### 3.624 $$\int \frac{(d+e x)^{7/2}}{(a-c x^2)^2} \, dx$$

Optimal. Leaf size=263 $-\frac{\left (5 \sqrt{a} e+2 \sqrt{c} d\right ) \left (\sqrt{c} d-\sqrt{a} e\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} c^{9/4}}+\frac{\left (2 \sqrt{c} d-5 \sqrt{a} e\right ) \left (\sqrt{a} e+\sqrt{c} d\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} c^{9/4}}+\frac{e \sqrt{d+e x} \left (5 a e^2+c d^2\right )}{2 a c^2}+\frac{(d+e x)^{5/2} (a e+c d x)}{2 a c \left (a-c x^2\right )}+\frac{d e (d+e x)^{3/2}}{2 a c}$

[Out]

(e*(c*d^2 + 5*a*e^2)*Sqrt[d + e*x])/(2*a*c^2) + (d*e*(d + e*x)^(3/2))/(2*a*c) + ((a*e + c*d*x)*(d + e*x)^(5/2)
)/(2*a*c*(a - c*x^2)) - ((Sqrt[c]*d - Sqrt[a]*e)^(5/2)*(2*Sqrt[c]*d + 5*Sqrt[a]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e
*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(4*a^(3/2)*c^(9/4)) + ((2*Sqrt[c]*d - 5*Sqrt[a]*e)*(Sqrt[c]*d + Sqrt[a]*e)^
(5/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(9/4))

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Rubi [A]  time = 0.534155, antiderivative size = 263, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {739, 825, 827, 1166, 208} $-\frac{\left (5 \sqrt{a} e+2 \sqrt{c} d\right ) \left (\sqrt{c} d-\sqrt{a} e\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} c^{9/4}}+\frac{\left (2 \sqrt{c} d-5 \sqrt{a} e\right ) \left (\sqrt{a} e+\sqrt{c} d\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} c^{9/4}}+\frac{e \sqrt{d+e x} \left (5 a e^2+c d^2\right )}{2 a c^2}+\frac{(d+e x)^{5/2} (a e+c d x)}{2 a c \left (a-c x^2\right )}+\frac{d e (d+e x)^{3/2}}{2 a c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(7/2)/(a - c*x^2)^2,x]

[Out]

(e*(c*d^2 + 5*a*e^2)*Sqrt[d + e*x])/(2*a*c^2) + (d*e*(d + e*x)^(3/2))/(2*a*c) + ((a*e + c*d*x)*(d + e*x)^(5/2)
)/(2*a*c*(a - c*x^2)) - ((Sqrt[c]*d - Sqrt[a]*e)^(5/2)*(2*Sqrt[c]*d + 5*Sqrt[a]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e
*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(4*a^(3/2)*c^(9/4)) + ((2*Sqrt[c]*d - 5*Sqrt[a]*e)*(Sqrt[c]*d + Sqrt[a]*e)^
(5/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(9/4))

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 825

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g*(d + e*x)^m)/
(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d + c*e*f)*x, x])/(a + c*x^2), x], x] /
; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && FractionQ[m] && GtQ[m, 0]

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)^{7/2}}{\left (a-c x^2\right )^2} \, dx &=\frac{(a e+c d x) (d+e x)^{5/2}}{2 a c \left (a-c x^2\right )}-\frac{\int \frac{(d+e x)^{3/2} \left (\frac{1}{2} \left (-2 c d^2+5 a e^2\right )+\frac{3}{2} c d e x\right )}{a-c x^2} \, dx}{2 a c}\\ &=\frac{d e (d+e x)^{3/2}}{2 a c}+\frac{(a e+c d x) (d+e x)^{5/2}}{2 a c \left (a-c x^2\right )}+\frac{\int \frac{\sqrt{d+e x} \left (c d \left (c d^2-4 a e^2\right )-\frac{1}{2} c e \left (c d^2+5 a e^2\right ) x\right )}{a-c x^2} \, dx}{2 a c^2}\\ &=\frac{e \left (c d^2+5 a e^2\right ) \sqrt{d+e x}}{2 a c^2}+\frac{d e (d+e x)^{3/2}}{2 a c}+\frac{(a e+c d x) (d+e x)^{5/2}}{2 a c \left (a-c x^2\right )}-\frac{\int \frac{-\frac{1}{2} c \left (c d^2-5 a e^2\right ) \left (2 c d^2+a e^2\right )-\frac{1}{2} c^2 d e \left (c d^2-13 a e^2\right ) x}{\sqrt{d+e x} \left (a-c x^2\right )} \, dx}{2 a c^3}\\ &=\frac{e \left (c d^2+5 a e^2\right ) \sqrt{d+e x}}{2 a c^2}+\frac{d e (d+e x)^{3/2}}{2 a c}+\frac{(a e+c d x) (d+e x)^{5/2}}{2 a c \left (a-c x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} c^2 d^2 e \left (c d^2-13 a e^2\right )-\frac{1}{2} c e \left (c d^2-5 a e^2\right ) \left (2 c d^2+a e^2\right )-\frac{1}{2} c^2 d e \left (c d^2-13 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 )}{a c^3}\\ &=\frac{e \left (c d^2+5 a e^2\right ) \sqrt{d+e x}}{2 a c^2}+\frac{d e (d+e x)^{3/2}}{2 a c}+\frac{(a e+c d x) (d+e x)^{5/2}}{2 a c \left (a-c x^2\right )}+\frac{\left (\left (2 \sqrt{c} d-5 \sqrt{a} e\right ) \left (\sqrt{c} d+\sqrt{a} e\right )^3\right ) \operatorname{Subst}\left (\int \frac{1}{c d+\sqrt{a} \sqrt{c} e-c x^2} \, dx,x,\sqrt{d+e x}\right )}{4 a^{3/2} c^{3/2}}-\frac{\left (\left (\sqrt{c} d-\sqrt{a} e\right )^3 \left (2 \sqrt{c} d+5 \sqrt{a} e\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 )}{4 a^{3/2} c^{3/2}}\\ &=\frac{e \left (c d^2+5 a e^2\right ) \sqrt{d+e x}}{2 a c^2}+\frac{d e (d+e x)^{3/2}}{2 a c}+\frac{(a e+c d x) (d+e x)^{5/2}}{2 a c \left (a-c x^2\right )}-\frac{\left (\sqrt{c} d-\sqrt{a} e\right )^{5/2} \left (2 \sqrt{c} d+5 \sqrt{a} e\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} c^{9/4}}+\frac{\left (2 \sqrt{c} d-5 \sqrt{a} e\right ) \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 )}{4 a^{3/2} c^{9/4}}\\ \end{align*}

Mathematica [A]  time = 0.512183, size = 251, normalized size = 0.95 $\frac{2 \sqrt{a} \sqrt [4]{c} \sqrt{d+e x} \left (5 a^2 e^3+a c e \left (3 d^2+3 d e x-4 e^2 x^2\right )+c^2 d^3 x\right )+\left (c x^2-a\right ) \left (5 \sqrt{a} e+2 \sqrt{c} d\right ) \left (\sqrt{c} d-\sqrt{a} e\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-\left (c x^2-a\right ) \left (2 \sqrt{c} d-5 \sqrt{a} e\right ) \left (\sqrt{a} e+\sqrt{c} d\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} c^{9/4} \left (a-c x^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^(7/2)/(a - c*x^2)^2,x]

[Out]

(2*Sqrt[a]*c^(1/4)*Sqrt[d + e*x]*(5*a^2*e^3 + c^2*d^3*x + a*c*e*(3*d^2 + 3*d*e*x - 4*e^2*x^2)) + (Sqrt[c]*d -
Sqrt[a]*e)^(5/2)*(2*Sqrt[c]*d + 5*Sqrt[a]*e)*(-a + c*x^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqr
t[a]*e]] - (2*Sqrt[c]*d - 5*Sqrt[a]*e)*(Sqrt[c]*d + Sqrt[a]*e)^(5/2)*(-a + c*x^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*
x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(9/4)*(a - c*x^2))

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Maple [B]  time = 0.226, size = 717, normalized size = 2.7 \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)^(7/2)/(-c*x^2+a)^2,x)

[Out]

