### 3.641 $$\int \frac{\sqrt{d+e x}}{(a-c x^2)^3} \, dx$$

Optimal. Leaf size=281 $-\frac{\left (-18 \sqrt{a} \sqrt{c} d e+5 a e^2+12 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2}}+\frac{\left (18 \sqrt{a} \sqrt{c} d e+5 a e^2+12 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{3/4} \left (\sqrt{a} e+\sqrt{c} d\right )^{3/2}}-\frac{\sqrt{d+e x} \left (a d e-x \left (6 c d^2-5 a e^2\right )\right )}{16 a^2 \left (a-c x^2\right ) \left (c d^2-a e^2\right )}+\frac{x \sqrt{d+e x}}{4 a \left (a-c x^2\right )^2}$

[Out]

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

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Rubi [A]  time = 0.490152, antiderivative size = 281, 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 = {737, 823, 827, 1166, 208} $-\frac{\left (-18 \sqrt{a} \sqrt{c} d e+5 a e^2+12 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{32 a^{5/2} c^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2}}+\frac{\left (18 \sqrt{a} \sqrt{c} d e+5 a e^2+12 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{32 a^{5/2} c^{3/4} \left (\sqrt{a} e+\sqrt{c} d\right )^{3/2}}-\frac{\sqrt{d+e x} \left (a d e-x \left (6 c d^2-5 a e^2\right )\right )}{16 a^2 \left (a-c x^2\right ) \left (c d^2-a e^2\right )}+\frac{x \sqrt{d+e x}}{4 a \left (a-c x^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[d + e*x]/(a - c*x^2)^3,x]

[Out]

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

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*a*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(d*(2*p + 3) + e*(m + 2*p + 3)*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, 0] && (LtQ[m, 1]
|| (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && 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{\sqrt{d+e x}}{\left (a-c x^2\right )^3} \, dx &=\frac{x \sqrt{d+e x}}{4 a \left (a-c x^2\right )^2}-\frac{\int \frac{-3 d-\frac{5 e x}{2}}{\sqrt{d+e x} \left (a-c x^2\right )^2} \, dx}{4 a}\\ &=\frac{x \sqrt{d+e x}}{4 a \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a d e-\left (6 c d^2-5 a e^2\right ) x\right )}{16 a^2 \left (c d^2-a e^2\right ) \left (a-c x^2\right )}+\frac{\int \frac{\frac{1}{4} c d \left (12 c d^2-13 a e^2\right )+\frac{1}{4} c e \left (6 c d^2-5 a e^2\right ) x}{\sqrt{d+e x} \left (a-c x^2\right )} \, dx}{8 a^2 c \left (c d^2-a e^2\right )}\\ &=\frac{x \sqrt{d+e x}}{4 a \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a d e-\left (6 c d^2-5 a e^2\right ) x\right )}{16 a^2 \left (c d^2-a e^2\right ) \left (a-c