3.305 \(\int (b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=118 \[ \frac{5 b^4 (b+2 c x) \sqrt{b x+c x^2}}{512 c^3}-\frac{5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}-\frac{5 b^6 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{512 c^{7/2}}+\frac{(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c} \]

[Out]

(5*b^4*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(512*c^3) - (5*b^2*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(192*c^2) + ((b + 2*
c*x)*(b*x + c*x^2)^(5/2))/(12*c) - (5*b^6*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(512*c^(7/2))

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Rubi [A]  time = 0.036886, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {612, 620, 206} \[ \frac{5 b^4 (b+2 c x) \sqrt{b x+c x^2}}{512 c^3}-\frac{5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}-\frac{5 b^6 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{512 c^{7/2}}+\frac{(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(5/2),x]

[Out]

(5*b^4*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(512*c^3) - (5*b^2*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(192*c^2) + ((b + 2*
c*x)*(b*x + c*x^2)^(5/2))/(12*c) - (5*b^6*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(512*c^(7/2))

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]

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])

Rubi steps

\begin{align*} \int \left (b x+c x^2\right )^{5/2} \, dx &=\frac{(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac{\left (5 b^2\right ) \int \left (b x+c x^2\right )^{3/2} \, dx}{24 c}\\ &=-\frac{5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}+\frac{\left (5 b^4\right ) \int \sqrt{b x+c x^2} \, dx}{128 c^2}\\ &=\frac{5 b^4 (b+2 c x) \sqrt{b x+c x^2}}{512 c^3}-\frac{5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac{\left (5 b^6\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{1024 c^3}\\ &=\frac{5 b^4 (b+2 c x) \sqrt{b x+c x^2}}{512 c^3}-\frac{5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac{\left (5 b^6\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{512 c^3}\\ &=\frac{5 b^4 (b+2 c x) \sqrt{b x+c x^2}}{512 c^3}-\frac{5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac{5 b^6 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{512 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.121959, size = 120, normalized size = 1.02 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (8 b^3 c^2 x^2+432 b^2 c^3 x^3-10 b^4 c x+15 b^5+640 b c^4 x^4+256 c^5 x^5\right )-\frac{15 b^{11/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{1536 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(5/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^5 - 10*b^4*c*x + 8*b^3*c^2*x^2 + 432*b^2*c^3*x^3 + 640*b*c^4*x^4 + 256*c^5*x
^5) - (15*b^(11/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(1536*c^(7/2))

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Maple [A]  time = 0.049, size = 134, normalized size = 1.1 \begin{align*}{\frac{2\,cx+b}{12\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{b}^{2}x}{96\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{b}^{3}}{192\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{b}^{4}x}{256\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{5\,{b}^{5}}{512\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{5\,{b}^{6}}{1024}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*c*x+b)*(c*x^2+b*x)^(5/2)/c-5/96*b^2/c*(c*x^2+b*x)^(3/2)*x-5/192*b^3/c^2*(c*x^2+b*x)^(3/2)+5/256*b^4/c^
2*(c*x^2+b*x)^(1/2)*x+5/512*b^5/c^3*(c*x^2+b*x)^(1/2)-5/1024*b^6/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1
/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.02157, size = 501, normalized size = 4.25 \begin{align*} \left [\frac{15 \, b^{6} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (256 \, c^{6} x^{5} + 640 \, b c^{5} x^{4} + 432 \, b^{2} c^{4} x^{3} + 8 \, b^{3} c^{3} x^{2} - 10 \, b^{4} c^{2} x + 15 \, b^{5} c\right )} \sqrt{c x^{2} + b x}}{3072 \, c^{4}}, \frac{15 \, b^{6} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (256 \, c^{6} x^{5} + 640 \, b c^{5} x^{4} + 432 \, b^{2} c^{4} x^{3} + 8 \, b^{3} c^{3} x^{2} - 10 \, b^{4} c^{2} x + 15 \, b^{5} c\right )} \sqrt{c x^{2} + b x}}{1536 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3072*(15*b^6*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(256*c^6*x^5 + 640*b*c^5*x^4 + 432*b^
2*c^4*x^3 + 8*b^3*c^3*x^2 - 10*b^4*c^2*x + 15*b^5*c)*sqrt(c*x^2 + b*x))/c^4, 1/1536*(15*b^6*sqrt(-c)*arctan(sq
rt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (256*c^6*x^5 + 640*b*c^5*x^4 + 432*b^2*c^4*x^3 + 8*b^3*c^3*x^2 - 10*b^4*c^2*
x + 15*b^5*c)*sqrt(c*x^2 + b*x))/c^4]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b x + c x^{2}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*x + c*x**2)**(5/2), x)

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Giac [A]  time = 1.35832, size = 144, normalized size = 1.22 \begin{align*} \frac{5 \, b^{6} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{1024 \, c^{\frac{7}{2}}} + \frac{1}{1536} \, \sqrt{c x^{2} + b x}{\left (\frac{15 \, b^{5}}{c^{3}} - 2 \,{\left (\frac{5 \, b^{4}}{c^{2}} - 4 \,{\left (\frac{b^{3}}{c} + 2 \,{\left (27 \, b^{2} + 8 \,{\left (2 \, c^{2} x + 5 \, b c\right )} x\right )} x\right )} x\right )} x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

5/1024*b^6*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2) + 1/1536*sqrt(c*x^2 + b*x)*(15*b^5
/c^3 - 2*(5*b^4/c^2 - 4*(b^3/c + 2*(27*b^2 + 8*(2*c^2*x + 5*b*c)*x)*x)*x)*x)