3.1617 \(\int \frac{b+2 c x}{(d+e x)^{5/2} (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=518 \[ \frac{2 \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\sqrt{d+e x} \left (a e^2-b d e+c d^2\right )^2}-\frac{\sqrt{2} \sqrt{c} \left (2 c^2 d \left (d \sqrt{b^2-4 a c}+4 a e\right )-2 c e \left (b d \sqrt{b^2-4 a c}+a e \sqrt{b^2-4 a c}+2 a b e+b^2 d\right )+b^2 e^2 \left (\sqrt{b^2-4 a c}+b\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )^2}+\frac{\sqrt{2} \sqrt{c} \left (-2 c^2 d \left (d \sqrt{b^2-4 a c}-4 a e\right )-2 c e \left (-b d \sqrt{b^2-4 a c}-a e \sqrt{b^2-4 a c}+2 a b e+b^2 d\right )+b^2 e^2 \left (b-\sqrt{b^2-4 a c}\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )^2}+\frac{2 (2 c d-b e)}{3 (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )} \]

[Out]

(2*(2*c*d - b*e))/(3*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^(3/2)) + (2*(2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e)))/
((c*d^2 - b*d*e + a*e^2)^2*Sqrt[d + e*x]) - (Sqrt[2]*Sqrt[c]*(b^2*(b + Sqrt[b^2 - 4*a*c])*e^2 + 2*c^2*d*(Sqrt[
b^2 - 4*a*c]*d + 4*a*e) - 2*c*e*(b^2*d + b*Sqrt[b^2 - 4*a*c]*d + 2*a*b*e + a*Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sq
rt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b - Sq
rt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)^2) + (Sqrt[2]*Sqrt[c]*(b^2*(b - Sqrt[b^2 - 4*a*c])*e^2 - 2*c^2*d*(
Sqrt[b^2 - 4*a*c]*d - 4*a*e) - 2*c*e*(b^2*d - b*Sqrt[b^2 - 4*a*c]*d + 2*a*b*e - a*Sqrt[b^2 - 4*a*c]*e))*ArcTan
h[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b
+ Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)^2)

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Rubi [A]  time = 1.90204, antiderivative size = 518, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {828, 826, 1166, 208} \[ \frac{2 \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\sqrt{d+e x} \left (a e^2-b d e+c d^2\right )^2}-\frac{\sqrt{2} \sqrt{c} \left (2 c^2 d \left (d \sqrt{b^2-4 a c}+4 a e\right )-2 c e \left (b d \sqrt{b^2-4 a c}+a e \sqrt{b^2-4 a c}+2 a b e+b^2 d\right )+b^2 e^2 \left (\sqrt{b^2-4 a c}+b\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )^2}+\frac{\sqrt{2} \sqrt{c} \left (-2 c^2 d \left (d \sqrt{b^2-4 a c}-4 a e\right )-2 c e \left (-b d \sqrt{b^2-4 a c}-a e \sqrt{b^2-4 a c}+2 a b e+b^2 d\right )+b^2 e^2 \left (b-\sqrt{b^2-4 a c}\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )^2}+\frac{2 (2 c d-b e)}{3 (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(2*c*d - b*e))/(3*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^(3/2)) + (2*(2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e)))/
((c*d^2 - b*d*e + a*e^2)^2*Sqrt[d + e*x]) - (Sqrt[2]*Sqrt[c]*(b^2*(b + Sqrt[b^2 - 4*a*c])*e^2 + 2*c^2*d*(Sqrt[
b^2 - 4*a*c]*d + 4*a*e) - 2*c*e*(b^2*d + b*Sqrt[b^2 - 4*a*c]*d + 2*a*b*e + a*Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sq
rt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b - Sq
rt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)^2) + (Sqrt[2]*Sqrt[c]*(b^2*(b - Sqrt[b^2 - 4*a*c])*e^2 - 2*c^2*d*(
Sqrt[b^2 - 4*a*c]*d - 4*a*e) - 2*c*e*(b^2*d - b*Sqrt[b^2 - 4*a*c]*d + 2*a*b*e - a*Sqrt[b^2 - 4*a*c]*e))*ArcTan
h[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b
+ Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)^2)