x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{4} c d e \left (12 c d^2-13 a e^2\right )-\frac{1}{4} c d e \left (6 c d^2-5 a e^2\right )+\frac{1}{4} c e \left (6 c d^2-5 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 \left (c d^2-a e^2\right )}\\ &=\frac{x \sqrt{d+e x}}{4 a \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a d e-\left (6 c d^2-5 a e^2\right ) x\right )}{16 a^2 \left (c d^2-a e^2\right ) \left (a-c x^2\right )}-\frac{\left (12 c d^2-18 \sqrt{a} \sqrt{c} d e+5 a e^2\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 )}+\frac{\left (12 c d^2+18 \sqrt{a} \sqrt{c} d e+5 a e^2\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 )}\\ &=\frac{x \sqrt{d+e x}}{4 a \left (a-c x^2\right )^2}-\frac{\sqrt{d+e x} \left (a d e-\left (6 c d^2-5 a e^2\right ) x\right )}{16 a^2 \left (c d^2-a e^2\right ) \left (a-c x^2\right )}-\frac{\left (12 c d^2-18 \sqrt{a} \sqrt{c} d e+5 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^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2}}+\frac{\left (12 c d^2+18 \sqrt{a} \sqrt{c} d e+5 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^{3/4} \left (\sqrt{c} d+\sqrt{a} e\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.735417, size = 368, normalized size = 1.31 $\frac{\frac{2 (d+e x)^{3/2} \left (5 a^2 e^3-a c d e (3 d+8 e x)+6 c^2 d^3 x\right )}{a-c x^2}+\frac{4 \sqrt{a} c^{3/4} d e \sqrt{d+e x} \left (3 c d^2-4 a e^2\right )+\sqrt{\sqrt{a} e+\sqrt{c} d} \left (18 \sqrt{a} \sqrt{c} d e+5 a e^2+12 c d^2\right ) \left (\sqrt{c} d-\sqrt{a} e\right )^2 \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )-\left (\sqrt{a} e+\sqrt{c} d\right )^2 \left (-18 \sqrt{a} \sqrt{c} d e+5 a e^2+12 c d^2\right ) \sqrt{\sqrt{c} d-\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{\sqrt{a} c^{3/4}}+\frac{8 a (d+e x)^{3/2} \left (c d^2-a e^2\right ) (c d x-a e)}{\left (a-c x^2\right )^2}}{32 a^2 \left (c d^2-a e^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((8*a*(c*d^2 - a*e^2)*(-(a*e) + c*d*x)*(d + e*x)^(3/2))/(a - c*x^2)^2 + (2*(d + e*x)^(3/2)*(5*a^2*e^3 + 6*c^2*
d^3*x - a*c*d*e*(3*d + 8*e*x)))/(a - c*x^2) + (4*Sqrt[a]*c^(3/4)*d*e*(3*c*d^2 - 4*a*e^2)*Sqrt[d + e*x] - Sqrt[
Sqrt[c]*d - Sqrt[a]*e]*(Sqrt[c]*d + Sqrt[a]*e)^2*(12*c*d^2 - 18*Sqrt[a]*Sqrt[c]*d*e + 5*a*e^2)*ArcTanh[(c^(1/4
)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]] + (Sqrt[c]*d - Sqrt[a]*e)^2*Sqrt[Sqrt[c]*d + Sqrt[a]*e]*(12*c*d^
2 + 18*Sqrt[a]*Sqrt[c]*d*e + 5*a*e^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(Sqrt[a]*c
^(3/4)))/(32*a^2*(c*d^2 - a*e^2)^2)

