3.1243 \(\int \frac{A+B x}{\sqrt{d+e x} (b x+c x^2)^2} \, dx\)

Optimal. Leaf size=188 \[ \frac{\sqrt{c} \left (5 A b c e-4 A c^2 d-3 b^2 B e+2 b B c d\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 (c d-b e)^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) (-A b e-4 A c d+2 b B d)}{b^3 d^{3/2}}-\frac{\sqrt{d+e x} (c x (2 A c d-b (A e+B d))+A b (c d-b e))}{b^2 d \left (b x+c x^2\right ) (c d-b e)} \]

[Out]

-((Sqrt[d + e*x]*(A*b*(c*d - b*e) + c*(2*A*c*d - b*(B*d + A*e))*x))/(b^2*d*(c*d - b*e)*(b*x + c*x^2))) - ((2*b
*B*d - 4*A*c*d - A*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b^3*d^(3/2)) + (Sqrt[c]*(2*b*B*c*d - 4*A*c^2*d - 3*b^
2*B*e + 5*A*b*c*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*(c*d - b*e)^(3/2))

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Rubi [A]  time = 0.338604, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {822, 826, 1166, 208} \[ \frac{\sqrt{c} \left (5 A b c e-4 A c^2 d-3 b^2 B e+2 b B c d\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 (c d-b e)^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) (-A b e-4 A c d+2 b B d)}{b^3 d^{3/2}}-\frac{\sqrt{d+e x} (c x (2 A c d-b (A e+B d))+A b (c d-b e))}{b^2 d \left (b x+c x^2\right ) (c d-b e)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(Sqrt[d + e*x]*(b*x + c*x^2)^2),x]

[Out]

-((Sqrt[d + e*x]*(A*b*(c*d - b*e) + c*(2*A*c*d - b*(B*d + A*e))*x))/(b^2*d*(c*d - b*e)*(b*x + c*x^2))) - ((2*b
*B*d - 4*A*c*d - A*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b^3*d^(3/2)) + (Sqrt[c]*(2*b*B*c*d - 4*A*c^2*d - 3*b^
2*B*e + 5*A*b*c*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*(c*d - b*e)^(3/2))

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], 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] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

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

Mathematica [A]  time = 0.491969, size = 231, normalized size = 1.23 \[ \frac{-\frac{2 \sqrt{c} d \left (-b c (5 A e+2 B d)+4 A c^2 d+3 b^2 B e\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^2 (c d-b e)^{3/2}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) (-A b e-4 A c d+2 b B d)}{b^2 \sqrt{d}}+\frac{c \sqrt{d+e x} (A b e+4 A c d-2 b B d)}{b (b+c x) (b e-c d)}+\frac{3 A c e \sqrt{d+e x}}{(b+c x) (c d-b e)}-\frac{2 A \sqrt{d+e x}}{x (b+c x)}}{2 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(Sqrt[d + e*x]*(b*x + c*x^2)^2),x]

[Out]

((3*A*c*e*Sqrt[d + e*x])/((c*d - b*e)*(b + c*x)) + (c*(-2*b*B*d + 4*A*c*d + A*b*e)*Sqrt[d + e*x])/(b*(-(c*d) +
b*e)*(b + c*x)) - (2*A*Sqrt[d + e*x])/(x*(b + c*x)) - (2*(2*b*B*d - 4*A*c*d - A*b*e)*ArcTanh[Sqrt[d + e*x]/Sq
rt[d]])/(b^2*Sqrt[d]) - (2*Sqrt[c]*d*(4*A*c^2*d + 3*b^2*B*e - b*c*(2*B*d + 5*A*e))*ArcTanh[(Sqrt[c]*Sqrt[d + e
*x])/Sqrt[c*d - b*e]])/(b^2*(c*d - b*e)^(3/2)))/(2*b*d)

