\(\int \frac {(a+b x^3+c x^6)^{3/2}}{x^7} \, dx\) [208]

Optimal result

Integrand size = 20, antiderivative size = 151 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^7} \, dx=-\frac {\left (b-2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{4 x^3}-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{6 x^6}-\frac {\left (b^2+4 a c\right ) \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{8 \sqrt {a}}+\frac {1}{2} b \sqrt {c} \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right ) \]

[Out]

-1/6*(c*x^6+b*x^3+a)^(3/2)/x^6-1/8*(4*a*c+b^2)*arctanh(1/2*(b*x^3+2*a)/a^(1/2)/(c*x^6+b*x^3+a)^(1/2))/a^(1/2)+
1/2*b*arctanh(1/2*(2*c*x^3+b)/c^(1/2)/(c*x^6+b*x^3+a)^(1/2))*c^(1/2)-1/4*(-2*c*x^3+b)*(c*x^6+b*x^3+a)^(1/2)/x^
3

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1371, 746, 826, 857, 635, 212, 738} \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^7} \, dx=-\frac {\left (4 a c+b^2\right ) \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{8 \sqrt {a}}+\frac {1}{2} b \sqrt {c} \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{6 x^6}-\frac {\left (b-2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{4 x^3} \]

[In]

Int[(a + b*x^3 + c*x^6)^(3/2)/x^7,x]

[Out]

-1/4*((b - 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6])/x^3 - (a + b*x^3 + c*x^6)^(3/2)/(6*x^6) - ((b^2 + 4*a*c)*ArcTanh[
(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + c*x^6])])/(8*Sqrt[a]) + (b*Sqrt[c]*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*
Sqrt[a + b*x^3 + c*x^6])])/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 635

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 738

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]

Rule 746

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 + 1))), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*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] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 826

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

Rule 857

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 1371

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]]

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^7} \, dx=\frac {1}{12} \left (\frac {\sqrt {a+b x^3+c x^6} \left (-2 a-5 b x^3+4 c x^6\right )}{x^6}+\frac {3 \left (b^2+4 a c\right ) \text {arctanh}\left (\frac {\sqrt {c} x^3-\sqrt {a+b x^3+c x^6}}{\sqrt {a}}\right )}{\sqrt {a}}-6 b \sqrt {c} \log \left (b+2 c x^3-2 \sqrt {c} \sqrt {a+b x^3+c x^6}\right )\right ) \]

[In]

Integrate[(a + b*x^3 + c*x^6)^(3/2)/x^7,x]

[Out]

((Sqrt[a + b*x^3 + c*x^6]*(-2*a - 5*b*x^3 + 4*c*x^6))/x^6 + (3*(b^2 + 4*a*c)*ArcTanh[(Sqrt[c]*x^3 - Sqrt[a + b
*x^3 + c*x^6])/Sqrt[a]])/Sqrt[a] - 6*b*Sqrt[c]*Log[b + 2*c*x^3 - 2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6]])/12

Maple [F]

\[\int \frac {\left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}}{x^{7}}d x\]

[In]

int((c*x^6+b*x^3+a)^(3/2)/x^7,x)

[Out]

int((c*x^6+b*x^3+a)^(3/2)/x^7,x)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 713, normalized size of antiderivative = 4.72 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^7} \, dx=\left [\frac {12 \, a b \sqrt {c} x^{6} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) + 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {a} x^{6} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) + 4 \, {\left (4 \, a c x^{6} - 5 \, a b x^{3} - 2 \, a^{2}\right )} \sqrt {c x^{6} + b x^{3} + a}}{48 \, a x^{6}}, -\frac {24 \, a b \sqrt {-c} x^{6} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) - 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {a} x^{6} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) - 4 \, {\left (4 \, a c x^{6} - 5 \, a b x^{3} - 2 \, a^{2}\right )} \sqrt {c x^{6} + b x^{3} + a}}{48 \, a x^{6}}, \frac {6 \, a b \sqrt {c} x^{6} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) + 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) + 2 \, {\left (4 \, a c x^{6} - 5 \, a b x^{3} - 2 \, a^{2}\right )} \sqrt {c x^{6} + b x^{3} + a}}{24 \, a x^{6}}, -\frac {12 \, a b \sqrt {-c} x^{6} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) - 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) - 2 \, {\left (4 \, a c x^{6} - 5 \, a b x^{3} - 2 \, a^{2}\right )} \sqrt {c x^{6} + b x^{3} + a}}{24 \, a x^{6}}\right ] \]

