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

Optimal. Leaf size=163 \[ \frac{45 a^6 (a+2 b x) \sqrt{a x+b x^2}}{16384 b^5}-\frac{15 a^4 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{2048 b^4}+\frac{3 a^2 (a+2 b x) \left (a x+b x^2\right )^{5/2}}{128 b^3}-\frac{45 a^8 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{16384 b^{11/2}}-\frac{9 a \left (a x+b x^2\right )^{7/2}}{112 b^2}+\frac{x \left (a x+b x^2\right )^{7/2}}{8 b} \]

[Out]

(45*a^6*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(16384*b^5) - (15*a^4*(a + 2*b*x)*(a*x + b*x^2)^(3/2))/(2048*b^4) + (3*
a^2*(a + 2*b*x)*(a*x + b*x^2)^(5/2))/(128*b^3) - (9*a*(a*x + b*x^2)^(7/2))/(112*b^2) + (x*(a*x + b*x^2)^(7/2))
/(8*b) - (45*a^8*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(16384*b^(11/2))

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Rubi [A]  time = 0.0736416, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {670, 640, 612, 620, 206} \[ \frac{45 a^6 (a+2 b x) \sqrt{a x+b x^2}}{16384 b^5}-\frac{15 a^4 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{2048 b^4}+\frac{3 a^2 (a+2 b x) \left (a x+b x^2\right )^{5/2}}{128 b^3}-\frac{45 a^8 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{16384 b^{11/2}}-\frac{9 a \left (a x+b x^2\right )^{7/2}}{112 b^2}+\frac{x \left (a x+b x^2\right )^{7/2}}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a*x + b*x^2)^(5/2),x]

[Out]

(45*a^6*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(16384*b^5) - (15*a^4*(a + 2*b*x)*(a*x + b*x^2)^(3/2))/(2048*b^4) + (3*
a^2*(a + 2*b*x)*(a*x + b*x^2)^(5/2))/(128*b^3) - (9*a*(a*x + b*x^2)^(7/2))/(112*b^2) + (x*(a*x + b*x^2)^(7/2))
/(8*b) - (45*a^8*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(16384*b^(11/2))

Rule 670

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

Rule 640

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

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 x^2 \left (a x+b x^2\right )^{5/2} \, dx &=\frac{x \left (a x+b x^2\right )^{7/2}}{8 b}-\frac{(9 a) \int x \left (a x+b x^2\right )^{5/2} \, dx}{16 b}\\ &=-\frac{9 a \left (a x+b x^2\right )^{7/2}}{112 b^2}+\frac{x \left (a x+b x^2\right )^{7/2}}{8 b}+\frac{\left (9 a^2\right ) \int \left (a x+b x^2\right )^{5/2} \, dx}{32 b^2}\\ &=\frac{3 a^2 (a+2 b x) \left (a x+b x^2\right )^{5/2}}{128 b^3}-\frac{9 a \left (a x+b x^2\right )^{7/2}}{112 b^2}+\frac{x \left (a x+b x^2\right )^{7/2}}{8 b}-\frac{\left (15 a^4\right ) \int \left (a x+b x^2\right )^{3/2} \, dx}{256 b^3}\\ &=-\frac{15 a^4 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{2048 b^4}+\frac{3 a^2 (a+2 b x) \left (a x+b x^2\right )^{5/2}}{128 b^3}-\frac{9 a \left (a x+b x^2\right )^{7/2}}{112 b^2}+\frac{x \left (a x+b x^2\right )^{7/2}}{8 b}+\frac{\left (45 a^6\right ) \int \sqrt{a x+b x^2} \, dx}{4096 b^4}\\ &=\frac{45 a^6 (a+2 b x) \sqrt{a x+b x^2}}{16384 b^5}-\frac{15 a^4 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{2048 b^4}+\frac{3 a^2 (a+2 b x) \left (a x+b x^2\right )^{5/2}}{128 b^3}-\frac{9 a \left (a x+b x^2\right )^{7/2}}{112 b^2}+\frac{x \left (a x+b x^2\right )^{7/2}}{8 b}-\frac{\left (45 a^8\right ) \int \frac{1}{\sqrt{a x+b x^2}} \, dx}{32768 b^5}\\ &=\frac{45 a^6 (a+2 b x) \sqrt{a x+b x^2}}{16384 b^5}-\frac{15 a^4 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{2048 b^4}+\frac{3 a^2 (a+2 b x) \left (a x+b x^2\right )^{5/2}}{128 b^3}-\frac{9 a \left (a x+b x^2\right )^{7/2}}{112 b^2}+\frac{x \left (a x+b x^2\right )^{7/2}}{8 b}-\frac{\left (45 a^8\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a x+b x^2}}\right )}{16384 b^5}\\ &=\frac{45 a^6 (a+2 b x) \sqrt{a x+b x^2}}{16384 b^5}-\frac{15 a^4 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{2048 b^4}+\frac{3 a^2 (a+2 b x) \left (a x+b x^2\right )^{5/2}}{128 b^3}-\frac{9 a \left (a x+b x^2\right )^{7/2}}{112 b^2}+\frac{x \left (a x+b x^2\right )^{7/2}}{8 b}-\frac{45 a^8 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{16384 b^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.206821, size = 142, normalized size = 0.87 \[ \frac{\sqrt{x (a+b x)} \left (\sqrt{b} \left (168 a^5 b^2 x^2-144 a^4 b^3 x^3+128 a^3 b^4 x^4+20736 a^2 b^5 x^5-210 a^6 b x+315 a^7+33792 a b^6 x^6+14336 b^7 x^7\right )-\frac{315 a^{15/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{x} \sqrt{\frac{b x}{a}+1}}\right )}{114688 b^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a*x + b*x^2)^(5/2),x]