Rule 828

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[((
e*f - d*g)*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d
+ e*x)^(m + 1)*Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x])/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && FractionQ[m] && LtQ[m, -1]

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + 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{b+2 c x}{(d+e x)^{5/2} \left (a+b x+c x^2\right )} \, dx &=\frac{2 (2 c d-b e)}{3 \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}+\frac{\int \frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{(d+e x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{c d^2-b d e+a e^2}\\ &=\frac{2 (2 c d-b e)}{3 \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}+\frac{2 \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right )}{\left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}+\frac{\int \frac{-2 b^2 c d e+4 a c^2 d e+b^3 e^2+b c \left (c d^2-3 a e^2\right )+c \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) x}{\sqrt{d+e x} \left (a+b x+c x^2\right )} \, dx}{\left (c d^2-b d e+a e^2\right )^2}\\ &=\frac{2 (2 c d-b e)}{3 \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}+\frac{2 \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right )}{\left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}+\frac{2 \operatorname{Subst}\left (\int \frac{-c d \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right )+e \left (-2 b^2 c d e+4 a c^2 d e+b^3 e^2+b c \left (c d^2-3 a e^2\right )\right )+c \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) x^2}{c d^2-b d e+a e^2+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x}\right )}{\left (c d^2-b d e+a e^2\right )^2}\\ &=\frac{2 (2 c d-b e)}{3 \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}+\frac{2 \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right )}{\left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}-\frac{\left (c \left (b^2 \left (b-\sqrt{b^2-4 a c}\right ) e^2-2 c^2 d \left (\sqrt{b^2-4 a c} d-4 a e\right )-2 c e \left (b^2 d-b \sqrt{b^2-4 a c} d+2 a b e-a \sqrt{b^2-4 a c} e\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{\sqrt{b^2-4 a c} \left (c d^2-b d e+a e^2\right )^2}+\frac{\left (c \left (b^2 \left (b+\sqrt{b^2-4 a c}\right ) e^2+2 c^2 d \left (\sqrt{b^2-4 a c} d+4 a e\right )-2 c e \left (b^2 d+b \sqrt{b^2-4 a c} d+2 a b e+a \sqrt{b^2-4 a c} e\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{\sqrt{b^2-4 a c} \left (c d^2-b d e+a e^2\right )^2}\\ &=\frac{2 (2 c d-b e)}{3 \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}+\frac{2 \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right )}{\left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}-\frac{\sqrt{2} \sqrt{c} \left (b^2 \left (b+\sqrt{b^2-4 a c}\right ) e^2+2 c^2 d \left (\sqrt{b^2-4 a c} d+4 a e\right )-2 c e \left (b^2 d+b \sqrt{b^2-4 a c} d+2 a b e+a \sqrt{b^2-4 a c} e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )^2}+\frac{\sqrt{2} \sqrt{c} \left (b^2 \left (b-\sqrt{b^2-4 a c}\right ) e^2-2 c^2 d \left (\sqrt{b^2-4 a c} d-4 a e\right )-2 c e \left (b^2 d-b \sqrt{b^2-4 a c} d+2 a b e-a \sqrt{b^2-4 a c} e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 1.82452, size = 481, normalized size = 0.93 \[ \frac{2 \left (\frac{-6 c e (a e+b d)+3 b^2 e^2+6 c^2 d^2}{\sqrt{d+e x} \left (e (a e-b d)+c d^2\right )}-\frac{3 \sqrt{c} \left (-\frac{\left (2 c^2 d \left (d \sqrt{b^2-4 a c}+4 a e\right )-2 c e \left (b d \sqrt{b^2-4 a c}+a e \sqrt{b^2-4 a c}+2 a b e+b^2 d\right )+b^2 e^2 \left (\sqrt{b^2-4 a c}+b\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{\left (2 c^2 d \left (d \sqrt{b^2-4 a c}-4 a e\right )-2 c e \left (b d \sqrt{b^2-4 a c}+a e \sqrt{b^2-4 a c}-2 a b e+b^2 (-d)\right )+b^2 e^2 \left (\sqrt{b^2-4 a c}-b\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} \sqrt{b^2-4 a c} \left (e (b d-a e)-c d^2\right )}+\frac{2 c d-b e}{(d+e x)^{3/2}}\right )}{3 \left (e (a e-b d)+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((2*c*d - b*e)/(d + e*x)^(3/2) + (6*c^2*d^2 + 3*b^2*e^2 - 6*c*e*(b*d + a*e))/((c*d^2 + e*(-(b*d) + a*e))*Sq
rt[d + e*x]) - (3*Sqrt[c]*(-(((b^2*(b + Sqrt[b^2 - 4*a*c])*e^2 + 2*c^2*d*(Sqrt[b^2 - 4*a*c]*d + 4*a*e) - 2*c*e
*(b^2*d + b*Sqrt[b^2 - 4*a*c]*d + 2*a*b*e + a*Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sq
rt[2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e]])/Sqrt[2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e]) - ((b^2*(-b + Sqrt[b^2 - 4*a
*c])*e^2 + 2*c^2*d*(Sqrt[b^2 - 4*a*c]*d - 4*a*e) - 2*c*e*(-(b^2*d) + b*Sqrt[b^2 - 4*a*c]*d - 2*a*b*e + a*Sqrt[
b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]])/Sqrt[2*c*d
- (b + Sqrt[b^2 - 4*a*c])*e]))/(Sqrt[2]*Sqrt[b^2 - 4*a*c]*(-(c*d^2) + e*(b*d - a*e)))))/(3*(c*d^2 + e*(-(b*d)
+ a*e)))

