3.1166 \(\int \frac{(A+B x) \sqrt{b x+c x^2}}{d+e x} \, dx\)

Optimal. Leaf size=200 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \left (4 A c e (2 c d-b e)-B \left (-b^2 e^2-4 b c d e+8 c^2 d^2\right )\right )}{4 c^{3/2} e^3}-\frac{\sqrt{b x+c x^2} (-4 A c e-b B e+4 B c d-2 B c e x)}{4 c e^2}-\frac{\sqrt{d} (B d-A e) \sqrt{c d-b e} \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{e^3} \]

[Out]

-((4*B*c*d - b*B*e - 4*A*c*e - 2*B*c*e*x)*Sqrt[b*x + c*x^2])/(4*c*e^2) - ((4*A*c*e*(2*c*d - b*e) - B*(8*c^2*d^
2 - 4*b*c*d*e - b^2*e^2))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2)*e^3) - (Sqrt[d]*(B*d - A*e)*Sqrt[
c*d - b*e]*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/e^3

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Rubi [A]  time = 0.276265, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {814, 843, 620, 206, 724} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \left (4 A c e (2 c d-b e)-B \left (-b^2 e^2-4 b c d e+8 c^2 d^2\right )\right )}{4 c^{3/2} e^3}-\frac{\sqrt{b x+c x^2} (-4 A c e-b B e+4 B c d-2 B c e x)}{4 c e^2}-\frac{\sqrt{d} (B d-A e) \sqrt{c d-b e} \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*B*c*d - b*B*e - 4*A*c*e - 2*B*c*e*x)*Sqrt[b*x + c*x^2])/(4*c*e^2) - ((4*A*c*e*(2*c*d - b*e) - B*(8*c^2*d^
2 - 4*b*c*d*e - b^2*e^2))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2)*e^3) - (Sqrt[d]*(B*d - A*e)*Sqrt[
c*d - b*e]*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/e^3

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

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

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{(A+B x) \sqrt{b x+c x^2}}{d+e x} \, dx &=-\frac{(4 B c d-b B e-4 A c e-2 B c e x) \sqrt{b x+c x^2}}{4 c e^2}-\frac{\int \frac{-\frac{1}{2} b d (4 B c d-b B e-4 A c e)+\frac{1}{2} \left (4 A c e (2 c d-b e)-B \left (8 c^2 d^2-4 b c d e-b^2 e^2\right )\right ) x}{(d+e x) \sqrt{b x+c x^2}} \, dx}{4 c e^2}\\ &=-\frac{(4 B c d-b B e-4 A c e-2 B c e x) \sqrt{b x+c x^2}}{4 c e^2}-\frac{(d (B d-A e) (c d-b e)) \int \frac{1}{(d+e x) \sqrt{b x+c x^2}} \, dx}{e^3}-\frac{\left (4 A c e (2 c d-b e)-B \left (8 c^2 d^2-4 b c d e-b^2 e^2\right )\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{8 c e^3}\\ &=-\frac{(4 B c d-b B e-4 A c e-2 B c e x) \sqrt{b x+c x^2}}{4 c e^2}+\frac{(2 d (B d-A e) (c d-b e)) \operatorname{Subst}\left (\int \frac{1}{4 c d^2-4 b d e-x^2} \, dx,x,\frac{-b d-(2 c d-b e) x}{\sqrt{b x+c x^2}}\right )}{e^3}-\frac{\left (4 A c e (2 c d-b e)-B \left (8 c^2 d^2-4 b c d e-b^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{4 c e^3}\\ &=-\frac{(4 B c d-b B e-4 A c e-2 B c e x) \sqrt{b x+c x^2}}{4 c e^2}-\frac{\left (4 A c e (2 c d-b e)-B \left (8 c^2 d^2-4 b c d e-b^2 e^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{3/2} e^3}-\frac{\sqrt{d} (B d-A e) \sqrt{c d-b e} \tanh ^{-1}\left (\frac{b d+(2 c d-b e) x}{2 \sqrt{d} \sqrt{c d-b e} \sqrt{b x+c x^2}}\right )}{e^3}\\ \end{align*}