[Out]

(Sqrt[x*(a + b*x)]*(Sqrt[b]*(315*a^7 - 210*a^6*b*x + 168*a^5*b^2*x^2 - 144*a^4*b^3*x^3 + 128*a^3*b^4*x^4 + 207
36*a^2*b^5*x^5 + 33792*a*b^6*x^6 + 14336*b^7*x^7) - (315*a^(15/2)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(Sqrt[x]
*Sqrt[1 + (b*x)/a])))/(114688*b^(11/2))

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Maple [A]  time = 0.053, size = 185, normalized size = 1.1 \begin{align*}{\frac{x}{8\,b} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}-{\frac{9\,a}{112\,{b}^{2}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}+{\frac{3\,{a}^{2}x}{64\,{b}^{2}} \left ( b{x}^{2}+ax \right ) ^{{\frac{5}{2}}}}+{\frac{3\,{a}^{3}}{128\,{b}^{3}} \left ( b{x}^{2}+ax \right ) ^{{\frac{5}{2}}}}-{\frac{15\,{a}^{4}x}{1024\,{b}^{3}} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}-{\frac{15\,{a}^{5}}{2048\,{b}^{4}} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}+{\frac{45\,{a}^{6}x}{8192\,{b}^{4}}\sqrt{b{x}^{2}+ax}}+{\frac{45\,{a}^{7}}{16384\,{b}^{5}}\sqrt{b{x}^{2}+ax}}-{\frac{45\,{a}^{8}}{32768}\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){b}^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x*(b*x^2+a*x)^(7/2)/b-9/112*a*(b*x^2+a*x)^(7/2)/b^2+3/64/b^2*a^2*(b*x^2+a*x)^(5/2)*x+3/128/b^3*a^3*(b*x^2+
a*x)^(5/2)-15/1024/b^3*a^4*(b*x^2+a*x)^(3/2)*x-15/2048/b^4*a^5*(b*x^2+a*x)^(3/2)+45/8192/b^4*a^6*(b*x^2+a*x)^(
1/2)*x+45/16384/b^5*a^7*(b*x^2+a*x)^(1/2)-45/32768/b^(11/2)*a^8*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*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(x^2*(b*x^2+a*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.10095, size = 633, normalized size = 3.88 \begin{align*} \left [\frac{315 \, a^{8} \sqrt{b} \log \left (2 \, b x + a - 2 \, \sqrt{b x^{2} + a x} \sqrt{b}\right ) + 2 \,{\left (14336 \, b^{8} x^{7} + 33792 \, a b^{7} x^{6} + 20736 \, a^{2} b^{6} x^{5} + 128 \, a^{3} b^{5} x^{4} - 144 \, a^{4} b^{4} x^{3} + 168 \, a^{5} b^{3} x^{2} - 210 \, a^{6} b^{2} x + 315 \, a^{7} b\right )} \sqrt{b x^{2} + a x}}{229376 \, b^{6}}, \frac{315 \, a^{8} \sqrt{-b} \arctan \left (\frac{\sqrt{b x^{2} + a x} \sqrt{-b}}{b x}\right ) +{\left (14336 \, b^{8} x^{7} + 33792 \, a b^{7} x^{6} + 20736 \, a^{2} b^{6} x^{5} + 128 \, a^{3} b^{5} x^{4} - 144 \, a^{4} b^{4} x^{3} + 168 \, a^{5} b^{3} x^{2} - 210 \, a^{6} b^{2} x + 315 \, a^{7} b\right )} \sqrt{b x^{2} + a x}}{114688 \, b^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/229376*(315*a^8*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(14336*b^8*x^7 + 33792*a*b^7*x^6 +
20736*a^2*b^6*x^5 + 128*a^3*b^5*x^4 - 144*a^4*b^4*x^3 + 168*a^5*b^3*x^2 - 210*a^6*b^2*x + 315*a^7*b)*sqrt(b*x
^2 + a*x))/b^6, 1/114688*(315*a^8*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) + (14336*b^8*x^7 + 33792*a
*b^7*x^6 + 20736*a^2*b^6*x^5 + 128*a^3*b^5*x^4 - 144*a^4*b^4*x^3 + 168*a^5*b^3*x^2 - 210*a^6*b^2*x + 315*a^7*b
)*sqrt(b*x^2 + a*x))/b^6]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22452, size = 177, normalized size = 1.09 \begin{align*} \frac{45 \, a^{8} \log \left ({\left | -2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} \sqrt{b} - a \right |}\right )}{32768 \, b^{\frac{11}{2}}} + \frac{1}{114688} \, \sqrt{b x^{2} + a x}{\left (\frac{315 \, a^{7}}{b^{5}} - 2 \,{\left (\frac{105 \, a^{6}}{b^{4}} - 4 \,{\left (\frac{21 \, a^{5}}{b^{3}} - 2 \,{\left (\frac{9 \, a^{4}}{b^{2}} - 8 \,{\left (\frac{a^{3}}{b} + 2 \,{\left (81 \, a^{2} + 4 \,{\left (14 \, b^{2} x + 33 \, a b\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

45/32768*a^8*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/b^(11/2) + 1/114688*sqrt(b*x^2 + a*x)*(3
15*a^7/b^5 - 2*(105*a^6/b^4 - 4*(21*a^5/b^3 - 2*(9*a^4/b^2 - 8*(a^3/b + 2*(81*a^2 + 4*(14*b^2*x + 33*a*b)*x)*x
)*x)*x)*x)*x)