### 3.26 $$\int (a x+b x^2)^{5/2} \, dx$$

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0359948, 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} $\frac{5 a^4 (a+2 b x) \sqrt{a x+b x^2}}{512 b^3}-\frac{5 a^2 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{192 b^2}-\frac{5 a^6 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{512 b^{7/2}}+\frac{(a+2 b x) \left (a x+b x^2\right )^{5/2}}{12 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*x + b*x^2)^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.183326, size = 120, normalized size = 1.02 $\frac{\sqrt{x (a+b x)} \left (\sqrt{b} \left (8 a^3 b^2 x^2+432 a^2 b^3 x^3-10 a^4 b x+15 a^5+640 a b^4 x^4+256 b^5 x^5\right )-\frac{15 a^{11/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{x} \sqrt{\frac{b x}{a}+1}}\right )}{1536 b^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*x + b*x^2)^(5/2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.047, size = 134, normalized size = 1.1 \begin{align*}{\frac{2\,bx+a}{12\,b} \left ( b{x}^{2}+ax \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{a}^{2}x}{96\,b} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{a}^{3}}{192\,{b}^{2}} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{a}^{4}x}{256\,{b}^{2}}\sqrt{b{x}^{2}+ax}}+{\frac{5\,{a}^{5}}{512\,{b}^{3}}\sqrt{b{x}^{2}+ax}}-{\frac{5\,{a}^{6}}{1024}\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){b}^{-{\frac{7}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.93928, size = 501, normalized size = 4.25 \begin{align*} \left [\frac{15 \, a^{6} \sqrt{b} \log \left (2 \, b x + a - 2 \, \sqrt{b x^{2} + a x} \sqrt{b}\right ) + 2 \,{\left (256 \, b^{6} x^{5} + 640 \, a b^{5} x^{4} + 432 \, a^{2} b^{4} x^{3} + 8 \, a^{3} b^{3} x^{2} - 10 \, a^{4} b^{2} x + 15 \, a^{5} b\right )} \sqrt{b x^{2} + a x}}{3072 \, b^{4}}, \frac{15 \, a^{6} \sqrt{-b} \arctan \left (\frac{\sqrt{b x^{2} + a x} \sqrt{-b}}{b x}\right ) +{\left (256 \, b^{6} x^{5} + 640 \, a b^{5} x^{4} + 432 \, a^{2} b^{4} x^{3} + 8 \, a^{3} b^{3} x^{2} - 10 \, a^{4} b^{2} x + 15 \, a^{5} b\right )} \sqrt{b x^{2} + a x}}{1536 \, b^{4}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x + b*x**2)**(5/2), x)

________________________________________________________________________________________

Giac [A]  time = 1.27501, size = 144, normalized size = 1.22 \begin{align*} \frac{5 \, a^{6} \log \left ({\left | -2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} \sqrt{b} - a \right |}\right )}{1024 \, b^{\frac{7}{2}}} + \frac{1}{1536} \, \sqrt{b x^{2} + a x}{\left (\frac{15 \, a^{5}}{b^{3}} - 2 \,{\left (\frac{5 \, a^{4}}{b^{2}} - 4 \,{\left (\frac{a^{3}}{b} + 2 \,{\left (27 \, a^{2} + 8 \,{\left (2 \, b^{2} x + 5 \, a b\right )} x\right )} x\right )} x\right )} x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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