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

Optimal. Leaf size=101 \[ -\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{64 b^{3/2}}+\frac {5 a^2 (a+2 b x) \sqrt {a x+b x^2}}{64 b}+\frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x} \]

[Out]

5/24*a*(b*x^2+a*x)^(3/2)+1/4*(b*x^2+a*x)^(5/2)/x-5/64*a^4*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))/b^(3/2)+5/64*a^
2*(2*b*x+a)*(b*x^2+a*x)^(1/2)/b

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 101, 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 = {664, 612, 620, 206} \[ -\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{64 b^{3/2}}+\frac {5 a^2 (a+2 b x) \sqrt {a x+b x^2}}{64 b}+\frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x^2,x]

[Out]

(5*a^2*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(64*b) + (5*a*(a*x + b*x^2)^(3/2))/24 + (a*x + b*x^2)^(5/2)/(4*x) - (5*a
^4*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(64*b^(3/2))

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])

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 664

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^2} \, dx &=\frac {\left (a x+b x^2\right )^{5/2}}{4 x}+\frac {1}{8} (5 a) \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx\\ &=\frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x}+\frac {1}{16} \left (5 a^2\right ) \int \sqrt {a x+b x^2} \, dx\\ &=\frac {5 a^2 (a+2 b x) \sqrt {a x+b x^2}}{64 b}+\frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x}-\frac {\left (5 a^4\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx}{128 b}\\ &=\frac {5 a^2 (a+2 b x) \sqrt {a x+b x^2}}{64 b}+\frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x}-\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )}{64 b}\\ &=\frac {5 a^2 (a+2 b x) \sqrt {a x+b x^2}}{64 b}+\frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x}-\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{64 b^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 98, normalized size = 0.97 \[ \frac {\sqrt {x (a+b x)} \left (\sqrt {b} \left (15 a^3+118 a^2 b x+136 a b^2 x^2+48 b^3 x^3\right )-\frac {15 a^{7/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {x} \sqrt {\frac {b x}{a}+1}}\right )}{192 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^2,x]

[Out]

(Sqrt[x*(a + b*x)]*(Sqrt[b]*(15*a^3 + 118*a^2*b*x + 136*a*b^2*x^2 + 48*b^3*x^3) - (15*a^(7/2)*ArcSinh[(Sqrt[b]
*Sqrt[x])/Sqrt[a]])/(Sqrt[x]*Sqrt[1 + (b*x)/a])))/(192*b^(3/2))

________________________________________________________________________________________

fricas [A]  time = 1.01, size = 169, normalized size = 1.67 \[ \left [\frac {15 \, a^{4} \sqrt {b} \log \left (2 \, b x + a - 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (48 \, b^{4} x^{3} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x^{2} + a x}}{384 \, b^{2}}, \frac {15 \, a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) + {\left (48 \, b^{4} x^{3} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x^{2} + a x}}{192 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(15*a^4*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(48*b^4*x^3 + 136*a*b^3*x^2 + 118*a^2*
b^2*x + 15*a^3*b)*sqrt(b*x^2 + a*x))/b^2, 1/192*(15*a^4*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) + (4
8*b^4*x^3 + 136*a*b^3*x^2 + 118*a^2*b^2*x + 15*a^3*b)*sqrt(b*x^2 + a*x))/b^2]

________________________________________________________________________________________

giac [A]  time = 0.21, size = 84, normalized size = 0.83 \[ \frac {5 \, a^{4} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right )}{128 \, b^{\frac {3}{2}}} + \frac {1}{192} \, \sqrt {b x^{2} + a x} {\left (\frac {15 \, a^{3}}{b} + 2 \, {\left (59 \, a^{2} + 4 \, {\left (6 \, b^{2} x + 17 \, a b\right )} x\right )} x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/128*a^4*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/b^(3/2) + 1/192*sqrt(b*x^2 + a*x)*(15*a^3/b
+ 2*(59*a^2 + 4*(6*b^2*x + 17*a*b)*x)*x)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 135, normalized size = 1.34 \[ -\frac {5 a^{4} \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{128 b^{\frac {3}{2}}}+\frac {5 \sqrt {b \,x^{2}+a x}\, a^{2} x}{32}+\frac {5 \sqrt {b \,x^{2}+a x}\, a^{3}}{64 b}-\frac {5 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b x}{12}-\frac {5 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} a}{24}-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {5}{2}} b}{3 a}+\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{3 a \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/a/x^2*(b*x^2+a*x)^(7/2)-2/3/a*b*(b*x^2+a*x)^(5/2)-5/12*b*(b*x^2+a*x)^(3/2)*x-5/24*a*(b*x^2+a*x)^(3/2)+5/32
*a^2*(b*x^2+a*x)^(1/2)*x+5/64*a^3/b*(b*x^2+a*x)^(1/2)-5/128*a^4/b^(3/2)*ln((b*x+1/2*a)/b^(1/2)+(b*x^2+a*x)^(1/
2))

________________________________________________________________________________________

maxima [A]  time = 1.35, size = 98, normalized size = 0.97 \[ \frac {5}{32} \, \sqrt {b x^{2} + a x} a^{2} x - \frac {5 \, a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {3}{2}}} + \frac {5}{24} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a + \frac {5 \, \sqrt {b x^{2} + a x} a^{3}}{64 \, b} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{4 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/32*sqrt(b*x^2 + a*x)*a^2*x - 5/128*a^4*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) + 5/24*(b*x^2 +
a*x)^(3/2)*a + 5/64*sqrt(b*x^2 + a*x)*a^3/b + 1/4*(b*x^2 + a*x)^(5/2)/x

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^2)^(5/2)/x^2,x)

[Out]

int((a*x + b*x^2)^(5/2)/x^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________