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

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

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 200, normalized size of antiderivative = 1.00, 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 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 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 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]

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]

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*}

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Mathematica [A]  time = 0.69, size = 208, normalized size = 1.04 \[ \frac {\sqrt {x (b+c x)} \left (\frac {\sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right ) \left (4 A c e (b e-2 c d)+B \left (-b^2 e^2-4 b c d e+8 c^2 d^2\right )\right )}{\sqrt {b} \sqrt {\frac {c x}{b}+1}}+\sqrt {c} \left (e \sqrt {x} (4 A c e+B (b e-4 c d+2 c e x))-\frac {8 c \sqrt {d} (B d-A e) \sqrt {c d-b e} \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {b+c x}}\right )\right )}{4 c^{3/2} e^3 \sqrt {x}} \]

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]*ArcTanh[(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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fricas [A]  time = 1.92, size = 800, normalized size = 4.00 \[ \left [\frac {{\left (8 \, B c^{2} d^{2} - 4 \, {\left (B b c + 2 \, A c^{2}\right )} d e - {\left (B b^{2} - 4 \, A b c\right )} e^{2}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 8 \, {\left (B c^{2} d - A c^{2} e\right )} \sqrt {c d^{2} - b d e} \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right ) + 2 \, {\left (2 \, B c^{2} e^{2} x - 4 \, B c^{2} d e + {\left (B b c + 4 \, A c^{2}\right )} e^{2}\right )} \sqrt {c x^{2} + b x}}{8 \, c^{2} e^{3}}, -\frac {16 \, {\left (B c^{2} d - A c^{2} e\right )} \sqrt {-c d^{2} + b d e} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) - {\left (8 \, B c^{2} d^{2} - 4 \, {\left (B b c + 2 \, A c^{2}\right )} d e - {\left (B b^{2} - 4 \, A b c\right )} e^{2}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (2 \, B c^{2} e^{2} x - 4 \, B c^{2} d e + {\left (B b c + 4 \, A c^{2}\right )} e^{2}\right )} \sqrt {c x^{2} + b x}}{8 \, c^{2} e^{3}}, -\frac {{\left (8 \, B c^{2} d^{2} - 4 \, {\left (B b c + 2 \, A c^{2}\right )} d e - {\left (B b^{2} - 4 \, A b c\right )} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + 4 \, {\left (B c^{2} d - A c^{2} e\right )} \sqrt {c d^{2} - b d e} \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right ) - {\left (2 \, B c^{2} e^{2} x - 4 \, B c^{2} d e + {\left (B b c + 4 \, A c^{2}\right )} e^{2}\right )} \sqrt {c x^{2} + b x}}{4 \, c^{2} e^{3}}, -\frac {8 \, {\left (B c^{2} d - A c^{2} e\right )} \sqrt {-c d^{2} + b d e} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) + {\left (8 \, B c^{2} d^{2} - 4 \, {\left (B b c + 2 \, A c^{2}\right )} d e - {\left (B b^{2} - 4 \, A b c\right )} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, B c^{2} e^{2} x - 4 \, B c^{2} d e + {\left (B b c + 4 \, A c^{2}\right )} e^{2}\right )} \sqrt {c x^{2} + b x}}{4 \, c^{2} e^{3}}\right ] \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [B]  time = 0.06, size = 1069, normalized size = 5.34 \[ \text {result too large to display} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-2*c*d>0)', see `assume?` f
or more details)Is b*e-2*c*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right )}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{d + e x}\, dx \]

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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