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Maple [B]  time = 0.276, size = 803, normalized size = 2.9 \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)^(1/2)/(-c*x^2+a)^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{\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((e*x+d)^(1/2)/(-c*x^2+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 4.57217, size = 7752, normalized size = 27.59 \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)^(1/2)/(-c*x^2+a)^3,x, algorithm="fricas")

[Out]

-1/64*((a^4*c*d^2 - a^5*e^2 + (a^2*c^3*d^2 - a^3*c^2*e^2)*x^4 - 2*(a^3*c^2*d^2 - a^4*c*e^2)*x^2)*sqrt((144*c^3
*d^7 - 420*a*c^2*d^5*e^2 + 385*a^2*c*d^3*e^4 - 105*a^3*d*e^6 + (a^5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*c^2*d^
2*e^4 - a^8*c*e^6)*sqrt((441*c^2*d^4*e^10 - 1050*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5*c^9*d^12 - 6*a^6*c^8*d^10*e
^2 + 15*a^7*c^7*d^8*e^4 - 20*a^8*c^6*d^6*e^6 + 15*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2*e^10 + a^11*c^3*e^12)))/(a^
5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^6))*log(-(3024*c^3*d^6*e^5 - 7884*a*c^2*d^4*e^7 +
5625*a^2*c*d^2*e^9 - 625*a^3*e^11)*sqrt(e*x + d) + (126*a^3*c^3*d^5*e^6 - 318*a^4*c^2*d^3*e^8 + 200*a^5*c*d*e^
10 + (12*a^5*c^7*d^10 - 55*a^6*c^6*d^8*e^2 + 98*a^7*c^5*d^6*e^4 - 84*a^8*c^4*d^4*e^6 + 34*a^9*c^3*d^2*e^8 - 5*
a^10*c^2*e^10)*sqrt((441*c^2*d^4*e^10 - 1050*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5*c^9*d^12 - 6*a^6*c^8*d^10*e^2 +
15*a^7*c^7*d^8*e^4 - 20*a^8*c^6*d^6*e^6 + 15*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2*e^10 + a^11*c^3*e^12)))*sqrt((1
44*c^3*d^7 - 420*a*c^2*d^5*e^2 + 385*a^2*c*d^3*e^4 - 105*a^3*d*e^6 + (a^5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*
c^2*d^2*e^4 - a^8*c*e^6)*sqrt((441*c^2*d^4*e^10 - 1050*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5*c^9*d^12 - 6*a^6*c^8*
d^10*e^2 + 15*a^7*c^7*d^8*e^4 - 20*a^8*c^6*d^6*e^6 + 15*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2*e^10 + a^11*c^3*e^12)
))/(a^5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^6))) - (a^4*c*d^2 - a^5*e^2 + (a^2*c^3*d^2 -
a^3*c^2*e^2)*x^4 - 2*(a^3*c^2*d^2 - a^4*c*e^2)*x^2)*sqrt((144*c^3*d^7 - 420*a*c^2*d^5*e^2 + 385*a^2*c*d^3*e^4
- 105*a^3*d*e^6 + (a^5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^6)*sqrt((441*c^2*d^4*e^10 -
1050*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5*c^9*d^12 - 6*a^6*c^8*d^10*e^2 + 15*a^7*c^7*d^8*e^4 - 20*a^8*c^6*d^6*e^6
+ 15*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2*e^10 + a^11*c^3*e^12)))/(a^5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*c^2*d^
2*e^4 - a^8*c*e^6))*log(-(3024*c^3*d^6*e^5 - 7884*a*c^2*d^4*e^7 + 5625*a^2*c*d^2*e^9 - 625*a^3*e^11)*sqrt(e*x
+ d) - (126*a^3*c^3*d^5*e^6 - 318*a^4*c^2*d^3*e^8 + 200*a^5*c*d*e^10 + (12*a^5*c^7*d^10 - 55*a^6*c^6*d^8*e^2 +
98*a^7*c^5*d^6*e^4 - 84*a^8*c^4*d^4*e^6 + 34*a^9*c^3*d^2*e^8 - 5*a^10*c^2*e^10)*sqrt((441*c^2*d^4*e^10 - 1050
*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5*c^9*d^12 - 6*a^6*c^8*d^10*e^2 + 15*a^7*c^7*d^8*e^4 - 20*a^8*c^6*d^6*e^6 + 1
5*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2*e^10 + a^11*c^3*e^12)))*sqrt((144*c^3*d^7 - 420*a*c^2*d^5*e^2 + 385*a^2*c*d
^3*e^4 - 105*a^3*d*e^6 + (a^5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^6)*sqrt((441*c^2*d^4*e
^10 - 1050*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5*c^9*d^12 - 6*a^6*c^8*d^10*e^2 + 15*a^7*c^7*d^8*e^4 - 20*a^8*c^6*d
