^3*B*a*c^3)*d^3*e^6 - 4*(B^4*a^3*c - 6*A
^2*B^2*a^2*c^2 - 3*A^4*a*c^3)*d^2*e^7 - 2*(5*A*B^3*a^3*c + 3*A^3*B*a^2*c^2)*d*e^8 + (B^4*a^4 - A^4*a^2*c^2)*e^
9)*sqrt(e*x + d) + 27*(4*A^2*B*a^3*c^4*d^3*e^5 - 2*(2*A*B^2*a^4*c^3 + A^3*a^3*c^4)*d^2*e^6 + (B^3*a^5*c^2 - A^
2*B*a^4*c^3)*d*e^7 + (A*B^2*a^5*c^2 + A^3*a^4*c^3)*e^8 + (4*A*a^5*c^8*d^5 - 2*B*a^6*c^7*d^4*e - 7*A*a^6*c^7*d^
3*e^2 + 3*B*a^7*c^6*d^2*e^3 + 3*A*a^7*c^6*d*e^4 - B*a^8*c^5*e^5)*sqrt((4*A^2*B^2*c^2*d^2*e^8 - 4*(A*B^3*a*c +
A^3*B*c^2)*d*e^9 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^10)/(a^5*c^9*d^4 - 2*a^6*c^8*d^2*e^2 + a^7*c^7*e^4)))
*sqrt((16*A^2*c^3*d^5 - 16*A*B*a*c^2*d^4*e + 16*A*B*a^2*c*d^2*e^3 - 2*A*B*a^3*e^5 + 4*(B^2*a^2*c - 5*A^2*a*c^2
)*d^3*e^2 - (3*B^2*a^3 - 5*A^2*a^2*c)*d*e^4 + (a^5*c^4*d^2 - a^6*c^3*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^8 - 4*(A*B
^3*a*c + A^3*B*c^2)*d*e^9 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^10)/(a^5*c^9*d^4 - 2*a^6*c^8*d^2*e^2 + a^7*c
^7*e^4)))/(a^5*c^4*d^2 - a^6*c^3*e^2))) - 3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)*sqrt((16*A^2*c^3*d^5 - 16*A*
B*a*c^2*d^4*e + 16*A*B*a^2*c*d^2*e^3 - 2*A*B*a^3*e^5 + 4*(B^2*a^2*c - 5*A^2*a*c^2)*d^3*e^2 - (3*B^2*a^3 - 5*A^
2*a^2*c)*d*e^4 + (a^5*c^4*d^2 - a^6*c^3*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^8 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^9 + (
B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^10)/(a^5*c^9*d^4 - 2*a^6*c^8*d^2*e^2 + a^7*c^7*e^4)))/(a^5*c^4*d^2 - a^6*
c^3*e^2))*log(-27*(32*A^3*B*c^4*d^5*e^4 - 16*(3*A^2*B^2*a*c^3 + A^4*c^4)*d^4*e^5 + 8*(3*A*B^3*a^2*c^2 - A^3*B*
a*c^3)*d^3*e^6 - 4*(B^4*a^3*c - 6*A^2*B^2*a^2*c^2 - 3*A^4*a*c^3)*d^2*e^7 - 2*(5*A*B^3*a^3*c + 3*A^3*B*a^2*c^2)
*d*e^8 + (B^4*a^4 - A^4*a^2*c^2)*e^9)*sqrt(e*x + d) - 27*(4*A^2*B*a^3*c^4*d^3*e^5 - 2*(2*A*B^2*a^4*c^3 + A^3*a
^3*c^4)*d^2*e^6 + (B^3*a^5*c^2 - A^2*B*a^4*c^3)*d*e^7 + (A*B^2*a^5*c^2 + A^3*a^4*c^3)*e^8 + (4*A*a^5*c^8*d^5 -
2*B*a^6*c^7*d^4*e - 7*A*a^6*c^7*d^3*e^2 + 3*B*a^7*c^6*d^2*e^3 + 3*A*a^7*c^6*d*e^4 - B*a^8*c^5*e^5)*sqrt((4*A^
2*B^2*c^2*d^2*e^8 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^9 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^10)/(a^5*c^9*d^4 -
2*a^6*c^8*d^2*e^2 + a^7*c^7*e^4)))*sqrt((16*A^2*c^3*d^5 - 16*A*B*a*c^2*d^4*e + 16*A*B*a^2*c*d^2*e^3 - 2*A*B*a
^3*e^5 + 4*(B^2*a^2*c - 5*A^2*a*c^2)*d^3*e^2 - (3*B^2*a^3 - 5*A^2*a^2*c)*d*e^4 + (a^5*c^4*d^2 - a^6*c^3*e^2)*s
qrt((4*A^2*B^2*c^2*d^2*e^8 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^9 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^10)/(a^5*
c^9*d^4 - 2*a^6*c^8*d^2*e^2 + a^7*c^7*e^4)))/(a^5*c^4*d^2 - a^6*c^3*e^2))) + 3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 +
a^4*c)*sqrt((16*A^2*c^3*d^5 - 16*A*B*a*c^2*d^4*e + 16*A*B*a^2*c*d^2*e^3 - 2*A*B*a^3*e^5 + 4*(B^2*a^2*c - 5*A^2
*a*c^2)*d^3*e^2 - (3*B^2*a^3 - 5*A^2*a^2*c)*d*e^4 - (a^5*c^4*d^2 - a^6*c^3*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^8 -
4*(A*B^3*a*c + A^3*B*c^2)*d*e^9 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^10)/(a^5*c^9*d^4 - 2*a^6*c^8*d^2*e^2 +
