3.1.26 \(\int \frac {(A+B x^2) (b x^2+c x^4)^3}{x^3} \, dx\) [26]

Optimal. Leaf size=68 \[ \frac {b (b B-A c) \left (b+c x^2\right )^4}{8 c^3}-\frac {(2 b B-A c) \left (b+c x^2\right )^5}{10 c^3}+\frac {B \left (b+c x^2\right )^6}{12 c^3} \]

[Out]

1/8*b*(-A*c+B*b)*(c*x^2+b)^4/c^3-1/10*(-A*c+2*B*b)*(c*x^2+b)^5/c^3+1/12*B*(c*x^2+b)^6/c^3

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Rubi [A]
time = 0.09, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1598, 457, 77} \begin {gather*} -\frac {\left (b+c x^2\right )^5 (2 b B-A c)}{10 c^3}+\frac {b \left (b+c x^2\right )^4 (b B-A c)}{8 c^3}+\frac {B \left (b+c x^2\right )^6}{12 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^3,x]

[Out]

(b*(b*B - A*c)*(b + c*x^2)^4)/(8*c^3) - ((2*b*B - A*c)*(b + c*x^2)^5)/(10*c^3) + (B*(b + c*x^2)^6)/(12*c^3)

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^3} \, dx &=\int x^3 \left (A+B x^2\right ) \left (b+c x^2\right )^3 \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int x (A+B x) (b+c x)^3 \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (\frac {b (b B-A c) (b+c x)^3}{c^2}+\frac {(-2 b B+A c) (b+c x)^4}{c^2}+\frac {B (b+c x)^5}{c^2}\right ) \, dx,x,x^2\right )\\ &=\frac {b (b B-A c) \left (b+c x^2\right )^4}{8 c^3}-\frac {(2 b B-A c) \left (b+c x^2\right )^5}{10 c^3}+\frac {B \left (b+c x^2\right )^6}{12 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 69, normalized size = 1.01 \begin {gather*} \frac {1}{120} x^4 \left (30 A b^3+20 b^2 (b B+3 A c) x^2+45 b c (b B+A c) x^4+12 c^2 (3 b B+A c) x^6+10 B c^3 x^8\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^3,x]

[Out]

(x^4*(30*A*b^3 + 20*b^2*(b*B + 3*A*c)*x^2 + 45*b*c*(b*B + A*c)*x^4 + 12*c^2*(3*b*B + A*c)*x^6 + 10*B*c^3*x^8))
/120

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Maple [A]
time = 0.40, size = 76, normalized size = 1.12

method	result	size
default	\(\frac {B \,c^{3} x^{12}}{12}+\frac {\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{10}}{10}+\frac {\left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{8}}{8}+\frac {\left (3 A \,b^{2} c +B \,b^{3}\right ) x^{6}}{6}+\frac {A \,b^{3} x^{4}}{4}\)	\(76\)
risch	\(\frac {1}{12} B \,c^{3} x^{12}+\frac {1}{10} x^{10} A \,c^{3}+\frac {3}{10} x^{10} B b \,c^{2}+\frac {3}{8} x^{8} A b \,c^{2}+\frac {3}{8} x^{8} B \,b^{2} c +\frac {1}{2} x^{6} A \,b^{2} c +\frac {1}{6} b^{3} B \,x^{6}+\frac {1}{4} A \,b^{3} x^{4}\)	\(78\)
norman	\(\frac {\left (\frac {1}{10} A \,c^{3}+\frac {3}{10} B b \,c^{2}\right ) x^{12}+\left (\frac {3}{8} A b \,c^{2}+\frac {3}{8} B \,b^{2} c \right ) x^{10}+\left (\frac {1}{2} A \,b^{2} c +\frac {1}{6} B \,b^{3}\right ) x^{8}+\frac {A \,b^{3} x^{6}}{4}+\frac {B \,c^{3} x^{14}}{12}}{x^{2}}\)	\(79\)
gosper	\(\frac {x^{4} \left (10 B \,c^{3} x^{8}+12 A \,c^{3} x^{6}+36 x^{6} B b \,c^{2}+45 A b \,c^{2} x^{4}+45 x^{4} B \,b^{2} c +60 A \,b^{2} c \,x^{2}+20 x^{2} B \,b^{3}+30 A \,b^{3}\right )}{120}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^3/x^3,x,method=_RETURNVERBOSE)

