3.1.27 \(\int \frac {(a x+b x^2)^{5/2}}{x} \, dx\) [27]

Optimal. Leaf size=107 \[ -\frac {3 a^3 (a+2 b x) \sqrt {a x+b x^2}}{128 b^2}+\frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2}+\frac {3 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{128 b^{5/2}} \]

[Out]

1/16*a*(2*b*x+a)*(b*x^2+a*x)^(3/2)/b+1/5*(b*x^2+a*x)^(5/2)+3/128*a^5*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))/b^(5
/2)-3/128*a^3*(2*b*x+a)*(b*x^2+a*x)^(1/2)/b^2

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Rubi [A]
time = 0.03, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {678, 626, 634, 212} \begin {gather*} \frac {3 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{128 b^{5/2}}-\frac {3 a^3 (a+2 b x) \sqrt {a x+b x^2}}{128 b^2}+\frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x,x]

[Out]

(-3*a^3*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(128*b^2) + (a*(a + 2*b*x)*(a*x + b*x^2)^(3/2))/(16*b) + (a*x + b*x^2)^
(5/2)/5 + (3*a^5*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(128*b^(5/2))

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 678

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\left (a x+b x^2\right )^{5/2}}{x} \, dx &=\frac {1}{5} \left (a x+b x^2\right )^{5/2}+\frac {1}{2} a \int \left (a x+b x^2\right )^{3/2} \, dx\\ &=\frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2}-\frac {\left (3 a^3\right ) \int \sqrt {a x+b x^2} \, dx}{32 b}\\ &=-\frac {3 a^3 (a+2 b x) \sqrt {a x+b x^2}}{128 b^2}+\frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2}+\frac {\left (3 a^5\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx}{256 b^2}\\ &=-\frac {3 a^3 (a+2 b x) \sqrt {a x+b x^2}}{128 b^2}+\frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2}+\frac {\left (3 a^5\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )}{128 b^2}\\ &=-\frac {3 a^3 (a+2 b x) \sqrt {a x+b x^2}}{128 b^2}+\frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2}+\frac {3 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{128 b^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 110, normalized size = 1.03 \begin {gather*} \frac {\sqrt {x (a+b x)} \left (\sqrt {b} \left (-15 a^4+10 a^3 b x+248 a^2 b^2 x^2+336 a b^3 x^3+128 b^4 x^4\right )-\frac {15 a^5 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{\sqrt {x} \sqrt {a+b x}}\right )}{640 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x,x]

[Out]

(Sqrt[x*(a + b*x)]*(Sqrt[b]*(-15*a^4 + 10*a^3*b*x + 248*a^2*b^2*x^2 + 336*a*b^3*x^3 + 128*b^4*x^4) - (15*a^5*L
og[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(Sqrt[x]*Sqrt[a + b*x])))/(640*b^(5/2))

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Maple [A]
time = 0.41, size = 104, normalized size = 0.97

method	result	size
risch	\(-\frac {\left (-128 b^{4} x^{4}-336 a \,b^{3} x^{3}-248 a^{2} b^{2} x^{2}-10 a^{3} x b +15 a^{4}\right ) x \left (b x +a \right )}{640 b^{2} \sqrt {x \left (b x +a \right )}}+\frac {3 a^{5} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{256 b^{\frac {5}{2}}}\)	\(95\)
default	\(\frac {\left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{5}+\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{2}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b*x^2+a*x)^(5/2)+1/2*a*(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 = 0.30, size = 118, normalized size = 1.10 \begin {gather*} \frac {1}{8} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x - \frac {3 \, \sqrt {b x^{2} + a x} a^{3} x}{64 \, b} + \frac {3 \, a^{5} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{256 \, b^{\frac {5}{2}}} + \frac {1}{5} \, {\left (b x^{2} + a x\right )}^{\frac {5}{2}} - \frac {3 \, \sqrt {b x^{2} + a x} a^{4}}{128 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x^2 + a*x)^(3/2)*a*x - 3/64*sqrt(b*x^2 + a*x)*a^3*x/b + 3/256*a^5*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*s
qrt(b))/b^(5/2) + 1/5*(b*x^2 + a*x)^(5/2) - 3/128*sqrt(b*x^2 + a*x)*a^4/b^2 + 1/16*(b*x^2 + a*x)^(3/2)*a^2/b

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Fricas [A]
time = 1.53, size = 192, normalized size = 1.79 \begin {gather*} \left [\frac {15 \, a^{5} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (128 \, b^{5} x^{4} + 336 \, a b^{4} x^{3} + 248 \, a^{2} b^{3} x^{2} + 10 \, a^{3} b^{2} x - 15 \, a^{4} b\right )} \sqrt {b x^{2} + a x}}{1280 \, b^{3}}, -\frac {15 \, a^{5} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) - {\left (128 \, b^{5} x^{4} + 336 \, a b^{4} x^{3} + 248 \, a^{2} b^{3} x^{2} + 10 \, a^{3} b^{2} x - 15 \, a^{4} b\right )} \sqrt {b x^{2} + a x}}{640 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1280*(15*a^5*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(128*b^5*x^4 + 336*a*b^4*x^3 + 248*a^
2*b^3*x^2 + 10*a^3*b^2*x - 15*a^4*b)*sqrt(b*x^2 + a*x))/b^3, -1/640*(15*a^5*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*
sqrt(-b)/(b*x)) - (128*b^5*x^4 + 336*a*b^4*x^3 + 248*a^2*b^3*x^2 + 10*a^3*b^2*x - 15*a^4*b)*sqrt(b*x^2 + a*x))
/b^3]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x*(a + b*x))**(5/2)/x, x)

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Giac [A]
time = 0.98, size = 96, normalized size = 0.90 \begin {gather*} -\frac {3 \, a^{5} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right )}{256 \, b^{\frac {5}{2}}} - \frac {1}{640} \, \sqrt {b x^{2} + a x} {\left (\frac {15 \, a^{4}}{b^{2}} - 2 \, {\left (\frac {5 \, a^{3}}{b} + 4 \, {\left (31 \, a^{2} + 2 \, {\left (8 \, b^{2} x + 21 \, a b\right )} x\right )} x\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/256*a^5*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/b^(5/2) - 1/640*sqrt(b*x^2 + a*x)*(15*a^4/
b^2 - 2*(5*a^3/b + 4*(31*a^2 + 2*(8*b^2*x + 21*a*b)*x)*x)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^2)^(5/2)/x,x)

[Out]

int((a*x + b*x^2)^(5/2)/x, x)

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