Mathematica [A]  time = 0.974671, size = 208, normalized size = 1.04 \[ \frac{\sqrt{x (b+c x)} \left (\frac{\sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right ) \left (4 A c e (b e-2 c d)+B \left (-b^2 e^2-4 b c d e+8 c^2 d^2\right )\right )}{\sqrt{b} \sqrt{\frac{c x}{b}+1}}+\sqrt{c} \left (e \sqrt{x} (4 A c e+B (b e-4 c d+2 c e x))+\frac{8 c \sqrt{d} (B d-A e) \sqrt{b e-c d} \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )}{\sqrt{b+c x}}\right )\right )}{4 c^{3/2} e^3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(((4*A*c*e*(-2*c*d + b*e) + B*(8*c^2*d^2 - 4*b*c*d*e - b^2*e^2))*ArcSinh[(Sqrt[c]*Sqrt[x])/
Sqrt[b]])/(Sqrt[b]*Sqrt[1 + (c*x)/b]) + Sqrt[c]*(e*Sqrt[x]*(4*A*c*e + B*(-4*c*d + b*e + 2*c*e*x)) + (8*c*Sqrt[
d]*(B*d - A*e)*Sqrt[-(c*d) + b*e]*ArcTan[(Sqrt[-(c*d) + b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/Sqrt[b + c*x])
))/(4*c^(3/2)*e^3*Sqrt[x])

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Maple [B]  time = 0.02, size = 1069, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B/e*(c*x^2+b*x)^(1/2)*x+1/4*B/e/c*(c*x^2+b*x)^(1/2)*b-1/8*B/e*b^2/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*
x)^(1/2))+1/e*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2)*A-1/e^2*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+
d/e)-d*(b*e-c*d)/e^2)^(1/2)*B*d+1/2/e*ln((1/2*(b*e-2*c*d)/e+(x+d/e)*c)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d
/e)-d*(b*e-c*d)/e^2)^(1/2))/c^(1/2)*b*A-1/2/e^2*ln((1/2*(b*e-2*c*d)/e+(x+d/e)*c)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c
*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/c^(1/2)*b*B*d-1/e^2*ln((1/2*(b*e-2*c*d)/e+(x+d/e)*c)/c^(1/2)+((x+d/e)^2*
c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))*c^(1/2)*d*A+1/e^3*ln((1/2*(b*e-2*c*d)/e+(x+d/e)*c)/c^(1/2)+((x
+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))*c^(1/2)*d^2*B+1/e^2*d/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*
d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d
)/e^2)^(1/2))/(x+d/e))*b*A-1/e^3*d^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*(
-d*(b*e-c*d)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/(x+d/e))*b*B-1/e^3*d^2/(-d*
(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*((x+d/e)^2*c+(b*e
-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/(x+d/e))*c*A+1/e^4*d^3/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e
^2+(b*e-2*c*d)/e*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))
/(x+d/e))*c*B

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.19908, size = 1769, normalized size = 8.84 \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)*(c*x^2+b*x)^(1/2)/(e*x+d),x, algorithm="fricas")

[Out]

[1/8*((8*B*c^2*d^2 - 4*(B*b*c + 2*A*c^2)*d*e - (B*b^2 - 4*A*b*c)*e^2)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b
*x)*sqrt(c)) - 8*(B*c^2*d - A*c^2*e)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sq
rt(c*x^2 + b*x))/(e*x + d)) + 2*(2*B*c^2*e^2*x - 4*B*c^2*d*e + (B*b*c + 4*A*c^2)*e^2)*sqrt(c*x^2 + b*x))/(c^2*
e^3), -1/8*(16*(B*c^2*d - A*c^2*e)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/((c*d -
b*e)*x)) - (8*B*c^2*d^2 - 4*(B*b*c + 2*A*c^2)*d*e - (B*b^2 - 4*A*b*c)*e^2)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x
^2 + b*x)*sqrt(c)) - 2*(2*B*c^2*e^2*x - 4*B*c^2*d*e + (B*b*c + 4*A*c^2)*e^2)*sqrt(c*x^2 + b*x))/(c^2*e^3), -1/
4*((8*B*c^2*d^2 - 4*(B*b*c + 2*A*c^2)*d*e - (B*b^2 - 4*A*b*c)*e^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/
(c*x)) + 4*(B*c^2*d - A*c^2*e)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x
^2 + b*x))/(e*x + d)) - (2*B*c^2*e^2*x - 4*B*c^2*d*e + (B*b*c + 4*A*c^2)*e^2)*sqrt(c*x^2 + b*x))/(c^2*e^3), -1
/4*(8*(B*c^2*d - A*c^2*e)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/((c*d - b*e)*x))
+ (8*B*c^2*d^2 - 4*(B*b*c + 2*A*c^2)*d*e - (B*b^2 - 4*A*b*c)*e^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/
(c*x)) - (2*B*c^2*e^2*x - 4*B*c^2*d*e + (B*b*c + 4*A*c^2)*e^2)*sqrt(c*x^2 + b*x))/(c^2*e^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x \left (b + c x\right )} \left (A + B x\right )}{d + e x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError