3.1.30 \(\int \frac {(a x+b x^2)^{5/2}}{x^4} \, dx\) [30]

Optimal. Leaf size=92 \[ \frac {15}{4} a b \sqrt {a x+b x^2}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {15}{4} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) \]

[Out]

5/2*b*(b*x^2+a*x)^(3/2)/x-2*(b*x^2+a*x)^(5/2)/x^3+15/4*a^2*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))*b^(1/2)+15/4*a
*b*(b*x^2+a*x)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 92, 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 = {676, 678, 634, 212} \begin {gather*} \frac {15}{4} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}+\frac {15}{4} a b \sqrt {a x+b x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x^4,x]

[Out]

(15*a*b*Sqrt[a*x + b*x^2])/4 + (5*b*(a*x + b*x^2)^(3/2))/(2*x) - (2*(a*x + b*x^2)^(5/2))/x^3 + (15*a^2*Sqrt[b]
*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/4

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 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 676

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 + p + 1))), x] - Dist[c*(p/(e^2*(m + p + 1))), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

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^4} \, dx &=-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+(5 b) \int \frac {\left (a x+b x^2\right )^{3/2}}{x^2} \, dx\\ &=\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {1}{4} (15 a b) \int \frac {\sqrt {a x+b x^2}}{x} \, dx\\ &=\frac {15}{4} a b \sqrt {a x+b x^2}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {1}{8} \left (15 a^2 b\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx\\ &=\frac {15}{4} a b \sqrt {a x+b x^2}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {1}{4} \left (15 a^2 b\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )\\ &=\frac {15}{4} a b \sqrt {a x+b x^2}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {15}{4} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 92, normalized size = 1.00 \begin {gather*} \frac {\sqrt {a+b x} \left (\sqrt {a+b x} \left (-8 a^2+9 a b x+2 b^2 x^2\right )-15 a^2 \sqrt {b} \sqrt {x} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )\right )}{4 \sqrt {x (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^4,x]

[Out]

(Sqrt[a + b*x]*(Sqrt[a + b*x]*(-8*a^2 + 9*a*b*x + 2*b^2*x^2) - 15*a^2*Sqrt[b]*Sqrt[x]*Log[-(Sqrt[b]*Sqrt[x]) +
Sqrt[a + b*x]]))/(4*Sqrt[x*(a + b*x)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(181\) vs. \(2(74)=148\).
time = 0.40, size = 182, normalized size = 1.98

method	result	size
risch	\(-\frac {\left (b x +a \right ) \left (-2 b^{2} x^{2}-9 a b x +8 a^{2}\right )}{4 \sqrt {x \left (b x +a \right )}}+\frac {15 a^{2} \sqrt {b}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8}\)	\(69\)
default	\(-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{a \,x^{4}}+\frac {6 b \left (\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{a \,x^{3}}-\frac {8 b \left (\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{3 a \,x^{2}}-\frac {10 b \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{5}+\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{2}\right )}{3 a}\right )}{a}\right )}{a}\)	\(182\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a*x)^(5/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

-2/a/x^4*(b*x^2+a*x)^(7/2)+6*b/a*(2/a/x^3*(b*x^2+a*x)^(7/2)-8*b/a*(2/3/a/x^2*(b*x^2+a*x)^(7/2)-10/3*b/a*(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 = 0.28, size = 84, normalized size = 0.91 \begin {gather*} \frac {15}{8} \, a^{2} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {15 \, \sqrt {b x^{2} + a x} a^{2}}{4 \, x} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a}{4 \, x^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{2 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^4,x, algorithm="maxima")

[Out]

15/8*a^2*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 15/4*sqrt(b*x^2 + a*x)*a^2/x + 5/4*(b*x^2 + a*
x)^(3/2)*a/x^2 + 1/2*(b*x^2 + a*x)^(5/2)/x^3

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Fricas [A]
time = 2.18, size = 144, normalized size = 1.57 \begin {gather*} \left [\frac {15 \, a^{2} \sqrt {b} x \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{8 \, x}, -\frac {15 \, a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) - {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{4 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^4,x, algorithm="fricas")

[Out]

[1/8*(15*a^2*sqrt(b)*x*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(2*b^2*x^2 + 9*a*b*x - 8*a^2)*sqrt(b*x
^2 + a*x))/x, -1/4*(15*a^2*sqrt(-b)*x*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) - (2*b^2*x^2 + 9*a*b*x - 8*a^2)
*sqrt(b*x^2 + a*x))/x]

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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^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a*x)**(5/2)/x**4,x)

[Out]

Integral((x*(a + b*x))**(5/2)/x**4, x)

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Giac [A]
time = 0.81, size = 89, normalized size = 0.97 \begin {gather*} -\frac {15}{8} \, a^{2} \sqrt {b} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right ) + \frac {2 \, a^{3}}{\sqrt {b} x - \sqrt {b x^{2} + a x}} + \frac {1}{4} \, {\left (2 \, b^{2} x + 9 \, a b\right )} \sqrt {b x^{2} + a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^4,x, algorithm="giac")

[Out]

-15/8*a^2*sqrt(b)*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a)) + 2*a^3/(sqrt(b)*x - sqrt(b*x^2 + a
*x)) + 1/4*(2*b^2*x + 9*a*b)*sqrt(b*x^2 + a*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^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^2)^(5/2)/x^4,x)

[Out]

int((a*x + b*x^2)^(5/2)/x^4, x)

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