3.625 \(\int \frac{(d+e x)^{5/2}}{(a-c x^2)^2} \, dx\)

Optimal. Leaf size=231 \[ -\frac{\left (3 \sqrt{a} e+2 \sqrt{c} d\right ) \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} c^{7/4}}+\frac{\left (2 \sqrt{c} d-3 \sqrt{a} e\right ) \left (\sqrt{a} e+\sqrt{c} d\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} c^{7/4}}+\frac{(d+e x)^{3/2} (a e+c d x)}{2 a c \left (a-c x^2\right )}+\frac{d e \sqrt{d+e x}}{2 a c} \]

[Out]

(d*e*Sqrt[d + e*x])/(2*a*c) + ((a*e + c*d*x)*(d + e*x)^(3/2))/(2*a*c*(a - c*x^2)) - ((Sqrt[c]*d - Sqrt[a]*e)^(
3/2)*(2*Sqrt[c]*d + 3*Sqrt[a]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(4*a^(3/2)*c^(7
/4)) + ((2*Sqrt[c]*d - 3*Sqrt[a]*e)*(Sqrt[c]*d + Sqrt[a]*e)^(3/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]
*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(7/4))

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Rubi [A]  time = 0.374772, antiderivative size = 231, 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, 825, 827, 1166, 208} \[ -\frac{\left (3 \sqrt{a} e+2 \sqrt{c} d\right ) \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} c^{7/4}}+\frac{\left (2 \sqrt{c} d-3 \sqrt{a} e\right ) \left (\sqrt{a} e+\sqrt{c} d\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} c^{7/4}}+\frac{(d+e x)^{3/2} (a e+c d x)}{2 a c \left (a-c x^2\right )}+\frac{d e \sqrt{d+e x}}{2 a c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(5/2)/(a - c*x^2)^2,x]

[Out]

(d*e*Sqrt[d + e*x])/(2*a*c) + ((a*e + c*d*x)*(d + e*x)^(3/2))/(2*a*c*(a - c*x^2)) - ((Sqrt[c]*d - Sqrt[a]*e)^(
3/2)*(2*Sqrt[c]*d + 3*Sqrt[a]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(4*a^(3/2)*c^(7
/4)) + ((2*Sqrt[c]*d - 3*Sqrt[a]*e)*(Sqrt[c]*d + Sqrt[a]*e)^(3/2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]
*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(7/4))

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

Mathematica [A]  time = 0.348846, size = 247, normalized size = 1.07 \[ \frac{2 \sqrt{a} c^{3/4} \sqrt{d+e x} \left (a e (2 d+e x)+c d^2 x\right )+\left (c x^2-a\right ) \sqrt{\sqrt{c} d-\sqrt{a} e} \left (\sqrt{a} \sqrt{c} d e-3 a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-\left (c x^2-a\right ) \sqrt{\sqrt{a} e+\sqrt{c} d} \left (-\sqrt{a} \sqrt{c} d e-3 a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} c^{7/4} \left (a-c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(5/2)/(a - c*x^2)^2,x]

[Out]

(2*Sqrt[a]*c^(3/4)*Sqrt[d + e*x]*(c*d^2*x + a*e*(2*d + e*x)) + Sqrt[Sqrt[c]*d - Sqrt[a]*e]*(2*c*d^2 + Sqrt[a]*
Sqrt[c]*d*e - 3*a*e^2)*(-a + c*x^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]] - Sqrt[Sqrt[c
]*d + Sqrt[a]*e]*(2*c*d^2 - Sqrt[a]*Sqrt[c]*d*e - 3*a*e^2)*(-a + c*x^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[S
qrt[c]*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(7/4)*(a - c*x^2))

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Maple [B]  time = 0.221, size = 575, normalized size = 2.5 \begin{align*} -{\frac{{e}^{3}}{ \left ( 2\,c{e}^{2}{x}^{2}-2\,a{e}^{2} \right ) c} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{e{d}^{2}}{ \left ( 2\,c{e}^{2}{x}^{2}-2\,a{e}^{2} \right ) a} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{{e}^{3}d}{ \left ( 2\,c{e}^{2}{x}^{2}-2\,a{e}^{2} \right ) c}\sqrt{ex+d}}+{\frac{e{d}^{3}}{ \left ( 2\,c{e}^{2}{x}^{2}-2\,a{e}^{2} \right ) a}\sqrt{ex+d}}-{{e}^{3}d\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ac{e}^{2}}}}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{\frac{ce{d}^{3}}{2\,a}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ac{e}^{2}}}}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{\frac{3\,{e}^{3}}{4\,c}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}}-{\frac{e{d}^{2}}{4\,a}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}}-{{e}^{3}d{\it Artanh} \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ac{e}^{2}}}}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{\frac{ce{d}^{3}}{2\,a}{\it Artanh} \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ac{e}^{2}}}}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}}-{\frac{3\,{e}^{3}}{4\,c}{\it Artanh} \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{\frac{e{d}^{2}}{4\,a}{\it Artanh} \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(-c*x^2+a)^2,x)

