3.366 \(\int \frac{1}{(d+e x)^{3/2} (b x+c x^2)} \, dx\)

Optimal. Leaf size=102 \[ \frac{2 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{3/2}}-\frac{2 e}{d \sqrt{d+e x} (c d-b e)}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.31775, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {709, 826, 1166, 208} \[ \frac{2 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b (c d-b e)^{3/2}}-\frac{2 e}{d \sqrt{d+e x} (c d-b e)}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b d^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(3/2)*(b*x + c*x^2)),x]

[Out]

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

Rule 709

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

Mathematica [C]  time = 0.0270691, size = 81, normalized size = 0.79 \[ -\frac{2 \left (c d \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c (d+e x)}{c d-b e}\right )+(b e-c d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{e x}{d}+1\right )\right )}{b d \sqrt{d+e x} (c d-b e)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(3/2)*(b*x + c*x^2)),x]

[Out]

(-2*(c*d*Hypergeometric2F1[-1/2, 1, 1/2, (c*(d + e*x))/(c*d - b*e)] + (-(c*d) + b*e)*Hypergeometric2F1[-1/2, 1
, 1/2, 1 + (e*x)/d]))/(b*d*(c*d - b*e)*Sqrt[d + e*x])

________________________________________________________________________________________

Maple [A]  time = 0.231, size = 97, normalized size = 1. \begin{align*} 2\,{\frac{{c}^{2}}{ \left ( be-cd \right ) b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{1}{b{d}^{3/2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+2\,{\frac{e}{d \left ( be-cd \right ) \sqrt{ex+d}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(3/2)/(c*x^2+b*x),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.73322, size = 1553, normalized size = 15.23 \begin{align*} \left [-\frac{2 \, \sqrt{e x + d} b d e +{\left (c d^{2} e x + c d^{3}\right )} \sqrt{\frac{c}{c d - b e}} \log \left (\frac{c e x + 2 \, c d - b e - 2 \,{\left (c d - b e\right )} \sqrt{e x + d} \sqrt{\frac{c}{c d - b e}}}{c x + b}\right ) -{\left (c d^{2} - b d e +{\left (c d e - b e^{2}\right )} x\right )} \sqrt{d} \log \left (\frac{e x - 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right )}{b c d^{4} - b^{2} d^{3} e +{\left (b c d^{3} e - b^{2} d^{2} e^{2}\right )} x}, -\frac{2 \, \sqrt{e x + d} b d e - 2 \,{\left (c d^{2} e x + c d^{3}\right )} \sqrt{-\frac{c}{c d - b e}} \arctan \left (-\frac{{\left (c d - b e\right )} \sqrt{e x + d} \sqrt{-\frac{c}{c d - b e}}}{c e x + c d}\right ) -{\left (c d^{2} - b d e +{\left (c d e - b e^{2}\right )} x\right )} \sqrt{d} \log \left (\frac{e x - 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right )}{b c d^{4} - b^{2} d^{3} e +{\left (b c d^{3} e - b^{2} d^{2} e^{2}\right )} x}, -\frac{2 \, \sqrt{e x + d} b d e - 2 \,{\left (c d^{2} - b d e +{\left (c d e - b e^{2}\right )} x\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) +{\left (c d^{2} e x + c d^{3}\right )} \sqrt{\frac{c}{c d - b e}} \log \left (\frac{c e x + 2 \, c d - b e - 2 \,{\left (c d - b e\right )} \sqrt{e x + d} \sqrt{\frac{c}{c d - b e}}}{c x + b}\right )}{b c d^{4} - b^{2} d^{3} e +{\left (b c d^{3} e - b^{2} d^{2} e^{2}\right )} x}, -\frac{2 \,{\left (\sqrt{e x + d} b d e -{\left (c d^{2} e x + c d^{3}\right )} \sqrt{-\frac{c}{c d - b e}} \arctan \left (-\frac{{\left (c d - b e\right )} \sqrt{e x + d} \sqrt{-\frac{c}{c d - b e}}}{c e x + c d}\right ) -{\left (c d^{2} - b d e +{\left (c d e - b e^{2}\right )} x\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right )\right )}}{b c d^{4} - b^{2} d^{3} e +{\left (b c d^{3} e - b^{2} d^{2} e^{2}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