^6*e^6 + 15*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2*e^10 + a^11*c^3*e^12)))/(a^5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*
c^2*d^2*e^4 - a^8*c*e^6))) + (a^4*c*d^2 - a^5*e^2 + (a^2*c^3*d^2 - a^3*c^2*e^2)*x^4 - 2*(a^3*c^2*d^2 - a^4*c*e
^2)*x^2)*sqrt((144*c^3*d^7 - 420*a*c^2*d^5*e^2 + 385*a^2*c*d^3*e^4 - 105*a^3*d*e^6 - (a^5*c^4*d^6 - 3*a^6*c^3*
d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^6)*sqrt((441*c^2*d^4*e^10 - 1050*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5*c^9*d
^12 - 6*a^6*c^8*d^10*e^2 + 15*a^7*c^7*d^8*e^4 - 20*a^8*c^6*d^6*e^6 + 15*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2*e^10
+ a^11*c^3*e^12)))/(a^5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^6))*log(-(3024*c^3*d^6*e^5 -
7884*a*c^2*d^4*e^7 + 5625*a^2*c*d^2*e^9 - 625*a^3*e^11)*sqrt(e*x + d) + (126*a^3*c^3*d^5*e^6 - 318*a^4*c^2*d^
3*e^8 + 200*a^5*c*d*e^10 - (12*a^5*c^7*d^10 - 55*a^6*c^6*d^8*e^2 + 98*a^7*c^5*d^6*e^4 - 84*a^8*c^4*d^4*e^6 + 3
4*a^9*c^3*d^2*e^8 - 5*a^10*c^2*e^10)*sqrt((441*c^2*d^4*e^10 - 1050*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5*c^9*d^12
- 6*a^6*c^8*d^10*e^2 + 15*a^7*c^7*d^8*e^4 - 20*a^8*c^6*d^6*e^6 + 15*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2*e^10 + a^
11*c^3*e^12)))*sqrt((144*c^3*d^7 - 420*a*c^2*d^5*e^2 + 385*a^2*c*d^3*e^4 - 105*a^3*d*e^6 - (a^5*c^4*d^6 - 3*a^
6*c^3*d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^6)*sqrt((441*c^2*d^4*e^10 - 1050*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5
*c^9*d^12 - 6*a^6*c^8*d^10*e^2 + 15*a^7*c^7*d^8*e^4 - 20*a^8*c^6*d^6*e^6 + 15*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2
*e^10 + a^11*c^3*e^12)))/(a^5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^6))) - (a^4*c*d^2 - a^
5*e^2 + (a^2*c^3*d^2 - a^3*c^2*e^2)*x^4 - 2*(a^3*c^2*d^2 - a^4*c*e^2)*x^2)*sqrt((144*c^3*d^7 - 420*a*c^2*d^5*e
^2 + 385*a^2*c*d^3*e^4 - 105*a^3*d*e^6 - (a^5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^6)*sqr
t((441*c^2*d^4*e^10 - 1050*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5*c^9*d^12 - 6*a^6*c^8*d^10*e^2 + 15*a^7*c^7*d^8*e^
4 - 20*a^8*c^6*d^6*e^6 + 15*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2*e^10 + a^11*c^3*e^12)))/(a^5*c^4*d^6 - 3*a^6*c^3*
d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^6))*log(-(3024*c^3*d^6*e^5 - 7884*a*c^2*d^4*e^7 + 5625*a^2*c*d^2*e^9 - 6
25*a^3*e^11)*sqrt(e*x + d) - (126*a^3*c^3*d^5*e^6 - 318*a^4*c^2*d^3*e^8 + 200*a^5*c*d*e^10 - (12*a^5*c^7*d^10
- 55*a^6*c^6*d^8*e^2 + 98*a^7*c^5*d^6*e^4 - 84*a^8*c^4*d^4*e^6 + 34*a^9*c^3*d^2*e^8 - 5*a^10*c^2*e^10)*sqrt((4
41*c^2*d^4*e^10 - 1050*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5*c^9*d^12 - 6*a^6*c^8*d^10*e^2 + 15*a^7*c^7*d^8*e^4 -
20*a^8*c^6*d^6*e^6 + 15*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2*e^10 + a^11*c^3*e^12)))*sqrt((144*c^3*d^7 - 420*a*c^2
*d^5*e^2 + 385*a^2*c*d^3*e^4 - 105*a^3*d*e^6 - (a^5*c^4*d^6 - 3*a^6*c^3*d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^
6)*sqrt((441*c^2*d^4*e^10 - 1050*a*c*d^2*e^12 + 625*a^2*e^14)/(a^5*c^9*d^12 - 6*a^6*c^8*d^10*e^2 + 15*a^7*c^7*
d^8*e^4 - 20*a^8*c^6*d^6*e^6 + 15*a^9*c^5*d^4*e^8 - 6*a^10*c^4*d^2*e^10 + a^11*c^3*e^12)))/(a^5*c^4*d^6 - 3*a^
6*c^3*d^4*e^2 + 3*a^7*c^2*d^2*e^4 - a^8*c*e^6))) - 4*(a*c*d*e*x^2 - a^2*d*e - (6*c^2*d^2 - 5*a*c*e^2)*x^3 + (1
0*a*c*d^2 - 9*a^2*e^2)*x)*sqrt(e*x + d))/(a^4*c*d^2 - a^5*e^2 + (a^2*c^3*d^2 - a^3*c^2*e^2)*x^4 - 2*(a^3*c^2*d
^2 - a^4*c*e^2)*x^2)

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

[Out]

Timed out