[Out]

1/12*B*c^3*x^12+1/10*(A*c^3+3*B*b*c^2)*x^10+1/8*(3*A*b*c^2+3*B*b^2*c)*x^8+1/6*(3*A*b^2*c+B*b^3)*x^6+1/4*A*b^3*
x^4

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Maxima [A]
time = 0.27, size = 73, normalized size = 1.07 \begin {gather*} \frac {1}{12} \, B c^{3} x^{12} + \frac {1}{10} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{10} + \frac {3}{8} \, {\left (B b^{2} c + A b c^{2}\right )} x^{8} + \frac {1}{4} \, A b^{3} x^{4} + \frac {1}{6} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^3,x, algorithm="maxima")

[Out]

1/12*B*c^3*x^12 + 1/10*(3*B*b*c^2 + A*c^3)*x^10 + 3/8*(B*b^2*c + A*b*c^2)*x^8 + 1/4*A*b^3*x^4 + 1/6*(B*b^3 + 3
*A*b^2*c)*x^6

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Fricas [A]
time = 1.84, size = 73, normalized size = 1.07 \begin {gather*} \frac {1}{12} \, B c^{3} x^{12} + \frac {1}{10} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{10} + \frac {3}{8} \, {\left (B b^{2} c + A b c^{2}\right )} x^{8} + \frac {1}{4} \, A b^{3} x^{4} + \frac {1}{6} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^3,x, algorithm="fricas")

[Out]

1/12*B*c^3*x^12 + 1/10*(3*B*b*c^2 + A*c^3)*x^10 + 3/8*(B*b^2*c + A*b*c^2)*x^8 + 1/4*A*b^3*x^4 + 1/6*(B*b^3 + 3
*A*b^2*c)*x^6

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Sympy [A]
time = 0.02, size = 82, normalized size = 1.21 \begin {gather*} \frac {A b^{3} x^{4}}{4} + \frac {B c^{3} x^{12}}{12} + x^{10} \left (\frac {A c^{3}}{10} + \frac {3 B b c^{2}}{10}\right ) + x^{8} \cdot \left (\frac {3 A b c^{2}}{8} + \frac {3 B b^{2} c}{8}\right ) + x^{6} \left (\frac {A b^{2} c}{2} + \frac {B b^{3}}{6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**3,x)

[Out]

A*b**3*x**4/4 + B*c**3*x**12/12 + x**10*(A*c**3/10 + 3*B*b*c**2/10) + x**8*(3*A*b*c**2/8 + 3*B*b**2*c/8) + x**
6*(A*b**2*c/2 + B*b**3/6)

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Giac [A]
time = 0.93, size = 77, normalized size = 1.13 \begin {gather*} \frac {1}{12} \, B c^{3} x^{12} + \frac {3}{10} \, B b c^{2} x^{10} + \frac {1}{10} \, A c^{3} x^{10} + \frac {3}{8} \, B b^{2} c x^{8} + \frac {3}{8} \, A b c^{2} x^{8} + \frac {1}{6} \, B b^{3} x^{6} + \frac {1}{2} \, A b^{2} c x^{6} + \frac {1}{4} \, A b^{3} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^3,x, algorithm="giac")

[Out]

1/12*B*c^3*x^12 + 3/10*B*b*c^2*x^10 + 1/10*A*c^3*x^10 + 3/8*B*b^2*c*x^8 + 3/8*A*b*c^2*x^8 + 1/6*B*b^3*x^6 + 1/
2*A*b^2*c*x^6 + 1/4*A*b^3*x^4

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Mupad [B]
time = 0.03, size = 69, normalized size = 1.01 \begin {gather*} x^6\,\left (\frac {B\,b^3}{6}+\frac {A\,c\,b^2}{2}\right )+x^{10}\,\left (\frac {A\,c^3}{10}+\frac {3\,B\,b\,c^2}{10}\right )+\frac {A\,b^3\,x^4}{4}+\frac {B\,c^3\,x^{12}}{12}+\frac {3\,b\,c\,x^8\,\left (A\,c+B\,b\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^3,x)

[Out]

x^6*((B*b^3)/6 + (A*b^2*c)/2) + x^10*((A*c^3)/10 + (3*B*b*c^2)/10) + (A*b^3*x^4)/4 + (B*c^3*x^12)/12 + (3*b*c*
x^8*(A*c + B*b))/8

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