\(\int \frac {(a x^2+b x^3)^{3/2}}{x^8} \, dx\) [250]

Optimal result

Integrand size = 19, antiderivative size = 137 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}+\frac {3 b^3 \sqrt {a x^2+b x^3}}{64 a^2 x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{64 a^{5/2}} \]

[Out]

-1/4*(b*x^3+a*x^2)^(3/2)/x^7-3/64*b^4*arctanh(x*a^(1/2)/(b*x^3+a*x^2)^(1/2))/a^(5/2)-1/8*b*(b*x^3+a*x^2)^(1/2)
/x^4-1/32*b^2*(b*x^3+a*x^2)^(1/2)/a/x^3+3/64*b^3*(b*x^3+a*x^2)^(1/2)/a^2/x^2

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2045, 2050, 2033, 212} \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{64 a^{5/2}}+\frac {3 b^3 \sqrt {a x^2+b x^3}}{64 a^2 x^2}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac {b \sqrt {a x^2+b x^3}}{8 x^4} \]

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^8,x]

[Out]

-1/8*(b*Sqrt[a*x^2 + b*x^3])/x^4 - (b^2*Sqrt[a*x^2 + b*x^3])/(32*a*x^3) + (3*b^3*Sqrt[a*x^2 + b*x^3])/(64*a^2*
x^2) - (a*x^2 + b*x^3)^(3/2)/(4*x^7) - (3*b^4*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(64*a^(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 2033

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2045

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b*
x^n)^p/(c*(m + j*p + 1))), x] - Dist[b*p*((n - j)/(c^n*(m + j*p + 1))), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2050

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}+\frac {1}{8} (3 b) \int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx \\ & = -\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}+\frac {1}{16} b^2 \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx \\ & = -\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac {\left (3 b^3\right ) \int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx}{64 a} \\ & = -\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}+\frac {3 b^3 \sqrt {a x^2+b x^3}}{64 a^2 x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}+\frac {\left (3 b^4\right ) \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx}{128 a^2} \\ & = -\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}+\frac {3 b^3 \sqrt {a x^2+b x^3}}{64 a^2 x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac {\left (3 b^4\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x}{\sqrt {a x^2+b x^3}}\right )}{64 a^2} \\ & = -\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}+\frac {3 b^3 \sqrt {a x^2+b x^3}}{64 a^2 x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac {3 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{64 a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a} \sqrt {a+b x} \left (16 a^3+24 a^2 b x+2 a b^2 x^2-3 b^3 x^3\right )+3 b^4 x^4 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{64 a^{5/2} x^5 \sqrt {a+b x}} \]

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x^8,x]

[Out]

-1/64*(Sqrt[x^2*(a + b*x)]*(Sqrt[a]*Sqrt[a + b*x]*(16*a^3 + 24*a^2*b*x + 2*a*b^2*x^2 - 3*b^3*x^3) + 3*b^4*x^4*
ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(a^(5/2)*x^5*Sqrt[a + b*x])

Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.67

method	result	size
risch	\(-\frac {\left (-3 b^{3} x^{3}+2 a \,b^{2} x^{2}+24 a^{2} b x +16 a^{3}\right ) \sqrt {x^{2} \left (b x +a \right )}}{64 x^{5} a^{2}}-\frac {3 b^{4} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{64 a^{\frac {5}{2}} x \sqrt {b x +a}}\)	\(92\)
default	\(\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (3 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {5}{2}}-11 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {7}{2}}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} x^{4} b^{4}-11 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {9}{2}}+3 \sqrt {b x +a}\, a^{\frac {11}{2}}\right )}{64 x^{7} \left (b x +a \right )^{\frac {3}{2}} a^{\frac {9}{2}}}\)	\(101\)
pseudoelliptic	\(-\frac {5 \left (-\frac {63 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) x^{7} b^{7}}{1280}+\sqrt {b x +a}\, \left (\frac {63 \sqrt {a}\, b^{6} x^{6}}{1280}-\frac {21 a^{\frac {3}{2}} b^{5} x^{5}}{640}+\frac {21 a^{\frac {5}{2}} b^{4} x^{4}}{800}-\frac {9 a^{\frac {7}{2}} b^{3} x^{3}}{400}+\frac {a^{\frac {9}{2}} b^{2} x^{2}}{50}+a^{\frac {11}{2}} b x +\frac {4 a^{\frac {13}{2}}}{5}\right )\right )}{28 a^{\frac {11}{2}} x^{7}}\)	\(105\)

[In]

int((b*x^3+a*x^2)^(3/2)/x^8,x,method=_RETURNVERBOSE)

[Out]

-1/64*(-3*b^3*x^3+2*a*b^2*x^2+24*a^2*b*x+16*a^3)/x^5/a^2*(x^2*(b*x+a))^(1/2)-3/64*b^4/a^(5/2)*arctanh((b*x+a)^
(1/2)/a^(1/2))*(x^2*(b*x+a))^(1/2)/x/(b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\left [\frac {3 \, \sqrt {a} b^{4} x^{5} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (3 \, a b^{3} x^{3} - 2 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{128 \, a^{3} x^{5}}, \frac {3 \, \sqrt {-a} b^{4} x^{5} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (3 \, a b^{3} x^{3} - 2 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{64 \, a^{3} x^{5}}\right ] \]

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="fricas")

[Out]

[1/128*(3*sqrt(a)*b^4*x^5*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) + 2*(3*a*b^3*x^3 - 2*a^2*b^
2*x^2 - 24*a^3*b*x - 16*a^4)*sqrt(b*x^3 + a*x^2))/(a^3*x^5), 1/64*(3*sqrt(-a)*b^4*x^5*arctan(sqrt(b*x^3 + a*x^
2)*sqrt(-a)/(a*x)) + (3*a*b^3*x^3 - 2*a^2*b^2*x^2 - 24*a^3*b*x - 16*a^4)*sqrt(b*x^3 + a*x^2))/(a^3*x^5)]

Sympy [F]

\[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}{x^{8}}\, dx \]

[In]

integrate((b*x**3+a*x**2)**(3/2)/x**8,x)

[Out]

Integral((x**2*(a + b*x))**(3/2)/x**8, x)

Maxima [F]

\[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{8}} \,d x } \]

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a*x^2)^(3/2)/x^8, x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\frac {\frac {3 \, b^{5} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} \mathrm {sgn}\left (x\right ) - 11 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} \mathrm {sgn}\left (x\right ) - 11 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} \mathrm {sgn}\left (x\right ) + 3 \, \sqrt {b x + a} a^{3} b^{5} \mathrm {sgn}\left (x\right )}{a^{2} b^{4} x^{4}}}{64 \, b} \]

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="giac")

[Out]

1/64*(3*b^5*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a^2) + (3*(b*x + a)^(7/2)*b^5*sgn(x) - 11*(b*x + a
)^(5/2)*a*b^5*sgn(x) - 11*(b*x + a)^(3/2)*a^2*b^5*sgn(x) + 3*sqrt(b*x + a)*a^3*b^5*sgn(x))/(a^2*b^4*x^4))/b

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}}{x^8} \,d x \]

[In]

int((a*x^2 + b*x^3)^(3/2)/x^8,x)

[Out]

int((a*x^2 + b*x^3)^(3/2)/x^8, x)