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Maple [B]  time = 0.02, size = 370, normalized size = 2. \begin{align*} -{\frac{A}{{b}^{2}dx}\sqrt{ex+d}}+{\frac{Ae}{{b}^{2}}{\it Artanh} \left ({\sqrt{ex+d}{\frac{1}{\sqrt{d}}}} \right ){d}^{-{\frac{3}{2}}}}+4\,{\frac{Ac}{{b}^{3}\sqrt{d}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }-2\,{\frac{B}{{b}^{2}\sqrt{d}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+{\frac{e{c}^{2}A}{{b}^{2} \left ( be-cd \right ) \left ( cex+be \right ) }\sqrt{ex+d}}-{\frac{Bce}{b \left ( be-cd \right ) \left ( cex+be \right ) }\sqrt{ex+d}}+5\,{\frac{e{c}^{2}A}{{b}^{2} \left ( be-cd \right ) \sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-4\,{\frac{{c}^{3}Ad}{{b}^{3} \left ( be-cd \right ) \sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-3\,{\frac{Bce}{b \left ( be-cd \right ) \sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+2\,{\frac{B{c}^{2}d}{{b}^{2} \left ( be-cd \right ) \sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b^2*A/d*(e*x+d)^(1/2)/x+e/b^2/d^(3/2)*arctanh((e*x+d)^(1/2)/d^(1/2))*A+4/b^3/d^(1/2)*arctanh((e*x+d)^(1/2)/
d^(1/2))*A*c-2/b^2/d^(1/2)*arctanh((e*x+d)^(1/2)/d^(1/2))*B+e*c^2/b^2/(b*e-c*d)*(e*x+d)^(1/2)/(c*e*x+b*e)*A-e*
c/b/(b*e-c*d)*(e*x+d)^(1/2)/(c*e*x+b*e)*B+5*e*c^2/b^2/(b*e-c*d)/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b
*e-c*d)*c)^(1/2))*A-4*c^3/b^3/(b*e-c*d)/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*A*d-3*
e*c/b/(b*e-c*d)/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B+2*c^2/b^2/(b*e-c*d)/((b*e-c*
d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B*d

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 19.5048, size = 3193, normalized size = 16.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(((2*(B*b*c^2 - 2*A*c^3)*d^3 - (3*B*b^2*c - 5*A*b*c^2)*d^2*e)*x^2 + (2*(B*b^2*c - 2*A*b*c^2)*d^3 - (3*B*
b^3 - 5*A*b^2*c)*d^2*e)*x)*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)) + ((A*b^2*c*e^2 + 2*(B*b*c^2 - 2*A*c^3)*d^2 - (2*B*b^2*c - 3*A*b*c^2)*d*e)*x^2 + (A*b^
3*e^2 + 2*(B*b^2*c - 2*A*b*c^2)*d^2 - (2*B*b^3 - 3*A*b^2*c)*d*e)*x)*sqrt(d)*log((e*x + 2*sqrt(e*x + d)*sqrt(d)
+ 2*d)/x) + 2*(A*b^2*c*d^2 - A*b^3*d*e - (A*b^2*c*d*e + (B*b^2*c - 2*A*b*c^2)*d^2)*x)*sqrt(e*x + d))/((b^3*c^
2*d^3 - b^4*c*d^2*e)*x^2 + (b^4*c*d^3 - b^5*d^2*e)*x), 1/2*(2*((2*(B*b*c^2 - 2*A*c^3)*d^3 - (3*B*b^2*c - 5*A*b
*c^2)*d^2*e)*x^2 + (2*(B*b^2*c - 2*A*b*c^2)*d^3 - (3*B*b^3 - 5*A*b^2*c)*d^2*e)*x)*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)) - ((A*b^2*c*e^2 + 2*(B*b*c^2 - 2*A*c^3)*d^2 - (
2*B*b^2*c - 3*A*b*c^2)*d*e)*x^2 + (A*b^3*e^2 + 2*(B*b^2*c - 2*A*b*c^2)*d^2 - (2*B*b^3 - 3*A*b^2*c)*d*e)*x)*sqr
t(d)*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 2*(A*b^2*c*d^2 - A*b^3*d*e - (A*b^2*c*d*e + (B*b^2*c - 2*A
*b*c^2)*d^2)*x)*sqrt(e*x + d))/((b^3*c^2*d^3 - b^4*c*d^2*e)*x^2 + (b^4*c*d^3 - b^5*d^2*e)*x), 1/2*(2*((A*b^2*c
*e^2 + 2*(B*b*c^2 - 2*A*c^3)*d^2 - (2*B*b^2*c - 3*A*b*c^2)*d*e)*x^2 + (A*b^3*e^2 + 2*(B*b^2*c - 2*A*b*c^2)*d^2
- (2*B*b^3 - 3*A*b^2*c)*d*e)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) - ((2*(B*b*c^2 - 2*A*c^3)*d^3 - (3*
B*b^2*c - 5*A*b*c^2)*d^2*e)*x^2 + (2*(B*b^2*c - 2*A*b*c^2)*d^3 - (3*B*b^3 - 5*A*b^2*c)*d^2*e)*x)*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)) - 2*(A*b^2*c*d^2
- A*b^3*d*e - (A*b^2*c*d*e + (B*b^2*c - 2*A*b*c^2)*d^2)*x)*sqrt(e*x + d))/((b^3*c^2*d^3 - b^4*c*d^2*e)*x^2 +
(b^4*c*d^3 - b^5*d^2*e)*x), (((2*(B*b*c^2 - 2*A*c^3)*d^3 - (3*B*b^2*c - 5*A*b*c^2)*d^2*e)*x^2 + (2*(B*b^2*c -
2*A*b*c^2)*d^3 - (3*B*b^3 - 5*A*b^2*c)*d^2*e)*x)*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)) + ((A*b^2*c*e^2 + 2*(B*b*c^2 - 2*A*c^3)*d^2 - (2*B*b^2*c - 3*A*b*c^2)*d*e)*x^2 +
(A*b^3*e^2 + 2*(B*b^2*c - 2*A*b*c^2)*d^2 - (2*B*b^3 - 3*A*b^2*c)*d*e)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-
d)/d) - (A*b^2*c*d^2 - A*b^3*d*e - (A*b^2*c*d*e + (B*b^2*c - 2*A*b*c^2)*d^2)*x)*sqrt(e*x + d))/((b^3*c^2*d^3 -
b^4*c*d^2*e)*x^2 + (b^4*c*d^3 - b^5*d^2*e)*x)]

