3.14 \(\int (b x+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=89 \[ \frac {3 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{5/2}}-\frac {3 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c} \]

[Out]

1/8*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c+3/64*b^4*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(5/2)-3/64*b^2*(2*c*x+b)*(c*
x^2+b*x)^(1/2)/c^2

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Rubi [A]  time = 0.02, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {612, 620, 206} \[ -\frac {3 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}+\frac {3 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{5/2}}+\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2),x]

[Out]

(-3*b^2*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^2) + ((b + 2*c*x)*(b*x + c*x^2)^(3/2))/(8*c) + (3*b^4*ArcTanh[(Sq
rt[c]*x)/Sqrt[b*x + c*x^2]])/(64*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 612

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 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]

Rubi steps

\begin {align*} \int \left (b x+c x^2\right )^{3/2} \, dx &=\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {\left (3 b^2\right ) \int \sqrt {b x+c x^2} \, dx}{16 c}\\ &=-\frac {3 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}+\frac {\left (3 b^4\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c^2}\\ &=-\frac {3 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}+\frac {\left (3 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{64 c^2}\\ &=-\frac {3 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}+\frac {3 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 98, normalized size = 1.10 \[ \frac {\sqrt {x (b+c x)} \left (\frac {3 b^{7/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}+\sqrt {c} \left (-3 b^3+2 b^2 c x+24 b c^2 x^2+16 c^3 x^3\right )\right )}{64 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-3*b^3 + 2*b^2*c*x + 24*b*c^2*x^2 + 16*c^3*x^3) + (3*b^(7/2)*ArcSinh[(Sqrt[c]*Sqr
t[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(64*c^(5/2))

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fricas [A]  time = 1.02, size = 170, normalized size = 1.91 \[ \left [\frac {3 \, b^{4} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} + 2 \, b^{2} c^{2} x - 3 \, b^{3} c\right )} \sqrt {c x^{2} + b x}}{128 \, c^{3}}, -\frac {3 \, b^{4} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} + 2 \, b^{2} c^{2} x - 3 \, b^{3} c\right )} \sqrt {c x^{2} + b x}}{64 \, c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/128*(3*b^4*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(16*c^4*x^3 + 24*b*c^3*x^2 + 2*b^2*c^2*
x - 3*b^3*c)*sqrt(c*x^2 + b*x))/c^3, -1/64*(3*b^4*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (16*c^4*
x^3 + 24*b*c^3*x^2 + 2*b^2*c^2*x - 3*b^3*c)*sqrt(c*x^2 + b*x))/c^3]

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giac [A]  time = 0.21, size = 83, normalized size = 0.93 \[ -\frac {3 \, b^{4} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {5}{2}}} + \frac {1}{64} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, c x + 3 \, b\right )} x + \frac {b^{2}}{c}\right )} x - \frac {3 \, b^{3}}{c^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/128*b^4*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2) + 1/64*sqrt(c*x^2 + b*x)*(2*(4*(2*
c*x + 3*b)*x + b^2/c)*x - 3*b^3/c^2)

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maple [A]  time = 0.04, size = 95, normalized size = 1.07 \[ \frac {3 b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {5}{2}}}-\frac {3 \sqrt {c \,x^{2}+b x}\, b^{2} x}{32 c}-\frac {3 \sqrt {c \,x^{2}+b x}\, b^{3}}{64 c^{2}}+\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c-3/32*b^2/c*(c*x^2+b*x)^(1/2)*x-3/64*b^3/c^2*(c*x^2+b*x)^(1/2)+3/128*b^4/c^(5
/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 1.36, size = 102, normalized size = 1.15 \[ \frac {1}{4} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} x - \frac {3 \, \sqrt {c x^{2} + b x} b^{2} x}{32 \, c} + \frac {3 \, b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {5}{2}}} - \frac {3 \, \sqrt {c x^{2} + b x} b^{3}}{64 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b}{8 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c*x^2 + b*x)^(3/2)*x - 3/32*sqrt(c*x^2 + b*x)*b^2*x/c + 3/128*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqr
t(c))/c^(5/2) - 3/64*sqrt(c*x^2 + b*x)*b^3/c^2 + 1/8*(c*x^2 + b*x)^(3/2)*b/c

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mupad [B]  time = 0.29, size = 87, normalized size = 0.98 \[ \frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (\frac {b}{2}+c\,x\right )}{4\,c}-\frac {3\,b^2\,\left (\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*x + c*x**2)**(3/2), x)

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