[Out]

-1/2*e^3/(c*e^2*x^2-a*e^2)/c*(e*x+d)^(3/2)-1/2*e/(c*e^2*x^2-a*e^2)/a*(e*x+d)^(3/2)*d^2-1/2*e^3/(c*e^2*x^2-a*e^
2)*d/c*(e*x+d)^(1/2)+1/2*e/(c*e^2*x^2-a*e^2)*d^3/a*(e*x+d)^(1/2)-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+1/2*e/a*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))*d^3+3/4*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))-1/4*e/a/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*ar
ctan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*d^2-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+1/2*e/a*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))*d^3-3/4*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))+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^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{5}{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)^(5/2)/(-c*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate((e*x + d)^(5/2)/(c*x^2 - a)^2, x)

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Fricas [B]  time = 2.34586, size = 2897, normalized size = 12.54 \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)^(5/2)/(-c*x^2+a)^2,x, algorithm="fricas")

[Out]

-1/8*((a*c^2*x^2 - a^2*c)*sqrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 + a^3*c^3*sqrt((25*c^2*d^4*e^6 - 90*
a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7)))/(a^3*c^3))*log(-(20*c^3*d^6*e^3 - 101*a*c^2*d^4*e^5 + 162*a^2*c*d^2*e^7
 - 81*a^3*e^9)*sqrt(e*x + d) + (5*a^2*c^3*d^3*e^4 - 9*a^3*c^2*d*e^6 - (2*a^3*c^6*d^2 - 3*a^4*c^5*e^2)*sqrt((25
*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7)))*sqrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 + a^3
*c^3*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7)))/(a^3*c^3))) - (a*c^2*x^2 - a^2*c)*sqrt((
4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 + a^3*c^3*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*
c^7)))/(a^3*c^3))*log(-(20*c^3*d^6*e^3 - 101*a*c^2*d^4*e^5 + 162*a^2*c*d^2*e^7 - 81*a^3*e^9)*sqrt(e*x + d) - (
5*a^2*c^3*d^3*e^4 - 9*a^3*c^2*d*e^6 - (2*a^3*c^6*d^2 - 3*a^4*c^5*e^2)*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 +
81*a^2*e^10)/(a^3*c^7)))*sqrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 + a^3*c^3*sqrt((25*c^2*d^4*e^6 - 90*a
*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7)))/(a^3*c^3))) + (a*c^2*x^2 - a^2*c)*sqrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*
a^2*d*e^4 - a^3*c^3*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7)))/(a^3*c^3))*log(-(20*c^3*d
^6*e^3 - 101*a*c^2*d^4*e^5 + 162*a^2*c*d^2*e^7 - 81*a^3*e^9)*sqrt(e*x + d) + (5*a^2*c^3*d^3*e^4 - 9*a^3*c^2*d*
e^6 + (2*a^3*c^6*d^2 - 3*a^4*c^5*e^2)*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7)))*sqrt((4
*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 - a^3*c^3*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c
^7)))/(a^3*c^3))) - (a*c^2*x^2 - a^2*c)*sqrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a^2*d*e^4 - a^3*c^3*sqrt((25*c^2
*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7)))/(a^3*c^3))*log(-(20*c^3*d^6*e^3 - 101*a*c^2*d^4*e^5 + 162
*a^2*c*d^2*e^7 - 81*a^3*e^9)*sqrt(e*x + d) - (5*a^2*c^3*d^3*e^4 - 9*a^3*c^2*d*e^6 + (2*a^3*c^6*d^2 - 3*a^4*c^5
*e^2)*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7)))*sqrt((4*c^2*d^5 - 15*a*c*d^3*e^2 + 15*a
^2*d*e^4 - a^3*c^3*sqrt((25*c^2*d^4*e^6 - 90*a*c*d^2*e^8 + 81*a^2*e^10)/(a^3*c^7)))/(a^3*c^3))) + 4*(2*a*d*e +
 (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(a*c^2*x^2 - a^2*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)**(5/2)/(-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)^(5/2)/(-c*x^2+a)^2,x, algorithm="giac")

[Out]

Timed out