3.12.65 \(\int (A+B x) \sqrt {b x+c x^2} \, dx\) [1165]

Optimal. Leaf size=97 \[ -\frac {(b B-2 A c) (b+2 c x) \sqrt {b x+c x^2}}{8 c^2}+\frac {B \left (b x+c x^2\right )^{3/2}}{3 c}+\frac {b^2 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{5/2}} \]

[Out]

1/3*B*(c*x^2+b*x)^(3/2)/c+1/8*b^2*(-2*A*c+B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(5/2)-1/8*(-2*A*c+B*b)*(
2*c*x+b)*(c*x^2+b*x)^(1/2)/c^2

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Rubi [A]
time = 0.03, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {654, 626, 634, 212} \begin {gather*} \frac {b^2 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{5/2}}-\frac {(b+2 c x) \sqrt {b x+c x^2} (b B-2 A c)}{8 c^2}+\frac {B \left (b x+c x^2\right )^{3/2}}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*Sqrt[b*x + c*x^2],x]

[Out]

-1/8*((b*B - 2*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/c^2 + (B*(b*x + c*x^2)^(3/2))/(3*c) + (b^2*(b*B - 2*A*c)*Ar
cTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*c^(5/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 634

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 654

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

Rubi steps

\begin {align*} \int (A+B x) \sqrt {b x+c x^2} \, dx &=\frac {B \left (b x+c x^2\right )^{3/2}}{3 c}+\frac {(-b B+2 A c) \int \sqrt {b x+c x^2} \, dx}{2 c}\\ &=-\frac {(b B-2 A c) (b+2 c x) \sqrt {b x+c x^2}}{8 c^2}+\frac {B \left (b x+c x^2\right )^{3/2}}{3 c}+\frac {\left (b^2 (b B-2 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{16 c^2}\\ &=-\frac {(b B-2 A c) (b+2 c x) \sqrt {b x+c x^2}}{8 c^2}+\frac {B \left (b x+c x^2\right )^{3/2}}{3 c}+\frac {\left (b^2 (b B-2 A c)\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{8 c^2}\\ &=-\frac {(b B-2 A c) (b+2 c x) \sqrt {b x+c x^2}}{8 c^2}+\frac {B \left (b x+c x^2\right )^{3/2}}{3 c}+\frac {b^2 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 109, normalized size = 1.12 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (-3 b^2 B+2 b c (3 A+B x)+4 c^2 x (3 A+2 B x)\right )-\frac {3 b^2 (b B-2 A c) \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{24 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-3*b^2*B + 2*b*c*(3*A + B*x) + 4*c^2*x*(3*A + 2*B*x)) - (3*b^2*(b*B - 2*A*c)*Log[
-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]])/(Sqrt[x]*Sqrt[b + c*x])))/(24*c^(5/2))

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Maple [A]
time = 0.54, size = 139, normalized size = 1.43

method	result	size
risch	\(\frac {\left (8 B \,c^{2} x^{2}+12 A \,c^{2} x +2 b B x c +6 A b c -3 b^{2} B \right ) x \left (c x +b \right )}{24 c^{2} \sqrt {x \left (c x +b \right )}}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) A}{8 c^{\frac {3}{2}}}+\frac {b^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) B}{16 c^{\frac {5}{2}}}\)	\(122\)
default	\(B \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )+A \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 154, normalized size = 1.59 \begin {gather*} \frac {1}{2} \, \sqrt {c x^{2} + b x} A x - \frac {\sqrt {c x^{2} + b x} B b x}{4 \, c} + \frac {B b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} - \frac {A b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} - \frac {\sqrt {c x^{2} + b x} B b^{2}}{8 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{3 \, c} + \frac {\sqrt {c x^{2} + b x} A b}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(c*x^2 + b*x)*A*x - 1/4*sqrt(c*x^2 + b*x)*B*b*x/c + 1/16*B*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqr
t(c))/c^(5/2) - 1/8*A*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) - 1/8*sqrt(c*x^2 + b*x)*B*b^2/c
^2 + 1/3*(c*x^2 + b*x)^(3/2)*B/c + 1/4*sqrt(c*x^2 + b*x)*A*b/c

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Fricas [A]
time = 2.48, size = 204, normalized size = 2.10 \begin {gather*} \left [-\frac {3 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{2} - 3 \, B b^{2} c + 6 \, A b c^{2} + 2 \, {\left (B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c^{3}}, -\frac {3 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (8 \, B c^{3} x^{2} - 3 \, B b^{2} c + 6 \, A b c^{2} + 2 \, {\left (B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*(B*b^3 - 2*A*b^2*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(8*B*c^3*x^2 - 3*B*b^2*
c + 6*A*b*c^2 + 2*(B*b*c^2 + 6*A*c^3)*x)*sqrt(c*x^2 + b*x))/c^3, -1/24*(3*(B*b^3 - 2*A*b^2*c)*sqrt(-c)*arctan(
sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (8*B*c^3*x^2 - 3*B*b^2*c + 6*A*b*c^2 + 2*(B*b*c^2 + 6*A*c^3)*x)*sqrt(c*x^2
+ b*x))/c^3]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {x \left (b + c x\right )} \left (A + B x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.71, size = 102, normalized size = 1.05 \begin {gather*} \frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, B x + \frac {B b c + 6 \, A c^{2}}{c^{2}}\right )} x - \frac {3 \, {\left (B b^{2} - 2 \, A b c\right )}}{c^{2}}\right )} - \frac {{\left (B b^{3} - 2 \, A b^{2} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x)*(2*(4*B*x + (B*b*c + 6*A*c^2)/c^2)*x - 3*(B*b^2 - 2*A*b*c)/c^2) - 1/16*(B*b^3 - 2*A*b^2
*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2)

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Mupad [B]
time = 1.69, size = 127, normalized size = 1.31 \begin {gather*} A\,\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )+\frac {B\,b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {B\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}-\frac {A\,b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/2)*(A + B*x),x)

[Out]

A*(b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) + (B*b^3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16*c^(5/2))
+ (B*(b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(24*c^2) - (A*b^2*log((b/2 + c*x)/c^(1/2) + (b*x + c*x
^2)^(1/2)))/(8*c^(3/2))

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