3.539 \(\int \frac{(d+e x)^{3/2}}{x^2 (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=403 \[ -\frac{\sqrt{2} \sqrt{c} \left (-2 a \left (e \left (d \sqrt{b^2-4 a c}-a e\right )+c d^2\right )+b d \left (d \sqrt{b^2-4 a c}-2 a e\right )+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\sqrt{2} \sqrt{c} \left (-2 a \left (c d^2-e \left (d \sqrt{b^2-4 a c}+a e\right )\right )-b d \left (d \sqrt{b^2-4 a c}+2 a e\right )+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{2 \sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{a^2}-\frac{d \sqrt{d+e x}}{a x}+\frac{\sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 3.07458, antiderivative size = 402, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {897, 1287, 199, 206, 1166, 208} \[ -\frac{\sqrt{2} \sqrt{c} \left (b d \left (d \sqrt{b^2-4 a c}-2 a e\right )-2 a e \left (d \sqrt{b^2-4 a c}-a e\right )-2 a c d^2+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\sqrt{2} \sqrt{c} \left (-b d \left (d \sqrt{b^2-4 a c}+2 a e\right )+2 a e \left (d \sqrt{b^2-4 a c}+a e\right )-2 a c d^2+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{2 \sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{a^2}-\frac{d \sqrt{d+e x}}{a x}+\frac{\sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1287

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[Ex
pandIntegrand[((f*x)^m*(d + e*x^2)^q)/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^
2 - 4*a*c, 0] && IntegerQ[q] && IntegerQ[m]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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}}{x^2 \left (a+b x+c x^2\right )} \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{x^4}{\left (-\frac{d}{e}+\frac{x^2}{e}\right )^2 \left (\frac{c d^2-b d e+a e^2}{e^2}-\frac{(2 c d-b e) x^2}{e^2}+\frac{c x^4}{e^2}\right )} \, dx,x,\sqrt{d+e x}\right )}{e}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (\frac{d^2 e^2}{a \left (d-x^2\right )^2}-\frac{d e (-b d+2 a e)}{a^2 \left (d-x^2\right )}+\frac{e \left (-(b d-a e) \left (c d^2-b d e+a e^2\right )+c d (b d-2 a e) x^2\right )}{a^2 \left (c d^2-b d e+a e^2-(2 c d-b e) x^2+c x^4\right )}\right ) \, dx,x,\sqrt{d+e x}\right )}{e}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{-(b d-a e) \left (c d^2-b d e+a e^2\right )+c d (b d-2 a e) 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 )}{a^2}+\frac{\left (2 d^2 e\right ) \operatorname{Subst}\left (\int \frac{1}{\left (d-x^2\right )^2} \, dx,x,\sqrt{d+e x}\right )}{a}+\frac{(2 d (b d-2 a e)) \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d+e x}\right )}{a^2}\\ &=-\frac{d \sqrt{d+e x}}{a x}+\frac{2 \sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{a^2}+\frac{(d e) \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d+e x}\right )}{a}+\frac{\left (c \left (b^2 d^2-2 a c d^2+b d \left (\sqrt{b^2-4 a c} d-2 a e\right )-2 a e \left (\sqrt{b^2-4 a c} d-a 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 )}{a^2 \sqrt{b^2-4 a c}}-\frac{\left (c \left (b^2 d^2-2 a c d^2+2 a e \left (\sqrt{b^2-4 a c} d+a e\right )-b d \left (\sqrt{b^2-4 a c} d+2 a 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 )}{a^2 \sqrt{b^2-4 a c}}\\ &=-\frac{d \sqrt{d+e x}}{a x}+\frac{\sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{a}+\frac{2 \sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{a^2}-\frac{\sqrt{2} \sqrt{c} \left (b^2 d^2-2 a c d^2+b d \left (\sqrt{b^2-4 a c} d-2 a e\right )-2 a e \left (\sqrt{b^2-4 a c} d-a 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 )}{a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}+\frac{\sqrt{2} \sqrt{c} \left (b^2 d^2-2 a c d^2+2 a e \left (\sqrt{b^2-4 a c} d+a e\right )-b d \left (\sqrt{b^2-4 a c} d+2 a 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 )}{a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\\ \end{align*}

Mathematica [A]  time = 1.58731, size = 393, normalized size = 0.98 \[ \frac{\frac{\sqrt{2} \sqrt{c} \left (2 a \left (e \left (d \sqrt{b^2-4 a c}-a e\right )+c d^2\right )+b d \left (2 a e-d \sqrt{b^2-4 a c}\right )-b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{b^2-4 a c} \sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{\sqrt{2} \sqrt{c} \left (b d \left (d \sqrt{b^2-4 a c}+2 a e\right )-2 a e \left (d \sqrt{b^2-4 a c}+a e\right )+2 a c d^2-b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+2 \sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )-\frac{a d \sqrt{d+e x}}{x}+a \sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.293, size = 1215, normalized size = 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)/x^2/(c*x^2+b*x+a),x)

[Out]

-d*(e*x+d)^(1/2)/a/x-3*e*arctanh((e*x+d)^(1/2)/d^(1/2))*d^(1/2)/a+2*d^(3/2)/a^2*arctanh((e*x+d)^(1/2)/d^(1/2))
*b-2*e^3*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))+2*e^2/a*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*d+2*e/a*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))*d^2-e/a^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^2*d^2-2*e/a*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))*d+1/a^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*d^2-2*e^3*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))+2*e^2/a*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*d+2*e/a*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))*d^2-e/a^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^2*d^2+2*e/a*c*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-1/a^2*c*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^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/x^2/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

[Out]

Timed out