3.2.31 \(\int \frac {x^3 (A+B x)}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=84 \[ -\frac {2 x^3 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}}-\frac {2 B x}{c^2 \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {788, 652, 620, 206} \begin {gather*} -\frac {2 x^3 (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {2 B x}{c^2 \sqrt {b x+c x^2}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

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

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 652

Int[((d_.) + (e_.)*(x_))^2*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + b*x +
c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fr
eeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p,
-1]

Rule 788

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

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 99, normalized size = 1.18 \begin {gather*} \frac {x \left (2 \sqrt {c} x \left (A c^2 x-3 b^2 B-4 b B c x\right )+6 b^{3/2} B \sqrt {x} (b+c x) \sqrt {\frac {c x}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )\right )}{3 b c^{5/2} (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

(x*(2*Sqrt[c]*x*(-3*b^2*B - 4*b*B*c*x + A*c^2*x) + 6*b^(3/2)*B*Sqrt[x]*(b + c*x)*Sqrt[1 + (c*x)/b]*ArcSinh[(Sq
rt[c]*Sqrt[x])/Sqrt[b]]))/(3*b*c^(5/2)*(x*(b + c*x))^(3/2))

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IntegrateAlgebraic [A]  time = 0.44, size = 92, normalized size = 1.10 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (-A c^2 x+3 b^2 B+4 b B c x\right )}{3 b c^2 (b+c x)^2}-\frac {B \log \left (-2 c^{5/2} \sqrt {b x+c x^2}+b c^2+2 c^3 x\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^3*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

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

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fricas [A]  time = 0.43, size = 239, normalized size = 2.85 \begin {gather*} \left [\frac {3 \, {\left (B b c^{2} x^{2} + 2 \, B b^{2} c x + B b^{3}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (3 \, B b^{2} c + {\left (4 \, B b c^{2} - A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b c^{5} x^{2} + 2 \, b^{2} c^{4} x + b^{3} c^{3}\right )}}, -\frac {2 \, {\left (3 \, {\left (B b c^{2} x^{2} + 2 \, B b^{2} c x + B b^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (3 \, B b^{2} c + {\left (4 \, B b c^{2} - A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{3 \, {\left (b c^{5} x^{2} + 2 \, b^{2} c^{4} x + b^{3} c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

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

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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^3*(B*x+A)/(c*x^2+b*x)^(5/2),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,[2]%%%},[4,4]%%%}+%%%{%%{[%%%{-4,[1]%%%},0]:[1,0,%%%{-1,[
1]%%%}]%%},[3,5]%%%}+%%%{%%%{6,[1]%%%},[2,6]%%%}+%%%{%%{[-4,0]:[1,0,%%%{-1,[1]%%%}]%%},[1,7]%%%}+%%%{1,[0,8]%%
%} / %%%{%%%{1,[4]%%%},[4,0]%%%}+%%%{%%{[%%%{-4,[3]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,1]%%%}+%%%{%%%{6,[3]%%%}
,[2,2]%%%}+%%%{%%{[%%%{-4,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,3]%%%}+%%%{%%%{1,[2]%%%},[0,4]%%%} Error: Bad
Argument Value

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maple [B]  time = 0.05, size = 206, normalized size = 2.45 \begin {gather*} -\frac {B \,x^{3}}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}-\frac {A \,x^{2}}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}+\frac {B b \,x^{2}}{2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}-\frac {A b x}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}+\frac {B \,b^{2} x}{6 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{3}}+\frac {2 A x}{3 \sqrt {c \,x^{2}+b x}\, b c}-\frac {7 B x}{3 \sqrt {c \,x^{2}+b x}\, c^{2}}+\frac {B \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {5}{2}}}+\frac {A}{3 \sqrt {c \,x^{2}+b x}\, c^{2}}-\frac {B b}{6 \sqrt {c \,x^{2}+b x}\, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)/(c*x^2+b*x)^(5/2),x)

[Out]

-1/3*B*x^3/c/(c*x^2+b*x)^(3/2)+1/2*B*b/c^2*x^2/(c*x^2+b*x)^(3/2)+1/6*B*b^2/c^3/(c*x^2+b*x)^(3/2)*x-7/3*B*x/c^2
/(c*x^2+b*x)^(1/2)-1/6*B*b/c^3/(c*x^2+b*x)^(1/2)+B/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))-A*x^2/c/(
c*x^2+b*x)^(3/2)-1/3*A*b/c^2/(c*x^2+b*x)^(3/2)*x+2/3*A/b/c/(c*x^2+b*x)^(1/2)*x+1/3*A/c^2/(c*x^2+b*x)^(1/2)

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maxima [B]  time = 0.67, size = 221, normalized size = 2.63 \begin {gather*} -\frac {1}{3} \, B x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {b x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} - \frac {2 \, x}{\sqrt {c x^{2} + b x} b c} - \frac {1}{\sqrt {c x^{2} + b x} c^{2}}\right )} - \frac {A x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {4 \, B x}{3 \, \sqrt {c x^{2} + b x} c^{2}} - \frac {A b x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} + \frac {2 \, A x}{3 \, \sqrt {c x^{2} + b x} b c} + \frac {B \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {5}{2}}} + \frac {A}{3 \, \sqrt {c x^{2} + b x} c^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} B}{3 \, b c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(A + B*x))/(b*x + c*x^2)^(5/2),x)

[Out]

int((x^3*(A + B*x))/(b*x + c*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x+A)/(c*x**2+b*x)**(5/2),x)

[Out]

Integral(x**3*(A + B*x)/(x*(b + c*x))**(5/2), x)

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