\(\int \frac {1}{x^{5/2} (b x+c x^2)^{3/2}} \, dx\) [111]

Optimal result

Integrand size = 19, antiderivative size = 145 \[ \int \frac {1}{x^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}+\frac {7 c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}-\frac {35 c^2}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}-\frac {35 c^3 \sqrt {x}}{8 b^4 \sqrt {b x+c x^2}}+\frac {35 c^3 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{9/2}} \]

[Out]

35/8*c^3*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(9/2)-1/3/b/x^(5/2)/(c*x^2+b*x)^(1/2)+7/12*c/b^2/x^(3/2)
/(c*x^2+b*x)^(1/2)-35/24*c^2/b^3/x^(1/2)/(c*x^2+b*x)^(1/2)-35/8*c^3*x^(1/2)/b^4/(c*x^2+b*x)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 145, 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 = {686, 680, 674, 213} \[ \int \frac {1}{x^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {35 c^3 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{9/2}}-\frac {35 c^3 \sqrt {x}}{8 b^4 \sqrt {b x+c x^2}}-\frac {35 c^2}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}+\frac {7 c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}-\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}} \]

[In]

Int[1/(x^(5/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

-1/3*1/(b*x^(5/2)*Sqrt[b*x + c*x^2]) + (7*c)/(12*b^2*x^(3/2)*Sqrt[b*x + c*x^2]) - (35*c^2)/(24*b^3*Sqrt[x]*Sqr
t[b*x + c*x^2]) - (35*c^3*Sqrt[x])/(8*b^4*Sqrt[b*x + c*x^2]) + (35*c^3*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt
[x])])/(8*b^(9/2))

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 680

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(2*c*d - b*e)*(d + e
*x)^m*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)*(b^2 - 4*a*c))), x] - Dist[(2*c*d - b*e)*((m + 2*p + 2)/((p + 1)*(
b^2 - 4*a*c))), 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] && LtQ[p, -1] && LtQ[0, m, 1] && IntegerQ[2*p]

Rule 686

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

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}-\frac {(7 c) \int \frac {1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx}{6 b} \\ & = -\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}+\frac {7 c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}+\frac {\left (35 c^2\right ) \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx}{24 b^2} \\ & = -\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}+\frac {7 c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}-\frac {35 c^2}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}-\frac {\left (35 c^3\right ) \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{16 b^3} \\ & = -\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}+\frac {7 c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}-\frac {35 c^2}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}-\frac {35 c^3 \sqrt {x}}{8 b^4 \sqrt {b x+c x^2}}-\frac {\left (35 c^3\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{16 b^4} \\ & = -\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}+\frac {7 c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}-\frac {35 c^2}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}-\frac {35 c^3 \sqrt {x}}{8 b^4 \sqrt {b x+c x^2}}-\frac {\left (35 c^3\right ) \text {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{8 b^4} \\ & = -\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}+\frac {7 c}{12 b^2 x^{3/2} \sqrt {b x+c x^2}}-\frac {35 c^2}{24 b^3 \sqrt {x} \sqrt {b x+c x^2}}-\frac {35 c^3 \sqrt {x}}{8 b^4 \sqrt {b x+c x^2}}+\frac {35 c^3 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {-\sqrt {b} \left (8 b^3-14 b^2 c x+35 b c^2 x^2+105 c^3 x^3\right )+105 c^3 x^3 \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{24 b^{9/2} x^{5/2} \sqrt {x (b+c x)}} \]

[In]

Integrate[1/(x^(5/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

(-(Sqrt[b]*(8*b^3 - 14*b^2*c*x + 35*b*c^2*x^2 + 105*c^3*x^3)) + 105*c^3*x^3*Sqrt[b + c*x]*ArcTanh[Sqrt[b + c*x
]/Sqrt[b]])/(24*b^(9/2)*x^(5/2)*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {\sqrt {x \left (c x +b \right )}\, \left (105 \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{3} x^{3}+14 b^{\frac {5}{2}} c x -35 b^{\frac {3}{2}} c^{2} x^{2}-105 c^{3} x^{3} \sqrt {b}-8 b^{\frac {7}{2}}\right )}{24 x^{\frac {7}{2}} \left (c x +b \right ) b^{\frac {9}{2}}}\)	\(87\)
risch	\(-\frac {\left (c x +b \right ) \left (57 c^{2} x^{2}-22 b c x +8 b^{2}\right )}{24 b^{4} x^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}-\frac {c^{3} \left (\frac {32}{\sqrt {c x +b}}-\frac {70 \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{16 b^{4} \sqrt {x \left (c x +b \right )}}\)	\(97\)

[In]

int(1/x^(5/2)/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/24/x^(7/2)*(x*(c*x+b))^(1/2)*(105*(c*x+b)^(1/2)*arctanh((c*x+b)^(1/2)/b^(1/2))*c^3*x^3+14*b^(5/2)*c*x-35*b^(
3/2)*c^2*x^2-105*c^3*x^3*b^(1/2)-8*b^(7/2))/(c*x+b)/b^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.66 \[ \int \frac {1}{x^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx=\left [\frac {105 \, {\left (c^{4} x^{5} + b c^{3} x^{4}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) - 2 \, {\left (105 \, b c^{3} x^{3} + 35 \, b^{2} c^{2} x^{2} - 14 \, b^{3} c x + 8 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, {\left (b^{5} c x^{5} + b^{6} x^{4}\right )}}, -\frac {105 \, {\left (c^{4} x^{5} + b c^{3} x^{4}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (105 \, b c^{3} x^{3} + 35 \, b^{2} c^{2} x^{2} - 14 \, b^{3} c x + 8 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, {\left (b^{5} c x^{5} + b^{6} x^{4}\right )}}\right ] \]

[In]

integrate(1/x^(5/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[1/48*(105*(c^4*x^5 + b*c^3*x^4)*sqrt(b)*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) - 2*(
105*b*c^3*x^3 + 35*b^2*c^2*x^2 - 14*b^3*c*x + 8*b^4)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^5*c*x^5 + b^6*x^4), -1/24*(
105*(c^4*x^5 + b*c^3*x^4)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (105*b*c^3*x^3 + 35*b^2*c^2*x^
2 - 14*b^3*c*x + 8*b^4)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^5*c*x^5 + b^6*x^4)]

Sympy [F]

\[ \int \frac {1}{x^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{x^{\frac {5}{2}} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/x**(5/2)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(1/(x**(5/2)*(x*(b + c*x))**(3/2)), x)

Maxima [F]

\[ \int \frac {1}{x^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/x^(5/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^2 + b*x)^(3/2)*x^(5/2)), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {35 \, c^{3} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{4}} - \frac {2 \, c^{3}}{\sqrt {c x + b} b^{4}} - \frac {57 \, {\left (c x + b\right )}^{\frac {5}{2}} c^{3} - 136 \, {\left (c x + b\right )}^{\frac {3}{2}} b c^{3} + 87 \, \sqrt {c x + b} b^{2} c^{3}}{24 \, b^{4} c^{3} x^{3}} \]

[In]

integrate(1/x^(5/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

-35/8*c^3*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^4) - 2*c^3/(sqrt(c*x + b)*b^4) - 1/24*(57*(c*x + b)^(5/2)
*c^3 - 136*(c*x + b)^(3/2)*b*c^3 + 87*sqrt(c*x + b)*b^2*c^3)/(b^4*c^3*x^3)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{x^{5/2}\,{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]

[In]

int(1/(x^(5/2)*(b*x + c*x^2)^(3/2)),x)

[Out]

int(1/(x^(5/2)*(b*x + c*x^2)^(3/2)), x)