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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((B*x+A)/(e*x+d)**(1/2)/(c*x**2+b*x)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.45262, size = 425, normalized size = 2.26 \begin{align*} -\frac{{\left (2 \, B b c^{2} d - 4 \, A c^{3} d - 3 \, B b^{2} c e + 5 \, A b c^{2} e\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{{\left (b^{3} c d - b^{4} e\right )} \sqrt{-c^{2} d + b c e}} + \frac{{\left (x e + d\right )}^{\frac{3}{2}} B b c d e - 2 \,{\left (x e + d\right )}^{\frac{3}{2}} A c^{2} d e - \sqrt{x e + d} B b c d^{2} e + 2 \, \sqrt{x e + d} A c^{2} d^{2} e +{\left (x e + d\right )}^{\frac{3}{2}} A b c e^{2} - 2 \, \sqrt{x e + d} A b c d e^{2} + \sqrt{x e + d} A b^{2} e^{3}}{{\left (b^{2} c d^{2} - b^{3} d e\right )}{\left ({\left (x e + d\right )}^{2} c - 2 \,{\left (x e + d\right )} c d + c d^{2} +{\left (x e + d\right )} b e - b d e\right )}} + \frac{{\left (2 \, B b d - 4 \, A c d - A b e\right )} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b^{3} \sqrt{-d} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*B*b*c^2*d - 4*A*c^3*d - 3*B*b^2*c*e + 5*A*b*c^2*e)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/((b^3*c*d
- b^4*e)*sqrt(-c^2*d + b*c*e)) + ((x*e + d)^(3/2)*B*b*c*d*e - 2*(x*e + d)^(3/2)*A*c^2*d*e - sqrt(x*e + d)*B*b*
c*d^2*e + 2*sqrt(x*e + d)*A*c^2*d^2*e + (x*e + d)^(3/2)*A*b*c*e^2 - 2*sqrt(x*e + d)*A*b*c*d*e^2 + sqrt(x*e + d
)*A*b^2*e^3)/((b^2*c*d^2 - b^3*d*e)*((x*e + d)^2*c - 2*(x*e + d)*c*d + c*d^2 + (x*e + d)*b*e - b*d*e)) + (2*B*
b*d - 4*A*c*d - A*b*e)*arctan(sqrt(x*e + d)/sqrt(-d))/(b^3*sqrt(-d)*d)