3.579 \(\int \frac{(a+b x^n+c x^{2 n})^{3/2}}{x} \, dx\)

Optimal. Leaf size=173 \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{2 a+b x^n}{2 \sqrt{a} \sqrt{a+b x^n+c x^{2 n}}}\right )}{n}-\frac{b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{2 \sqrt{c} \sqrt{a+b x^n+c x^{2 n}}}\right )}{16 c^{3/2} n}+\frac{\left (8 a c+b^2+2 b c x^n\right ) \sqrt{a+b x^n+c x^{2 n}}}{8 c n}+\frac{\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n} \]

[Out]

((b^2 + 8*a*c + 2*b*c*x^n)*Sqrt[a + b*x^n + c*x^(2*n)])/(8*c*n) + (a + b*x^n + c*x^(2*n))^(3/2)/(3*n) - (a^(3/
2)*ArcTanh[(2*a + b*x^n)/(2*Sqrt[a]*Sqrt[a + b*x^n + c*x^(2*n)])])/n - (b*(b^2 - 12*a*c)*ArcTanh[(b + 2*c*x^n)
/(2*Sqrt[c]*Sqrt[a + b*x^n + c*x^(2*n)])])/(16*c^(3/2)*n)

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Rubi [A]  time = 0.158621, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {1357, 734, 814, 843, 621, 206, 724} \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{2 a+b x^n}{2 \sqrt{a} \sqrt{a+b x^n+c x^{2 n}}}\right )}{n}-\frac{b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{2 \sqrt{c} \sqrt{a+b x^n+c x^{2 n}}}\right )}{16 c^{3/2} n}+\frac{\left (8 a c+b^2+2 b c x^n\right ) \sqrt{a+b x^n+c x^{2 n}}}{8 c n}+\frac{\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n + c*x^(2*n))^(3/2)/x,x]

[Out]

((b^2 + 8*a*c + 2*b*c*x^n)*Sqrt[a + b*x^n + c*x^(2*n)])/(8*c*n) + (a + b*x^n + c*x^(2*n))^(3/2)/(3*n) - (a^(3/
2)*ArcTanh[(2*a + b*x^n)/(2*Sqrt[a]*Sqrt[a + b*x^n + c*x^(2*n)])])/n - (b*(b^2 - 12*a*c)*ArcTanh[(b + 2*c*x^n)
/(2*Sqrt[c]*Sqrt[a + b*x^n + c*x^(2*n)])])/(16*c^(3/2)*n)

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 734

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 + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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])

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^n+c x^{2 n}\right )^{3/2}}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x+c x^2\right )^{3/2}}{x} \, dx,x,x^n\right )}{n}\\ &=\frac{\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n}-\frac{\operatorname{Subst}\left (\int \frac{(-2 a-b x) \sqrt{a+b x+c x^2}}{x} \, dx,x,x^n\right )}{2 n}\\ &=\frac{\left (b^2+8 a c+2 b c x^n\right ) \sqrt{a+b x^n+c x^{2 n}}}{8 c n}+\frac{\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n}+\frac{\operatorname{Subst}\left (\int \frac{8 a^2 c-\frac{1}{2} b \left (b^2-12 a c\right ) x}{x \sqrt{a+b x+c x^2}} \, dx,x,x^n\right )}{8 c n}\\ &=\frac{\left (b^2+8 a c+2 b c x^n\right ) \sqrt{a+b x^n+c x^{2 n}}}{8 c n}+\frac{\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,x^n\right )}{n}-\frac{\left (b \left (b^2-12 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^n\right )}{16 c n}\\ &=\frac{\left (b^2+8 a c+2 b c x^n\right ) \sqrt{a+b x^n+c x^{2 n}}}{8 c n}+\frac{\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x^n}{\sqrt{a+b x^n+c x^{2 n}}}\right )}{n}-\frac{\left (b \left (b^2-12 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^n}{\sqrt{a+b x^n+c x^{2 n}}}\right )}{8 c n}\\ &=\frac{\left (b^2+8 a c+2 b c x^n\right ) \sqrt{a+b x^n+c x^{2 n}}}{8 c n}+\frac{\left (a+b x^n+c x^{2 n}\right )^{3/2}}{3 n}-\frac{a^{3/2} \tanh ^{-1}\left (\frac{2 a+b x^n}{2 \sqrt{a} \sqrt{a+b x^n+c x^{2 n}}}\right )}{n}-\frac{b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{2 \sqrt{c} \sqrt{a+b x^n+c x^{2 n}}}\right )}{16 c^{3/2} n}\\ \end{align*}

Mathematica [A]  time = 0.292796, size = 158, normalized size = 0.91 \[ \frac{-48 a^{3/2} c^{3/2} \tanh ^{-1}\left (\frac{2 a+b x^n}{2 \sqrt{a} \sqrt{a+x^n \left (b+c x^n\right )}}\right )+2 \sqrt{c} \sqrt{a+x^n \left (b+c x^n\right )} \left (8 c \left (4 a+c x^{2 n}\right )+3 b^2+14 b c x^n\right )-3 b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^n}{2 \sqrt{c} \sqrt{a+x^n \left (b+c x^n\right )}}\right )}{48 c^{3/2} n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + x^n*(b + c*x^n)]*(3*b^2 + 14*b*c*x^n + 8*c*(4*a + c*x^(2*n))) - 48*a^(3/2)*c^(3/2)*ArcTanh
[(2*a + b*x^n)/(2*Sqrt[a]*Sqrt[a + x^n*(b + c*x^n)])] - 3*b*(b^2 - 12*a*c)*ArcTanh[(b + 2*c*x^n)/(2*Sqrt[c]*Sq
rt[a + x^n*(b + c*x^n)])])/(48*c^(3/2)*n)