2*e^3/c^2*(e*x+d)^(1/2)-3/2*e^3/c/(c*e^2*x^2-a*e^2)*d*(e*x+d)^(3/2)-1/2*e/(c*e^2*x^2-a*e^2)*d^3/a*(e*x+d)^(3/2
)-1/2*e^5/c^2/(c*e^2*x^2-a*e^2)*a*(e*x+d)^(1/2)+1/2*e/(c*e^2*x^2-a*e^2)/a*(e*x+d)^(1/2)*d^4-5/4*e^5/c*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))-9/4*e^3/(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+1/
2*e*c/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))*d^4+13/4*e^3/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))*d
-1/4*e/a/((-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^3-5/4*e^5
/c*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))-
9/4*e^3/(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+1/2*e*c/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))*d^4-13/4*e^3/c/((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+1/4*e/a/((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^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{7}{2}}}{{\left (c x^{2} - a\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(-c*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate((e*x + d)^(7/2)/(c*x^2 - a)^2, x)

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Fricas [B]  time = 3.51031, size = 4626, normalized size = 17.59 \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)^(7/2)/(-c*x^2+a)^2,x, algorithm="fricas")

[Out]

1/8*((a*c^3*x^2 - a^2*c^2)*sqrt((4*c^3*d^7 - 35*a*c^2*d^5*e^2 + 70*a^2*c*d^3*e^4 + 105*a^3*d*e^6 + a^3*c^4*sqr
t((1225*c^4*d^8*e^6 - 10780*a*c^3*d^6*e^8 + 21966*a^2*c^2*d^4*e^10 + 7700*a^3*c*d^2*e^12 + 625*a^4*e^14)/(a^3*
c^9)))/(a^3*c^4))*log((140*c^5*d^10*e^3 - 1771*a*c^4*d^8*e^5 + 6872*a^2*c^3*d^6*e^7 - 8366*a^3*c^2*d^4*e^9 + 2
500*a^4*c*d^2*e^11 + 625*a^5*e^13)*sqrt(e*x + d) + (35*a^2*c^5*d^6*e^4 + 21*a^3*c^4*d^4*e^6 - 795*a^4*c^3*d^2*
e^8 - 125*a^5*c^2*e^10 - 2*(a^3*c^8*d^3 - 4*a^4*c^7*d*e^2)*sqrt((1225*c^4*d^8*e^6 - 10780*a*c^3*d^6*e^8 + 2196
6*a^2*c^2*d^4*e^10 + 7700*a^3*c*d^2*e^12 + 625*a^4*e^14)/(a^3*c^9)))*sqrt((4*c^3*d^7 - 35*a*c^2*d^5*e^2 + 70*a
^2*c*d^3*e^4 + 105*a^3*d*e^6 + a^3*c^4*sqrt((1225*c^4*d^8*e^6 - 10780*a*c^3*d^6*e^8 + 21966*a^2*c^2*d^4*e^10 +
7700*a^3*c*d^2*e^12 + 625*a^4*e^14)/(a^3*c^9)))/(a^3*c^4))) - (a*c^3*x^2 - a^2*c^2)*sqrt((4*c^3*d^7 - 35*a*c^
2*d^5*e^2 + 70*a^2*c*d^3*e^4 + 105*a^3*d*e^6 + a^3*c^4*sqrt((1225*c^4*d^8*e^6 - 10780*a*c^3*d^6*e^8 + 21966*a^
2*c^2*d^4*e^10 + 7700*a^3*c*d^2*e^12 + 625*a^4*e^14)/(a^3*c^9)))/(a^3*c^4))*log((140*c^5*d^10*e^3 - 1771*a*c^4
