### 3.42 $$\int \frac{x^4}{\sqrt{b x+c x^2}} \, dx$$

Optimal. Leaf size=128 $-\frac{35 b^3 \sqrt{b x+c x^2}}{64 c^4}+\frac{35 b^2 x \sqrt{b x+c x^2}}{96 c^3}+\frac{35 b^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{9/2}}-\frac{7 b x^2 \sqrt{b x+c x^2}}{24 c^2}+\frac{x^3 \sqrt{b x+c x^2}}{4 c}$

[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))

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Rubi [A]  time = 0.0576066, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {670, 640, 620, 206} $-\frac{35 b^3 \sqrt{b x+c x^2}}{64 c^4}+\frac{35 b^2 x \sqrt{b x+c x^2}}{96 c^3}+\frac{35 b^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{9/2}}-\frac{7 b x^2 \sqrt{b x+c x^2}}{24 c^2}+\frac{x^3 \sqrt{b x+c x^2}}{4 c}$

Antiderivative was successfully veriﬁed.

[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 670

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]

Rule 640

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 620

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 206

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

Rubi steps

\begin{align*} \int \frac{x^4}{\sqrt{b x+c x^2}} \, dx &=\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 ) \operatorname{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]  time = 0.213593, size = 98, normalized size = 0.77 $\frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (70 b^2 c x-105 b^3-56 b c^2 x^2+48 c^3 x^3\right )+\frac{105 b^{7/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{192 c^{9/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^4/Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^3 + 70*b^2*c*x - 56*b*c^2*x^2 + 48*c^3*x^3) + (105*b^(7/2)*ArcSinh[(Sqrt[c
]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(192*c^(9/2))

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Maple [A]  time = 0.048, size = 112, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{7\,b{x}^{2}}{24\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{35\,{b}^{2}x}{96\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{35\,{b}^{3}}{64\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{35\,{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{9}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.97581, size = 408, normalized size = 3.19 \begin{align*} \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 ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{x \left (b + c x\right )}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/sqrt(x*(b + c*x)), x)

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Giac [A]  time = 1.22022, size = 120, normalized size = 0.94 \begin{align*} \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}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)