3.2.23 \(\int \frac {x (A+B x)}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=60 \[ \frac {2 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}-\frac {2 x (b B-A c)}{b c \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {777, 620, 206} \begin {gather*} \frac {2 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}-\frac {2 x (b B-A c)}{b c \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

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

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 777

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((2
*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x)*(a + b*x + c*x^2)^
(p + 1))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p
+ 3))/(c*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && N
eQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 79, normalized size = 1.32 \begin {gather*} \frac {2 \sqrt {c} x (A c-b B)+2 b^{3/2} B \sqrt {x} \sqrt {\frac {c x}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b c^{3/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

(2*Sqrt[c]*(-(b*B) + A*c)*x + 2*b^(3/2)*B*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(b*c^(
3/2)*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.37, size = 76, normalized size = 1.27 \begin {gather*} \frac {2 \sqrt {b x+c x^2} (A c-b B)}{b c (b+c x)}-\frac {B \log \left (-2 c^{3/2} \sqrt {b x+c x^2}+b c+2 c^2 x\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(A + B*x))/(b*x + c*x^2)^(3/2),x]

[Out]

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

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fricas [A]  time = 0.42, size = 164, normalized size = 2.73 \begin {gather*} \left [\frac {{\left (B b c x + B b^{2}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (B b c - A c^{2}\right )} \sqrt {c x^{2} + b x}}{b c^{3} x + b^{2} c^{2}}, -\frac {2 \, {\left ({\left (B b c x + B b^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (B b c - A c^{2}\right )} \sqrt {c x^{2} + b x}\right )}}{b c^{3} x + b^{2} c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{%%%{1,[1]%%%},[2,2]%%%}+%%%{%%{[-2,0]:[1,0,%%%{-1,[1]%%%}]%%},[
1,3]%%%}+%%%{1,[0,4]%%%} / %%%{%%%{1,[2]%%%},[2,0]%%%}+%%%{%%{[%%%{-2,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,1]
%%%}+%%%{%%%{1,[1]%%%},[0,2]%%%} Error: Bad Argument Value

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maple [A]  time = 0.05, size = 67, normalized size = 1.12 \begin {gather*} \frac {2 A x}{\sqrt {c \,x^{2}+b x}\, b}-\frac {2 B x}{\sqrt {c \,x^{2}+b x}\, c}+\frac {B \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x+A)/(c*x^2+b*x)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.87, size = 65, normalized size = 1.08 \begin {gather*} \frac {2 \, A x}{\sqrt {c x^{2} + b x} b} - \frac {2 \, B x}{\sqrt {c x^{2} + b x} c} + \frac {B \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

2*A*x/(sqrt(c*x^2 + b*x)*b) - 2*B*x/(sqrt(c*x^2 + b*x)*c) + B*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(
3/2)

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mupad [B]  time = 1.33, size = 64, normalized size = 1.07 \begin {gather*} \frac {B\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{c^{3/2}}+\frac {2\,A\,x}{b\,\sqrt {x\,\left (b+c\,x\right )}}-\frac {2\,B\,x}{c\,\sqrt {c\,x^2+b\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(A + B*x))/(b*x + c*x^2)^(3/2),x)

[Out]

(B*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/c^(3/2) + (2*A*x)/(b*(x*(b + c*x))^(1/2)) - (2*B*x)/(c*(b*x
+ c*x^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(x*(A + B*x)/(x*(b + c*x))**(3/2), x)

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