\(\int \frac {x^4}{\sqrt {b x+c x^2}} \, dx\) [42]

Optimal result

Integrand size = 17, antiderivative size = 128 \[ \int \frac {x^4}{\sqrt {b x+c x^2}} \, dx=-\frac {35 b^3 \sqrt {b x+c x^2}}{64 c^4}+\frac {35 b^2 x \sqrt {b x+c x^2}}{96 c^3}-\frac {7 b x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {x^3 \sqrt {b x+c x^2}}{4 c}+\frac {35 b^4 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{9/2}} \]

[Out]

35/64*b^4*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(9/2)-35/64*b^3*(c*x^2+b*x)^(1/2)/c^4+35/96*b^2*x*(c*x^2+b*x)
^(1/2)/c^3-7/24*b*x^2*(c*x^2+b*x)^(1/2)/c^2+1/4*x^3*(c*x^2+b*x)^(1/2)/c

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {684, 654, 634, 212} \[ \int \frac {x^4}{\sqrt {b x+c x^2}} \, dx=\frac {35 b^4 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{9/2}}-\frac {35 b^3 \sqrt {b x+c x^2}}{64 c^4}+\frac {35 b^2 x \sqrt {b x+c x^2}}{96 c^3}-\frac {7 b x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {x^3 \sqrt {b x+c x^2}}{4 c} \]

[In]

Int[x^4/Sqrt[b*x + c*x^2],x]

[Out]

(-35*b^3*Sqrt[b*x + c*x^2])/(64*c^4) + (35*b^2*x*Sqrt[b*x + c*x^2])/(96*c^3) - (7*b*x^2*Sqrt[b*x + c*x^2])/(24
*c^2) + (x^3*Sqrt[b*x + c*x^2])/(4*c) + (35*b^4*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(9/2))

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 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 654

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

Rule 684

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

Rubi steps \begin{align*} \text {integral}& = \frac {x^3 \sqrt {b x+c x^2}}{4 c}-\frac {(7 b) \int \frac {x^3}{\sqrt {b x+c x^2}} \, dx}{8 c} \\ & = -\frac {7 b x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {x^3 \sqrt {b x+c x^2}}{4 c}+\frac {\left (35 b^2\right ) \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx}{48 c^2} \\ & = \frac {35 b^2 x \sqrt {b x+c x^2}}{96 c^3}-\frac {7 b x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {x^3 \sqrt {b x+c x^2}}{4 c}-\frac {\left (35 b^3\right ) \int \frac {x}{\sqrt {b x+c x^2}} \, dx}{64 c^3} \\ & = -\frac {35 b^3 \sqrt {b x+c x^2}}{64 c^4}+\frac {35 b^2 x \sqrt {b x+c x^2}}{96 c^3}-\frac {7 b x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {x^3 \sqrt {b x+c x^2}}{4 c}+\frac {\left (35 b^4\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c^4} \\ & = -\frac {35 b^3 \sqrt {b x+c x^2}}{64 c^4}+\frac {35 b^2 x \sqrt {b x+c x^2}}{96 c^3}-\frac {7 b x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {x^3 \sqrt {b x+c x^2}}{4 c}+\frac {\left (35 b^4\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{64 c^4} \\ & = -\frac {35 b^3 \sqrt {b x+c x^2}}{64 c^4}+\frac {35 b^2 x \sqrt {b x+c x^2}}{96 c^3}-\frac {7 b x^2 \sqrt {b x+c x^2}}{24 c^2}+\frac {x^3 \sqrt {b x+c x^2}}{4 c}+\frac {35 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.93 \[ \int \frac {x^4}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c} x \left (-105 b^4-35 b^3 c x+14 b^2 c^2 x^2-8 b c^3 x^3+48 c^4 x^4\right )+210 b^4 \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{192 c^{9/2} \sqrt {x (b+c x)}} \]

[In]

Integrate[x^4/Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[c]*x*(-105*b^4 - 35*b^3*c*x + 14*b^2*c^2*x^2 - 8*b*c^3*x^3 + 48*c^4*x^4) + 210*b^4*Sqrt[x]*Sqrt[b + c*x]
*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])])/(192*c^(9/2)*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 2.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.57

method	result	size
pseudoelliptic	\(\frac {\frac {35 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) b^{4}}{64}-\frac {35 \sqrt {x \left (c x +b \right )}\, \left (\sqrt {c}\, b^{3}-\frac {2 c^{\frac {3}{2}} b^{2} x}{3}+\frac {8 c^{\frac {5}{2}} b \,x^{2}}{15}-\frac {16 c^{\frac {7}{2}} x^{3}}{35}\right )}{64}}{c^{\frac {9}{2}}}\)	\(73\)
risch	\(-\frac {\left (-48 c^{3} x^{3}+56 b \,c^{2} x^{2}-70 b^{2} c x +105 b^{3}\right ) x \left (c x +b \right )}{192 c^{4} \sqrt {x \left (c x +b \right )}}+\frac {35 b^{4} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {9}{2}}}\)	\(84\)
default	\(\frac {x^{3} \sqrt {c \,x^{2}+b x}}{4 c}-\frac {7 b \left (\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )}{8 c}\)	\(123\)

