\(\int \frac {(b x+c x^2)^{3/2}}{x^{9/2}} \, dx\) [90]

Optimal result

Integrand size = 19, antiderivative size = 83 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{9/2}} \, dx=-\frac {3 c \sqrt {b x+c x^2}}{4 x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{2 x^{7/2}}-\frac {3 c^2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 \sqrt {b}} \]

[Out]

-1/2*(c*x^2+b*x)^(3/2)/x^(7/2)-3/4*c^2*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(1/2)-3/4*c*(c*x^2+b*x)^(1
/2)/x^(3/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {676, 674, 213} \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{9/2}} \, dx=-\frac {3 c^2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 \sqrt {b}}-\frac {3 c \sqrt {b x+c x^2}}{4 x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{2 x^{7/2}} \]

[In]

Int[(b*x + c*x^2)^(3/2)/x^(9/2),x]

[Out]

(-3*c*Sqrt[b*x + c*x^2])/(4*x^(3/2)) - (b*x + c*x^2)^(3/2)/(2*x^(7/2)) - (3*c^2*ArcTanh[Sqrt[b*x + c*x^2]/(Sqr
t[b]*Sqrt[x])])/(4*Sqrt[b])

Rule 213

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

Rule 674

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*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]

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]

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{9/2}} \, dx=-\frac {\sqrt {x (b+c x)} \left (\sqrt {b} \sqrt {b+c x} (2 b+5 c x)+3 c^2 x^2 \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{4 \sqrt {b} x^{5/2} \sqrt {b+c x}} \]

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^(9/2),x]

[Out]

-1/4*(Sqrt[x*(b + c*x)]*(Sqrt[b]*Sqrt[b + c*x]*(2*b + 5*c*x) + 3*c^2*x^2*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(Sqr
t[b]*x^(5/2)*Sqrt[b + c*x])

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.82

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (5 c x +2 b \right )}{4 x^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}}-\frac {3 c^{2} \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{4 \sqrt {b}\, \sqrt {x \left (c x +b \right )}}\)	\(68\)
default	\(-\frac {\sqrt {x \left (c x +b \right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{2} x^{2}+5 c x \sqrt {b}\, \sqrt {c x +b}+2 b^{\frac {3}{2}} \sqrt {c x +b}\right )}{4 x^{\frac {5}{2}} \sqrt {c x +b}\, \sqrt {b}}\)	\(72\)

[In]

int((c*x^2+b*x)^(3/2)/x^(9/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.84 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{9/2}} \, dx=\left [\frac {3 \, \sqrt {b} c^{2} x^{3} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) - 2 \, {\left (5 \, b c x + 2 \, b^{2}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{8 \, b x^{3}}, \frac {3 \, \sqrt {-b} c^{2} x^{3} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - {\left (5 \, b c x + 2 \, b^{2}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{4 \, b x^{3}}\right ] \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^(9/2),x, algorithm="fricas")

[Out]

[1/8*(3*sqrt(b)*c^2*x^3*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) - 2*(5*b*c*x + 2*b^2)*
sqrt(c*x^2 + b*x)*sqrt(x))/(b*x^3), 1/4*(3*sqrt(-b)*c^2*x^3*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) - (5*b*
c*x + 2*b^2)*sqrt(c*x^2 + b*x)*sqrt(x))/(b*x^3)]

Sympy [F]

\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{9/2}} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{\frac {9}{2}}}\, dx \]

[In]

integrate((c*x**2+b*x)**(3/2)/x**(9/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**(9/2), x)

Maxima [F]

\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{9/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{x^{\frac {9}{2}}} \,d x } \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^(9/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)/x^(9/2), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{9/2}} \, dx=\frac {\frac {3 \, c^{3} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {5 \, {\left (c x + b\right )}^{\frac {3}{2}} c^{3} - 3 \, \sqrt {c x + b} b c^{3}}{c^{2} x^{2}}}{4 \, c} \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^(9/2),x, algorithm="giac")

[Out]

1/4*(3*c^3*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) - (5*(c*x + b)^(3/2)*c^3 - 3*sqrt(c*x + b)*b*c^3)/(c^2*x^2)
)/c

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{9/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^{9/2}} \,d x \]

[In]

int((b*x + c*x^2)^(3/2)/x^(9/2),x)

[Out]

int((b*x + c*x^2)^(3/2)/x^(9/2), x)