[-(2*sqrt(e*x + d)*b*d*e + (c*d^2*e*x + c*d^3)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e - 2*(c*d - b*e)*sq
rt(e*x + d)*sqrt(c/(c*d - b*e)))/(c*x + b)) - (c*d^2 - b*d*e + (c*d*e - b*e^2)*x)*sqrt(d)*log((e*x - 2*sqrt(e*
x + d)*sqrt(d) + 2*d)/x))/(b*c*d^4 - b^2*d^3*e + (b*c*d^3*e - b^2*d^2*e^2)*x), -(2*sqrt(e*x + d)*b*d*e - 2*(c*
d^2*e*x + c*d^3)*sqrt(-c/(c*d - b*e))*arctan(-(c*d - b*e)*sqrt(e*x + d)*sqrt(-c/(c*d - b*e))/(c*e*x + c*d)) -
(c*d^2 - b*d*e + (c*d*e - b*e^2)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x))/(b*c*d^4 - b^2*d^3*e
+ (b*c*d^3*e - b^2*d^2*e^2)*x), -(2*sqrt(e*x + d)*b*d*e - 2*(c*d^2 - b*d*e + (c*d*e - b*e^2)*x)*sqrt(-d)*arct
an(sqrt(e*x + d)*sqrt(-d)/d) + (c*d^2*e*x + c*d^3)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e - 2*(c*d - b*e
)*sqrt(e*x + d)*sqrt(c/(c*d - b*e)))/(c*x + b)))/(b*c*d^4 - b^2*d^3*e + (b*c*d^3*e - b^2*d^2*e^2)*x), -2*(sqrt
(e*x + d)*b*d*e - (c*d^2*e*x + c*d^3)*sqrt(-c/(c*d - b*e))*arctan(-(c*d - b*e)*sqrt(e*x + d)*sqrt(-c/(c*d - b*
e))/(c*e*x + c*d)) - (c*d^2 - b*d*e + (c*d*e - b*e^2)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d))/(b*c*d^4 -
b^2*d^3*e + (b*c*d^3*e - b^2*d^2*e^2)*x)]

________________________________________________________________________________________

Sympy [A]  time = 17.9726, size = 94, normalized size = 0.92 \begin{align*} \frac{2 e}{d \sqrt{d + e x} \left (b e - c d\right )} + \frac{2 c \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{b e - c d}{c}}} \right )}}{b \sqrt{\frac{b e - c d}{c}} \left (b e - c d\right )} + \frac{2 \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- d}} \right )}}{b d \sqrt{- d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(3/2)/(c*x**2+b*x),x)

[Out]

2*e/(d*sqrt(d + e*x)*(b*e - c*d)) + 2*c*atan(sqrt(d + e*x)/sqrt((b*e - c*d)/c))/(b*sqrt((b*e - c*d)/c)*(b*e -
c*d)) + 2*atan(sqrt(d + e*x)/sqrt(-d))/(b*d*sqrt(-d))

________________________________________________________________________________________

Giac [A]  time = 1.25844, size = 153, normalized size = 1.5 \begin{align*} -\frac{2 \, c^{2} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{{\left (b c d - b^{2} e\right )} \sqrt{-c^{2} d + b c e}} - \frac{2 \, e}{{\left (c d^{2} - b d e\right )} \sqrt{x e + d}} + \frac{2 \, \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b \sqrt{-d} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c^2*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/((b*c*d - b^2*e)*sqrt(-c^2*d + b*c*e)) - 2*e/((c*d^2 - b*d
*e)*sqrt(x*e + d)) + 2*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)*d)