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Maple [B]  time = 0.04, size = 1962, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(3/2)*b*e+4/3/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(3/2)*c*d-4/(a*e^2-b*d*e+c*d^2)^2/(
e*x+d)^(1/2)*a*c*e^2+2/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)^(1/2)*b^2*e^2-4/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)^(1/2)*b*c*d
*e+4/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)^(1/2)*c^2*d^2+4/(a*e^2-b*d*e+c*d^2)^2*c^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/
((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2)
)^(1/2))*c)^(1/2))*e^3*a*b-8/(a*e^2-b*d*e+c*d^2)^2*c^3/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*
a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*d
*e^2-1/(a*e^2-b*d*e+c*d^2)^2*c/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2
)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^3*e^3+2/(a*e^2-b*d*e+c*d^
2)^2*c^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2
)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*d*e^2+2/(a*e^2-b*d*e+c*d^2)^2*c^2*2^(1/2)/((-
b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(
1/2))*c)^(1/2))*a*e^2-1/(a*e^2-b*d*e+c*d^2)^2*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan
h((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*e^2+2/(a*e^2-b*d*e+c*d^2)^2*c^2
*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4
*a*c-b^2))^(1/2))*c)^(1/2))*b*d*e-2/(a*e^2-b*d*e+c*d^2)^2*c^3*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c
)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*d^2+4/(a*e^2-b*d*e+c*
d^2)^2*c^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2
)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*e^3*a*b-8/(a*e^2-b*d*e+c*d^2)^2*c^3/(-e^2*(4*a*c-b
^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(
-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*d*e^2-1/(a*e^2-b*d*e+c*d^2)^2*c/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*
c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)
^(1/2))*b^3*e^3+2/(a*e^2-b*d*e+c*d^2)^2*c^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1
/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*d*e^2-2/(a*e^
2-b*d*e+c*d^2)^2*c^2*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b
*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*e^2+1/(a*e^2-b*d*e+c*d^2)^2*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-
b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*e^2-2
/(a*e^2-b*d*e+c*d^2)^2*c^2*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/
2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b*d*e+2/(a*e^2-b*d*e+c*d^2)^2*c^3*2^(1/2)/((b*e-2*c*d+(-e^2
*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^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{2 \, c x + b}{{\left (c x^{2} + b x + a\right )}{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)/(e*x+d)^(5/2)/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [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((2*c*x+b)/(e*x+d)^(5/2)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

Timed out

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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((2*c*x+b)/(e*x+d)**(5/2)/(c*x**2+b*x+a),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((2*c*x+b)/(e*x+d)^(5/2)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

Timed out