3.640 $$\int \frac{(d+e x)^{3/2}}{(a-c x^2)^3} \, dx$$

Optimal. Leaf size=268 $-\frac{3 \left (-2 \sqrt{a} \sqrt{c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{5/4} \sqrt{\sqrt{c} d-\sqrt{a} e}}+\frac{3 \left (2 \sqrt{a} \sqrt{c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{5/4} \sqrt{\sqrt{a} e+\sqrt{c} d}}-\frac{\sqrt{d+e x} (a e-6 c d x)}{16 a^2 c \left (a-c x^2\right )}+\frac{\sqrt{d+e x} (a e+c d x)}{4 a c \left (a-c x^2\right )^2}$

[Out]

((a*e + c*d*x)*Sqrt[d + e*x])/(4*a*c*(a - c*x^2)^2) - ((a*e - 6*c*d*x)*Sqrt[d + e*x])/(16*a^2*c*(a - c*x^2)) -
(3*(4*c*d^2 - 2*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(3
2*a^(5/2)*c^(5/4)*Sqrt[Sqrt[c]*d - Sqrt[a]*e]) + (3*(4*c*d^2 + 2*Sqrt[a]*Sqrt[c]*d*e - a*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[Sqrt[c]*d + Sqrt[a]*e])

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Rubi [A]  time = 0.505557, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {739, 823, 827, 1166, 208} $-\frac{3 \left (-2 \sqrt{a} \sqrt{c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{5/4} \sqrt{\sqrt{c} d-\sqrt{a} e}}+\frac{3 \left (2 \sqrt{a} \sqrt{c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{5/4} \sqrt{\sqrt{a} e+\sqrt{c} d}}-\frac{\sqrt{d+e x} (a e-6 c d x)}{16 a^2 c \left (a-c x^2\right )}+\frac{\sqrt{d+e x} (a e+c d x)}{4 a c \left (a-c x^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(3/2)/(a - c*x^2)^3,x]

[Out]

((a*e + c*d*x)*Sqrt[d + e*x])/(4*a*c*(a - c*x^2)^2) - ((a*e - 6*c*d*x)*Sqrt[d + e*x])/(16*a^2*c*(a - c*x^2)) -
(3*(4*c*d^2 - 2*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(3
2*a^(5/2)*c^(5/4)*Sqrt[Sqrt[c]*d - Sqrt[a]*e]) + (3*(4*c*d^2 + 2*Sqrt[a]*Sqrt[c]*d*e - a*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[Sqrt[c]*d + Sqrt[a]*e])

Rule 739

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*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

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

Mathematica [A]  time = 0.994475, size = 428, normalized size = 1.6 $\frac{\frac{2 (d+e x)^{5/2} \left (3 a^2 e^3-a c d e (5 d+4 e x)+6 c^2 d^3 x\right )}{a-c x^2}+\frac{2 \sqrt{a} \sqrt [4]{c} e \sqrt{d+e x} \left (3 a^2 e^4-a c d e^2 (13 d+4 e x)+6 c^2 d^3 (2 d+e x)\right )+3 \sqrt{\sqrt{a} e+\sqrt{c} d} \left (-a^{3/2} e^3+6 \sqrt{a} c d^2 e+a \sqrt{c} d e^2+4 c^{3/2} d^3\right ) \left (\sqrt{c} d-\sqrt{a} e\right )^2 \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )-3 \left (\sqrt{a} e+\sqrt{c} d\right )^2 \left (a^{3/2} e^3-6 \sqrt{a} c d^2 e+a \sqrt{c} d e^2+4 c^{3/2} d^3\right ) \sqrt{\sqrt{c} d-\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{\sqrt{a} c^{5/4}}+\frac{8 a (d+e x)^{5/2} \left (c d^2-a e^2\right ) (c d x-a e)}{\left (a-c x^2\right )^2}}{32 a^2 \left (c d^2-a e^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((8*a*(c*d^2 - a*e^2)*(-(a*e) + c*d*x)*(d + e*x)^(5/2))/(a - c*x^2)^2 + (2*(d + e*x)^(5/2)*(3*a^2*e^3 + 6*c^2*
d^3*x - a*c*d*e*(5*d + 4*e*x)))/(a - c*x^2) + (2*Sqrt[a]*c^(1/4)*e*Sqrt[d + e*x]*(3*a^2*e^4 + 6*c^2*d^3*(2*d +
e*x) - a*c*d*e^2*(13*d + 4*e*x)) - 3*Sqrt[Sqrt[c]*d - Sqrt[a]*e]*(Sqrt[c]*d + Sqrt[a]*e)^2*(4*c^(3/2)*d^3 - 6
*Sqrt[a]*c*d^2*e + a*Sqrt[c]*d*e^2 + a^(3/2)*e^3)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]]
+ 3*(Sqrt[c]*d - Sqrt[a]*e)^2*Sqrt[Sqrt[c]*d + Sqrt[a]*e]*(4*c^(3/2)*d^3 + 6*Sqrt[a]*c*d^2*e + a*Sqrt[c]*d*e^
2 - a^(3/2)*e^3)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(Sqrt[a]*c^(5/4)))/(32*a^2*(c*d
^2 - a*e^2)^2)

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Maple [B]  time = 0.218, size = 608, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)/(-c*x^2+a)^3,x)

[Out]

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

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

[Out]

