3.1.24 \(\int x^2 (a x+b x^2)^{5/2} \, dx\) [24]

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

[Out]

-15/2048*a^4*(2*b*x+a)*(b*x^2+a*x)^(3/2)/b^4+3/128*a^2*(2*b*x+a)*(b*x^2+a*x)^(5/2)/b^3-9/112*a*(b*x^2+a*x)^(7/
2)/b^2+1/8*x*(b*x^2+a*x)^(7/2)/b-45/16384*a^8*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))/b^(11/2)+45/16384*a^6*(2*b*
x+a)*(b*x^2+a*x)^(1/2)/b^5

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Rubi [A]
time = 0.05, antiderivative size = 163, normalized size of antiderivative = 1.00, 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 = {684, 654, 626, 634, 212} \begin {gather*} -\frac {45 a^8 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{16384 b^{11/2}}+\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} \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 626

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 634

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 654

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 684

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.20, size = 155, normalized size = 0.95 \begin {gather*} \frac {\sqrt {b} x \left (315 a^8+105 a^7 b x-42 a^6 b^2 x^2+24 a^5 b^3 x^3-16 a^4 b^4 x^4+20864 a^3 b^5 x^5+54528 a^2 b^6 x^6+48128 a b^7 x^7+14336 b^8 x^8\right )+315 a^8 \sqrt {x} \sqrt {a+b x} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{114688 b^{11/2} \sqrt {x (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a*x + b*x^2)^(5/2),x]

[Out]

(Sqrt[b]*x*(315*a^8 + 105*a^7*b*x - 42*a^6*b^2*x^2 + 24*a^5*b^3*x^3 - 16*a^4*b^4*x^4 + 20864*a^3*b^5*x^5 + 545
28*a^2*b^6*x^6 + 48128*a*b^7*x^7 + 14336*b^8*x^8) + 315*a^8*Sqrt[x]*Sqrt[a + b*x]*Log[-(Sqrt[b]*Sqrt[x]) + Sqr
t[a + b*x]])/(114688*b^(11/2)*Sqrt[x*(a + b*x)])

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Maple [A]
time = 0.40, size = 165, normalized size = 1.01

method	result	size
risch	\(\frac {\left (14336 b^{7} x^{7}+33792 a \,b^{6} x^{6}+20736 a^{2} b^{5} x^{5}+128 a^{3} b^{4} x^{4}-144 a^{4} x^{3} b^{3}+168 a^{5} b^{2} x^{2}-210 a^{6} b x +315 a^{7}\right ) x \left (b x +a \right )}{114688 b^{5} \sqrt {x \left (b x +a \right )}}-\frac {45 a^{8} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{32768 b^{\frac {11}{2}}}\)	\(128\)
default	\(\frac {x \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{8 b}-\frac {9 a \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{7 b}-\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{12 b}-\frac {5 a^{2} \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{24 b}\right )}{2 b}\right )}{16 b}\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x*(b*x^2+a*x)^(7/2)/b-9/16*a/b*(1/7*(b*x^2+a*x)^(7/2)/b-1/2*a/b*(1/12*(2*b*x+a)*(b*x^2+a*x)^(5/2)/b-5/24*a
^2/b*(1/8*(2*b*x+a)/b*(b*x^2+a*x)^(3/2)-3/16*a^2/b*(1/4*(2*b*x+a)/b*(b*x^2+a*x)^(1/2)-1/8*a^2/b^(3/2)*ln((1/2*
a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))))))

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Maxima [A]
time = 0.27, size = 183, normalized size = 1.12 \begin {gather*} \frac {45 \, \sqrt {b x^{2} + a x} a^{6} x}{8192 \, b^{4}} - \frac {15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{4} x}{1024 \, b^{3}} + \frac {3 \, {\left (b x^{2} + a x\right )}^{\frac {5}{2}} a^{2} x}{64 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {7}{2}} x}{8 \, b} - \frac {45 \, a^{8} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{32768 \, b^{\frac {11}{2}}} + \frac {45 \, \sqrt {b x^{2} + a x} a^{7}}{16384 \, b^{5}} - \frac {15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{5}}{2048 \, b^{4}} + \frac {3 \, {\left (b x^{2} + a x\right )}^{\frac {5}{2}} a^{3}}{128 \, b^{3}} - \frac {9 \, {\left (b x^{2} + a x\right )}^{\frac {7}{2}} a}{112 \, b^{2}} \end {gather*}

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]

45/8192*sqrt(b*x^2 + a*x)*a^6*x/b^4 - 15/1024*(b*x^2 + a*x)^(3/2)*a^4*x/b^3 + 3/64*(b*x^2 + a*x)^(5/2)*a^2*x/b
^2 + 1/8*(b*x^2 + a*x)^(7/2)*x/b - 45/32768*a^8*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(11/2) + 45/163
84*sqrt(b*x^2 + a*x)*a^7/b^5 - 15/2048*(b*x^2 + a*x)^(3/2)*a^5/b^4 + 3/128*(b*x^2 + a*x)^(5/2)*a^3/b^3 - 9/112
*(b*x^2 + a*x)^(7/2)*a/b^2

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Fricas [A]
time = 1.39, size = 257, normalized size = 1.58 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \left (x \left (a + b x\right )\right )^{\frac {5}{2}}\, dx \end {gather*}

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 = 1.06, size = 131, normalized size = 0.80 \begin {gather*} \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 {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a*x + b*x^2)^(5/2),x)

[Out]

int(x^2*(a*x + b*x^2)^(5/2), x)

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