*d^8*e^5 + 6872*a^2*c^3*d^6*e^7 - 8366*a^3*c^2*d^4*e^9 + 2500*a^4*c*d^2*e^11 + 625*a^5*e^13)*sqrt(e*x + d) - (
35*a^2*c^5*d^6*e^4 + 21*a^3*c^4*d^4*e^6 - 795*a^4*c^3*d^2*e^8 - 125*a^5*c^2*e^10 - 2*(a^3*c^8*d^3 - 4*a^4*c^7*
d*e^2)*sqrt((1225*c^4*d^8*e^6 - 10780*a*c^3*d^6*e^8 + 21966*a^2*c^2*d^4*e^10 + 7700*a^3*c*d^2*e^12 + 625*a^4*e
^14)/(a^3*c^9)))*sqrt((4*c^3*d^7 - 35*a*c^2*d^5*e^2 + 70*a^2*c*d^3*e^4 + 105*a^3*d*e^6 + a^3*c^4*sqrt((1225*c^
4*d^8*e^6 - 10780*a*c^3*d^6*e^8 + 21966*a^2*c^2*d^4*e^10 + 7700*a^3*c*d^2*e^12 + 625*a^4*e^14)/(a^3*c^9)))/(a^
3*c^4))) + (a*c^3*x^2 - a^2*c^2)*sqrt((4*c^3*d^7 - 35*a*c^2*d^5*e^2 + 70*a^2*c*d^3*e^4 + 105*a^3*d*e^6 - a^3*c
^4*sqrt((1225*c^4*d^8*e^6 - 10780*a*c^3*d^6*e^8 + 21966*a^2*c^2*d^4*e^10 + 7700*a^3*c*d^2*e^12 + 625*a^4*e^14)
/(a^3*c^9)))/(a^3*c^4))*log((140*c^5*d^10*e^3 - 1771*a*c^4*d^8*e^5 + 6872*a^2*c^3*d^6*e^7 - 8366*a^3*c^2*d^4*e
^9 + 2500*a^4*c*d^2*e^11 + 625*a^5*e^13)*sqrt(e*x + d) + (35*a^2*c^5*d^6*e^4 + 21*a^3*c^4*d^4*e^6 - 795*a^4*c^
3*d^2*e^8 - 125*a^5*c^2*e^10 + 2*(a^3*c^8*d^3 - 4*a^4*c^7*d*e^2)*sqrt((1225*c^4*d^8*e^6 - 10780*a*c^3*d^6*e^8
+ 21966*a^2*c^2*d^4*e^10 + 7700*a^3*c*d^2*e^12 + 625*a^4*e^14)/(a^3*c^9)))*sqrt((4*c^3*d^7 - 35*a*c^2*d^5*e^2
+ 70*a^2*c*d^3*e^4 + 105*a^3*d*e^6 - a^3*c^4*sqrt((1225*c^4*d^8*e^6 - 10780*a*c^3*d^6*e^8 + 21966*a^2*c^2*d^4*
e^10 + 7700*a^3*c*d^2*e^12 + 625*a^4*e^14)/(a^3*c^9)))/(a^3*c^4))) - (a*c^3*x^2 - a^2*c^2)*sqrt((4*c^3*d^7 - 3
5*a*c^2*d^5*e^2 + 70*a^2*c*d^3*e^4 + 105*a^3*d*e^6 - a^3*c^4*sqrt((1225*c^4*d^8*e^6 - 10780*a*c^3*d^6*e^8 + 21
966*a^2*c^2*d^4*e^10 + 7700*a^3*c*d^2*e^12 + 625*a^4*e^14)/(a^3*c^9)))/(a^3*c^4))*log((140*c^5*d^10*e^3 - 1771
*a*c^4*d^8*e^5 + 6872*a^2*c^3*d^6*e^7 - 8366*a^3*c^2*d^4*e^9 + 2500*a^4*c*d^2*e^11 + 625*a^5*e^13)*sqrt(e*x +
d) - (35*a^2*c^5*d^6*e^4 + 21*a^3*c^4*d^4*e^6 - 795*a^4*c^3*d^2*e^8 - 125*a^5*c^2*e^10 + 2*(a^3*c^8*d^3 - 4*a^
4*c^7*d*e^2)*sqrt((1225*c^4*d^8*e^6 - 10780*a*c^3*d^6*e^8 + 21966*a^2*c^2*d^4*e^10 + 7700*a^3*c*d^2*e^12 + 625
*a^4*e^14)/(a^3*c^9)))*sqrt((4*c^3*d^7 - 35*a*c^2*d^5*e^2 + 70*a^2*c*d^3*e^4 + 105*a^3*d*e^6 - a^3*c^4*sqrt((1
225*c^4*d^8*e^6 - 10780*a*c^3*d^6*e^8 + 21966*a^2*c^2*d^4*e^10 + 7700*a^3*c*d^2*e^12 + 625*a^4*e^14)/(a^3*c^9)
))/(a^3*c^4))) + 4*(4*a*c*e^3*x^2 - 3*a*c*d^2*e - 5*a^2*e^3 - (c^2*d^3 + 3*a*c*d*e^2)*x)*sqrt(e*x + d))/(a*c^3
*x^2 - a^2*c^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)**(7/2)/(-c*x**2+a)**2,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)^(7/2)/(-c*x^2+a)^2,x, algorithm="giac")

[Out]

Timed out