[In]

int(x^4/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

35/64/c^(9/2)*(arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))*b^4-(x*(c*x+b))^(1/2)*(c^(1/2)*b^3-2/3*c^(3/2)*b^2*x+8/15*
c^(5/2)*b*x^2-16/35*c^(7/2)*x^3))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.33 \[ \int \frac {x^4}{\sqrt {b x+c x^2}} \, dx=\left [\frac {105 \, b^{4} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (48 \, c^{4} x^{3} - 56 \, b c^{3} x^{2} + 70 \, b^{2} c^{2} x - 105 \, b^{3} c\right )} \sqrt {c x^{2} + b x}}{384 \, c^{5}}, -\frac {105 \, b^{4} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (48 \, c^{4} x^{3} - 56 \, b c^{3} x^{2} + 70 \, b^{2} c^{2} x - 105 \, b^{3} c\right )} \sqrt {c x^{2} + b x}}{192 \, c^{5}}\right ] \]

[In]

integrate(x^4/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[1/384*(105*b^4*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(48*c^4*x^3 - 56*b*c^3*x^2 + 70*b^2*c
^2*x - 105*b^3*c)*sqrt(c*x^2 + b*x))/c^5, -1/192*(105*b^4*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) -
(48*c^4*x^3 - 56*b*c^3*x^2 + 70*b^2*c^2*x - 105*b^3*c)*sqrt(c*x^2 + b*x))/c^5]

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.12 \[ \int \frac {x^4}{\sqrt {b x+c x^2}} \, dx=\begin {cases} \frac {35 b^{4} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{128 c^{4}} + \sqrt {b x + c x^{2}} \left (- \frac {35 b^{3}}{64 c^{4}} + \frac {35 b^{2} x}{96 c^{3}} - \frac {7 b x^{2}}{24 c^{2}} + \frac {x^{3}}{4 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {9}{2}}}{9 b^{5}} & \text {for}\: b \neq 0 \\\tilde {\infty } x^{5} & \text {otherwise} \end {cases} \]

[In]

integrate(x**4/(c*x**2+b*x)**(1/2),x)

[Out]

Piecewise((35*b**4*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c)
+ x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(128*c**4) + sqrt(b*x + c*x**2)*(-35*b**3/(64*c**4) +
35*b**2*x/(96*c**3) - 7*b*x**2/(24*c**2) + x**3/(4*c)), Ne(c, 0)), (2*(b*x)**(9/2)/(9*b**5), Ne(b, 0)), (zoo*x
**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.86 \[ \int \frac {x^4}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x} x^{3}}{4 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} b x^{2}}{24 \, c^{2}} + \frac {35 \, \sqrt {c x^{2} + b x} b^{2} x}{96 \, c^{3}} + \frac {35 \, b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {9}{2}}} - \frac {35 \, \sqrt {c x^{2} + b x} b^{3}}{64 \, c^{4}} \]

[In]

integrate(x^4/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

1/4*sqrt(c*x^2 + b*x)*x^3/c - 7/24*sqrt(c*x^2 + b*x)*b*x^2/c^2 + 35/96*sqrt(c*x^2 + b*x)*b^2*x/c^3 + 35/128*b^
4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) - 35/64*sqrt(c*x^2 + b*x)*b^3/c^4

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.68 \[ \int \frac {x^4}{\sqrt {b x+c x^2}} \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, x {\left (\frac {6 \, x}{c} - \frac {7 \, b}{c^{2}}\right )} + \frac {35 \, b^{2}}{c^{3}}\right )} x - \frac {105 \, b^{3}}{c^{4}}\right )} - \frac {35 \, b^{4} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {9}{2}}} \]

[In]

integrate(x^4/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*x*(6*x/c - 7*b/c^2) + 35*b^2/c^3)*x - 105*b^3/c^4) - 35/128*b^4*log(abs(2*(sqrt(
c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(9/2)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^4}{\sqrt {c\,x^2+b\,x}} \,d x \]

[In]

int(x^4/(b*x + c*x^2)^(1/2),x)

[Out]

int(x^4/(b*x + c*x^2)^(1/2), x)