3.3.35 \(\int (c x)^m (b x^2+c x^4)^2 \, dx\)

Optimal. Leaf size=52 \[ \frac {b^2 x^5 (c x)^m}{m+5}+\frac {2 b c x^7 (c x)^m}{m+7}+\frac {c^2 x^9 (c x)^m}{m+9} \]

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Rubi [A]  time = 0.04, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1142, 1584, 270} \begin {gather*} \frac {b^2 x^5 (c x)^m}{m+5}+\frac {2 b c x^7 (c x)^m}{m+7}+\frac {c^2 x^9 (c x)^m}{m+9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x)^m*(b*x^2 + c*x^4)^2,x]

[Out]

(b^2*x^5*(c*x)^m)/(5 + m) + (2*b*c*x^7*(c*x)^m)/(7 + m) + (c^2*x^9*(c*x)^m)/(9 + m)

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1142

Int[(u_)^(m_.)*((a_.) + (b_.)*(v_)^2 + (c_.)*(v_)^4)^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m),
Subst[Int[x^m*(a + b*x^2 + c*x^(2*2))^p, x], x, v], x] /; FreeQ[{a, b, c, m, p}, x] && LinearPairQ[u, v, x]

Rule 1584

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 (c x)^m \left (b x^2+c x^4\right )^2 \, dx &=\left (x^{-m} (c x)^m\right ) \operatorname {Subst}\left (\int x^m \left (b x^2+c x^4\right )^2 \, dx,x,x\right )\\ &=\left (x^{-m} (c x)^m\right ) \operatorname {Subst}\left (\int x^{4+m} \left (b+c x^2\right )^2 \, dx,x,x\right )\\ &=\left (x^{-m} (c x)^m\right ) \operatorname {Subst}\left (\int \left (b^2 x^{4+m}+2 b c x^{6+m}+c^2 x^{8+m}\right ) \, dx,x,x\right )\\ &=\frac {b^2 x^5 (c x)^m}{5+m}+\frac {2 b c x^7 (c x)^m}{7+m}+\frac {c^2 x^9 (c x)^m}{9+m}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 43, normalized size = 0.83 \begin {gather*} x^5 (c x)^m \left (\frac {b^2}{m+5}+\frac {2 b c x^2}{m+7}+\frac {c^2 x^4}{m+9}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^m*(b*x^2 + c*x^4)^2,x]

[Out]

x^5*(c*x)^m*(b^2/(5 + m) + (2*b*c*x^2)/(7 + m) + (c^2*x^4)/(9 + m))

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IntegrateAlgebraic [F]  time = 0.10, size = 0, normalized size = 0.00 \begin {gather*} \int (c x)^m \left (b x^2+c x^4\right )^2 \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(c*x)^m*(b*x^2 + c*x^4)^2,x]

[Out]

Defer[IntegrateAlgebraic][(c*x)^m*(b*x^2 + c*x^4)^2, x]

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fricas [A]  time = 0.65, size = 89, normalized size = 1.71 \begin {gather*} \frac {{\left ({\left (c^{2} m^{2} + 12 \, c^{2} m + 35 \, c^{2}\right )} x^{9} + 2 \, {\left (b c m^{2} + 14 \, b c m + 45 \, b c\right )} x^{7} + {\left (b^{2} m^{2} + 16 \, b^{2} m + 63 \, b^{2}\right )} x^{5}\right )} \left (c x\right )^{m}}{m^{3} + 21 \, m^{2} + 143 \, m + 315} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c^2*m^2 + 12*c^2*m + 35*c^2)*x^9 + 2*(b*c*m^2 + 14*b*c*m + 45*b*c)*x^7 + (b^2*m^2 + 16*b^2*m + 63*b^2)*x^5)*
(c*x)^m/(m^3 + 21*m^2 + 143*m + 315)

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giac [B]  time = 0.17, size = 141, normalized size = 2.71 \begin {gather*} \frac {\left (c x\right )^{m} c^{2} m^{2} x^{9} + 12 \, \left (c x\right )^{m} c^{2} m x^{9} + 2 \, \left (c x\right )^{m} b c m^{2} x^{7} + 35 \, \left (c x\right )^{m} c^{2} x^{9} + 28 \, \left (c x\right )^{m} b c m x^{7} + \left (c x\right )^{m} b^{2} m^{2} x^{5} + 90 \, \left (c x\right )^{m} b c x^{7} + 16 \, \left (c x\right )^{m} b^{2} m x^{5} + 63 \, \left (c x\right )^{m} b^{2} x^{5}}{m^{3} + 21 \, m^{2} + 143 \, m + 315} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c*x)^m*c^2*m^2*x^9 + 12*(c*x)^m*c^2*m*x^9 + 2*(c*x)^m*b*c*m^2*x^7 + 35*(c*x)^m*c^2*x^9 + 28*(c*x)^m*b*c*m*x^
7 + (c*x)^m*b^2*m^2*x^5 + 90*(c*x)^m*b*c*x^7 + 16*(c*x)^m*b^2*m*x^5 + 63*(c*x)^m*b^2*x^5)/(m^3 + 21*m^2 + 143*
m + 315)