[In]

integrate((c*x^6+b*x^3+a)^(3/2)/x^7,x, algorithm="fricas")

[Out]

[1/48*(12*a*b*sqrt(c)*x^6*log(-8*c^2*x^6 - 8*b*c*x^3 - b^2 - 4*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sqrt(c) -
4*a*c) + 3*(b^2 + 4*a*c)*sqrt(a)*x^6*log(-((b^2 + 4*a*c)*x^6 + 8*a*b*x^3 - 4*sqrt(c*x^6 + b*x^3 + a)*(b*x^3 +
2*a)*sqrt(a) + 8*a^2)/x^6) + 4*(4*a*c*x^6 - 5*a*b*x^3 - 2*a^2)*sqrt(c*x^6 + b*x^3 + a))/(a*x^6), -1/48*(24*a*
b*sqrt(-c)*x^6*arctan(1/2*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sqrt(-c)/(c^2*x^6 + b*c*x^3 + a*c)) - 3*(b^2 +
4*a*c)*sqrt(a)*x^6*log(-((b^2 + 4*a*c)*x^6 + 8*a*b*x^3 - 4*sqrt(c*x^6 + b*x^3 + a)*(b*x^3 + 2*a)*sqrt(a) + 8*
a^2)/x^6) - 4*(4*a*c*x^6 - 5*a*b*x^3 - 2*a^2)*sqrt(c*x^6 + b*x^3 + a))/(a*x^6), 1/24*(6*a*b*sqrt(c)*x^6*log(-8
*c^2*x^6 - 8*b*c*x^3 - b^2 - 4*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sqrt(c) - 4*a*c) + 3*(b^2 + 4*a*c)*sqrt(-
a)*x^6*arctan(1/2*sqrt(c*x^6 + b*x^3 + a)*(b*x^3 + 2*a)*sqrt(-a)/(a*c*x^6 + a*b*x^3 + a^2)) + 2*(4*a*c*x^6 - 5
*a*b*x^3 - 2*a^2)*sqrt(c*x^6 + b*x^3 + a))/(a*x^6), -1/24*(12*a*b*sqrt(-c)*x^6*arctan(1/2*sqrt(c*x^6 + b*x^3 +
a)*(2*c*x^3 + b)*sqrt(-c)/(c^2*x^6 + b*c*x^3 + a*c)) - 3*(b^2 + 4*a*c)*sqrt(-a)*x^6*arctan(1/2*sqrt(c*x^6 + b
*x^3 + a)*(b*x^3 + 2*a)*sqrt(-a)/(a*c*x^6 + a*b*x^3 + a^2)) - 2*(4*a*c*x^6 - 5*a*b*x^3 - 2*a^2)*sqrt(c*x^6 + b
*x^3 + a))/(a*x^6)]

Sympy [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^7} \, dx=\int \frac {\left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}}{x^{7}}\, dx \]

[In]

integrate((c*x**6+b*x**3+a)**(3/2)/x**7,x)

[Out]

Integral((a + b*x**3 + c*x**6)**(3/2)/x**7, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^7} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c*x^6+b*x^3+a)^(3/2)/x^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^7} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}}}{x^{7}} \,d x } \]

[In]

integrate((c*x^6+b*x^3+a)^(3/2)/x^7,x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)^(3/2)/x^7, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^7} \, dx=\int \frac {{\left (c\,x^6+b\,x^3+a\right )}^{3/2}}{x^7} \,d x \]

[In]

int((a + b*x^3 + c*x^6)^(3/2)/x^7,x)

[Out]

int((a + b*x^3 + c*x^6)^(3/2)/x^7, x)