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

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Fricas [B]  time = 2.71421, size = 3357, normalized size = 12.53 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*c^2*d^5 - 20*a*c*d^3*e^2 + 5*a^2*d*e^4 + (a^5*c^3*d^2 -
a^6*c^2*e^2)*sqrt(e^10/(a^5*c^7*d^4 - 2*a^6*c^6*d^2*e^2 + a^7*c^5*e^4)))/(a^5*c^3*d^2 - a^6*c^2*e^2))*log(27*
(16*c^2*d^4*e^5 - 12*a*c*d^2*e^7 + a^2*e^9)*sqrt(e*x + d) + 27*(2*a^3*c^2*d^2*e^6 - a^4*c*e^8 - (4*a^5*c^6*d^5
- 7*a^6*c^5*d^3*e^2 + 3*a^7*c^4*d*e^4)*sqrt(e^10/(a^5*c^7*d^4 - 2*a^6*c^6*d^2*e^2 + a^7*c^5*e^4)))*sqrt((16*c
^2*d^5 - 20*a*c*d^3*e^2 + 5*a^2*d*e^4 + (a^5*c^3*d^2 - a^6*c^2*e^2)*sqrt(e^10/(a^5*c^7*d^4 - 2*a^6*c^6*d^2*e^2
+ a^7*c^5*e^4)))/(a^5*c^3*d^2 - a^6*c^2*e^2))) - 3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)*sqrt((16*c^2*d^5 - 2
0*a*c*d^3*e^2 + 5*a^2*d*e^4 + (a^5*c^3*d^2 - a^6*c^2*e^2)*sqrt(e^10/(a^5*c^7*d^4 - 2*a^6*c^6*d^2*e^2 + a^7*c^5
*e^4)))/(a^5*c^3*d^2 - a^6*c^2*e^2))*log(27*(16*c^2*d^4*e^5 - 12*a*c*d^2*e^7 + a^2*e^9)*sqrt(e*x + d) - 27*(2*
a^3*c^2*d^2*e^6 - a^4*c*e^8 - (4*a^5*c^6*d^5 - 7*a^6*c^5*d^3*e^2 + 3*a^7*c^4*d*e^4)*sqrt(e^10/(a^5*c^7*d^4 - 2
*a^6*c^6*d^2*e^2 + a^7*c^5*e^4)))*sqrt((16*c^2*d^5 - 20*a*c*d^3*e^2 + 5*a^2*d*e^4 + (a^5*c^3*d^2 - a^6*c^2*e^2
)*sqrt(e^10/(a^5*c^7*d^4 - 2*a^6*c^6*d^2*e^2 + a^7*c^5*e^4)))/(a^5*c^3*d^2 - a^6*c^2*e^2))) + 3*(a^2*c^3*x^4 -
2*a^3*c^2*x^2 + a^4*c)*sqrt((16*c^2*d^5 - 20*a*c*d^3*e^2 + 5*a^2*d*e^4 - (a^5*c^3*d^2 - a^6*c^2*e^2)*sqrt(e^1
0/(a^5*c^7*d^4 - 2*a^6*c^6*d^2*e^2 + a^7*c^5*e^4)))/(a^5*c^3*d^2 - a^6*c^2*e^2))*log(27*(16*c^2*d^4*e^5 - 12*a
*c*d^2*e^7 + a^2*e^9)*sqrt(e*x + d) + 27*(2*a^3*c^2*d^2*e^6 - a^4*c*e^8 + (4*a^5*c^6*d^5 - 7*a^6*c^5*d^3*e^2 +
3*a^7*c^4*d*e^4)*sqrt(e^10/(a^5*c^7*d^4 - 2*a^6*c^6*d^2*e^2 + a^7*c^5*e^4)))*sqrt((16*c^2*d^5 - 20*a*c*d^3*e^
2 + 5*a^2*d*e^4 - (a^5*c^3*d^2 - a^6*c^2*e^2)*sqrt(e^10/(a^5*c^7*d^4 - 2*a^6*c^6*d^2*e^2 + a^7*c^5*e^4)))/(a^5
*c^3*d^2 - a^6*c^2*e^2))) - 3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)*sqrt((16*c^2*d^5 - 20*a*c*d^3*e^2 + 5*a^2*
d*e^4 - (a^5*c^3*d^2 - a^6*c^2*e^2)*sqrt(e^10/(a^5*c^7*d^4 - 2*a^6*c^6*d^2*e^2 + a^7*c^5*e^4)))/(a^5*c^3*d^2 -
a^6*c^2*e^2))*log(27*(16*c^2*d^4*e^5 - 12*a*c*d^2*e^7 + a^2*e^9)*sqrt(e*x + d) - 27*(2*a^3*c^2*d^2*e^6 - a^4*
c*e^8 + (4*a^5*c^6*d^5 - 7*a^6*c^5*d^3*e^2 + 3*a^7*c^4*d*e^4)*sqrt(e^10/(a^5*c^7*d^4 - 2*a^6*c^6*d^2*e^2 + a^7
*c^5*e^4)))*sqrt((16*c^2*d^5 - 20*a*c*d^3*e^2 + 5*a^2*d*e^4 - (a^5*c^3*d^2 - a^6*c^2*e^2)*sqrt(e^10/(a^5*c^7*d
^4 - 2*a^6*c^6*d^2*e^2 + a^7*c^5*e^4)))/(a^5*c^3*d^2 - a^6*c^2*e^2))) - 4*(6*c^2*d*x^3 - a*c*e*x^2 - 10*a*c*d*
x - 3*a^2*e)*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((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((e*x+d)^(3/2)/(-c*x^2+a)^3,x, algorithm="giac")

[Out]

Timed out