a^7*c^7*e^4)))/(a^5*c^4*d^2 - a^6*c^3*e^2))*log(-27*(32*A^3*B*c^4*d^5*e^4 - 16*(3*A^2*B^2*a*c^3 + A^4*c^4)*d^
4*e^5 + 8*(3*A*B^3*a^2*c^2 - A^3*B*a*c^3)*d^3*e^6 - 4*(B^4*a^3*c - 6*A^2*B^2*a^2*c^2 - 3*A^4*a*c^3)*d^2*e^7 -
2*(5*A*B^3*a^3*c + 3*A^3*B*a^2*c^2)*d*e^8 + (B^4*a^4 - A^4*a^2*c^2)*e^9)*sqrt(e*x + d) + 27*(4*A^2*B*a^3*c^4*d
^3*e^5 - 2*(2*A*B^2*a^4*c^3 + A^3*a^3*c^4)*d^2*e^6 + (B^3*a^5*c^2 - A^2*B*a^4*c^3)*d*e^7 + (A*B^2*a^5*c^2 + A^
3*a^4*c^3)*e^8 - (4*A*a^5*c^8*d^5 - 2*B*a^6*c^7*d^4*e - 7*A*a^6*c^7*d^3*e^2 + 3*B*a^7*c^6*d^2*e^3 + 3*A*a^7*c^
6*d*e^4 - B*a^8*c^5*e^5)*sqrt((4*A^2*B^2*c^2*d^2*e^8 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^9 + (B^4*a^2 + 2*A^2*B^2*
a*c + A^4*c^2)*e^10)/(a^5*c^9*d^4 - 2*a^6*c^8*d^2*e^2 + a^7*c^7*e^4)))*sqrt((16*A^2*c^3*d^5 - 16*A*B*a*c^2*d^4
*e + 16*A*B*a^2*c*d^2*e^3 - 2*A*B*a^3*e^5 + 4*(B^2*a^2*c - 5*A^2*a*c^2)*d^3*e^2 - (3*B^2*a^3 - 5*A^2*a^2*c)*d*
e^4 - (a^5*c^4*d^2 - a^6*c^3*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^8 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^9 + (B^4*a^2 + 2
*A^2*B^2*a*c + A^4*c^2)*e^10)/(a^5*c^9*d^4 - 2*a^6*c^8*d^2*e^2 + a^7*c^7*e^4)))/(a^5*c^4*d^2 - a^6*c^3*e^2)))
- 3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)*sqrt((16*A^2*c^3*d^5 - 16*A*B*a*c^2*d^4*e + 16*A*B*a^2*c*d^2*e^3 - 2
*A*B*a^3*e^5 + 4*(B^2*a^2*c - 5*A^2*a*c^2)*d^3*e^2 - (3*B^2*a^3 - 5*A^2*a^2*c)*d*e^4 - (a^5*c^4*d^2 - a^6*c^3*
e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^8 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e^9 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^10)
/(a^5*c^9*d^4 - 2*a^6*c^8*d^2*e^2 + a^7*c^7*e^4)))/(a^5*c^4*d^2 - a^6*c^3*e^2))*log(-27*(32*A^3*B*c^4*d^5*e^4
- 16*(3*A^2*B^2*a*c^3 + A^4*c^4)*d^4*e^5 + 8*(3*A*B^3*a^2*c^2 - A^3*B*a*c^3)*d^3*e^6 - 4*(B^4*a^3*c - 6*A^2*B^
2*a^2*c^2 - 3*A^4*a*c^3)*d^2*e^7 - 2*(5*A*B^3*a^3*c + 3*A^3*B*a^2*c^2)*d*e^8 + (B^4*a^4 - A^4*a^2*c^2)*e^9)*sq
rt(e*x + d) - 27*(4*A^2*B*a^3*c^4*d^3*e^5 - 2*(2*A*B^2*a^4*c^3 + A^3*a^3*c^4)*d^2*e^6 + (B^3*a^5*c^2 - A^2*B*a
^4*c^3)*d*e^7 + (A*B^2*a^5*c^2 + A^3*a^4*c^3)*e^8 - (4*A*a^5*c^8*d^5 - 2*B*a^6*c^7*d^4*e - 7*A*a^6*c^7*d^3*e^2
+ 3*B*a^7*c^6*d^2*e^3 + 3*A*a^7*c^6*d*e^4 - B*a^8*c^5*e^5)*sqrt((4*A^2*B^2*c^2*d^2*e^8 - 4*(A*B^3*a*c + A^3*B
*c^2)*d*e^9 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^10)/(a^5*c^9*d^4 - 2*a^6*c^8*d^2*e^2 + a^7*c^7*e^4)))*sqrt
((16*A^2*c^3*d^5 - 16*A*B*a*c^2*d^4*e + 16*A*B*a^2*c*d^2*e^3 - 2*A*B*a^3*e^5 + 4*(B^2*a^2*c - 5*A^2*a*c^2)*d^3
*e^2 - (3*B^2*a^3 - 5*A^2*a^2*c)*d*e^4 - (a^5*c^4*d^2 - a^6*c^3*e^2)*sqrt((4*A^2*B^2*c^2*d^2*e^8 - 4*(A*B^3*a*
c + A^3*B*c^2)*d*e^9 + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^10)/(a^5*c^9*d^4 - 2*a^6*c^8*d^2*e^2 + a^7*c^7*e^
4)))/(a^5*c^4*d^2 - a^6*c^3*e^2))) - 4*(A*a*c*e*x^2 + 4*B*a^2*d + 3*A*a^2*e - 3*(2*A*c^2*d - B*a*c*e)*x^3 + (1
0*A*a*c*d + B*a^2*e)*x)*sqrt(e*x + d))/(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)

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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)**(3/2)/(-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)^(3/2)/(-c*x^2+a)^3,x, algorithm="giac")

[Out]

Timed out