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Maple [A]  time = 0.045, size = 209, normalized size = 1.2 \begin{align*}{\frac{8\,{c}^{2} \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{2}+14\,b{{\rm e}^{n\ln \left ( x \right ) }}c+32\,ac+3\,{b}^{2}}{24\,cn}\sqrt{a+b{{\rm e}^{n\ln \left ( x \right ) }}+c \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{2}}}+{\frac{3\,ab}{4\,n}\ln \left ({ \left ({\frac{b}{2}}+c{{\rm e}^{n\ln \left ( x \right ) }} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{a+b{{\rm e}^{n\ln \left ( x \right ) }}+c \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{2}} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{b}^{3}}{16\,n}\ln \left ({ \left ({\frac{b}{2}}+c{{\rm e}^{n\ln \left ( x \right ) }} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{a+b{{\rm e}^{n\ln \left ( x \right ) }}+c \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{2}} \right ){c}^{-{\frac{3}{2}}}}-{\frac{1}{n}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{{{\rm e}^{n\ln \left ( x \right ) }}} \left ( 2\,a+b{{\rm e}^{n\ln \left ( x \right ) }}+2\,\sqrt{a}\sqrt{a+b{{\rm e}^{n\ln \left ( x \right ) }}+c \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*x^n+c*x^(2*n))^(3/2)/x,x)

[Out]

1/24*(8*c^2*exp(n*ln(x))^2+14*b*exp(n*ln(x))*c+32*a*c+3*b^2)*(a+b*exp(n*ln(x))+c*exp(n*ln(x))^2)^(1/2)/c/n+3/4
/c^(1/2)/n*a*b*ln((1/2*b+c*exp(n*ln(x)))/c^(1/2)+(a+b*exp(n*ln(x))+c*exp(n*ln(x))^2)^(1/2))-1/16/c^(3/2)/n*b^3
*ln((1/2*b+c*exp(n*ln(x)))/c^(1/2)+(a+b*exp(n*ln(x))+c*exp(n*ln(x))^2)^(1/2))-1/n*a^(3/2)*ln((2*a+b*exp(n*ln(x
))+2*a^(1/2)*(a+b*exp(n*ln(x))+c*exp(n*ln(x))^2)^(1/2))/exp(n*ln(x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac{3}{2}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 3.16363, size = 1955, normalized size = 11.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(48*a^(3/2)*c^2*log(-(8*a*b*x^n + 8*a^2 + (b^2 + 4*a*c)*x^(2*n) - 4*(sqrt(a)*b*x^n + 2*a^(3/2))*sqrt(c*x
^(2*n) + b*x^n + a))/x^(2*n)) - 3*(b^3 - 12*a*b*c)*sqrt(c)*log(-8*c^2*x^(2*n) - 8*b*c*x^n - b^2 - 4*a*c - 4*(2
*c^(3/2)*x^n + b*sqrt(c))*sqrt(c*x^(2*n) + b*x^n + a)) + 4*(8*c^3*x^(2*n) + 14*b*c^2*x^n + 3*b^2*c + 32*a*c^2)
*sqrt(c*x^(2*n) + b*x^n + a))/(c^2*n), 1/48*(24*a^(3/2)*c^2*log(-(8*a*b*x^n + 8*a^2 + (b^2 + 4*a*c)*x^(2*n) -
4*(sqrt(a)*b*x^n + 2*a^(3/2))*sqrt(c*x^(2*n) + b*x^n + a))/x^(2*n)) + 3*(b^3 - 12*a*b*c)*sqrt(-c)*arctan(1/2*(
2*sqrt(-c)*c*x^n + b*sqrt(-c))*sqrt(c*x^(2*n) + b*x^n + a)/(c^2*x^(2*n) + b*c*x^n + a*c)) + 2*(8*c^3*x^(2*n) +
14*b*c^2*x^n + 3*b^2*c + 32*a*c^2)*sqrt(c*x^(2*n) + b*x^n + a))/(c^2*n), 1/96*(96*sqrt(-a)*a*c^2*arctan(1/2*(
sqrt(-a)*b*x^n + 2*sqrt(-a)*a)*sqrt(c*x^(2*n) + b*x^n + a)/(a*c*x^(2*n) + a*b*x^n + a^2)) - 3*(b^3 - 12*a*b*c)
*sqrt(c)*log(-8*c^2*x^(2*n) - 8*b*c*x^n - b^2 - 4*a*c - 4*(2*c^(3/2)*x^n + b*sqrt(c))*sqrt(c*x^(2*n) + b*x^n +
a)) + 4*(8*c^3*x^(2*n) + 14*b*c^2*x^n + 3*b^2*c + 32*a*c^2)*sqrt(c*x^(2*n) + b*x^n + a))/(c^2*n), 1/48*(48*sq
rt(-a)*a*c^2*arctan(1/2*(sqrt(-a)*b*x^n + 2*sqrt(-a)*a)*sqrt(c*x^(2*n) + b*x^n + a)/(a*c*x^(2*n) + a*b*x^n + a
^2)) + 3*(b^3 - 12*a*b*c)*sqrt(-c)*arctan(1/2*(2*sqrt(-c)*c*x^n + b*sqrt(-c))*sqrt(c*x^(2*n) + b*x^n + a)/(c^2
*x^(2*n) + b*c*x^n + a*c)) + 2*(8*c^3*x^(2*n) + 14*b*c^2*x^n + 3*b^2*c + 32*a*c^2)*sqrt(c*x^(2*n) + b*x^n + a)
)/(c^2*n)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{n} + c x^{2 n}\right )^{\frac{3}{2}}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac{3}{2}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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