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maple [A]  time = 0.01, size = 96, normalized size = 1.85 \begin {gather*} \frac {\left (c^{2} m^{2} x^{4}+12 c^{2} m \,x^{4}+2 b c \,m^{2} x^{2}+35 c^{2} x^{4}+28 b c m \,x^{2}+b^{2} m^{2}+90 b c \,x^{2}+16 b^{2} m +63 b^{2}\right ) x^{5} \left (c x \right )^{m}}{\left (m +9\right ) \left (m +7\right ) \left (m +5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*x)^m*(c^2*m^2*x^4+12*c^2*m*x^4+2*b*c*m^2*x^2+35*c^2*x^4+28*b*c*m*x^2+b^2*m^2+90*b*c*x^2+16*b^2*m+63*b^2)*x^
5/(m+9)/(m+7)/(5+m)

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maxima [A]  time = 1.45, size = 55, normalized size = 1.06 \begin {gather*} \frac {c^{m + 2} x^{9} x^{m}}{m + 9} + \frac {2 \, b c^{m + 1} x^{7} x^{m}}{m + 7} + \frac {b^{2} c^{m} x^{5} x^{m}}{m + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^(m + 2)*x^9*x^m/(m + 9) + 2*b*c^(m + 1)*x^7*x^m/(m + 7) + b^2*c^m*x^5*x^m/(m + 5)

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mupad [B]  time = 4.19, size = 97, normalized size = 1.87 \begin {gather*} {\left (c\,x\right )}^m\,\left (\frac {b^2\,x^5\,\left (m^2+16\,m+63\right )}{m^3+21\,m^2+143\,m+315}+\frac {c^2\,x^9\,\left (m^2+12\,m+35\right )}{m^3+21\,m^2+143\,m+315}+\frac {2\,b\,c\,x^7\,\left (m^2+14\,m+45\right )}{m^3+21\,m^2+143\,m+315}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^m*(b*x^2 + c*x^4)^2,x)

[Out]

(c*x)^m*((b^2*x^5*(16*m + m^2 + 63))/(143*m + 21*m^2 + m^3 + 315) + (c^2*x^9*(12*m + m^2 + 35))/(143*m + 21*m^
2 + m^3 + 315) + (2*b*c*x^7*(14*m + m^2 + 45))/(143*m + 21*m^2 + m^3 + 315))

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sympy [A]  time = 2.29, size = 352, normalized size = 6.77 \begin {gather*} \begin {cases} \frac {- \frac {b^{2}}{4 x^{4}} - \frac {b c}{x^{2}} + c^{2} \log {\relax (x )}}{c^{9}} & \text {for}\: m = -9 \\\frac {- \frac {b^{2}}{2 x^{2}} + 2 b c \log {\relax (x )} + \frac {c^{2} x^{2}}{2}}{c^{7}} & \text {for}\: m = -7 \\\frac {b^{2} \log {\relax (x )} + b c x^{2} + \frac {c^{2} x^{4}}{4}}{c^{5}} & \text {for}\: m = -5 \\\frac {b^{2} c^{m} m^{2} x^{5} x^{m}}{m^{3} + 21 m^{2} + 143 m + 315} + \frac {16 b^{2} c^{m} m x^{5} x^{m}}{m^{3} + 21 m^{2} + 143 m + 315} + \frac {63 b^{2} c^{m} x^{5} x^{m}}{m^{3} + 21 m^{2} + 143 m + 315} + \frac {2 b c c^{m} m^{2} x^{7} x^{m}}{m^{3} + 21 m^{2} + 143 m + 315} + \frac {28 b c c^{m} m x^{7} x^{m}}{m^{3} + 21 m^{2} + 143 m + 315} + \frac {90 b c c^{m} x^{7} x^{m}}{m^{3} + 21 m^{2} + 143 m + 315} + \frac {c^{2} c^{m} m^{2} x^{9} x^{m}}{m^{3} + 21 m^{2} + 143 m + 315} + \frac {12 c^{2} c^{m} m x^{9} x^{m}}{m^{3} + 21 m^{2} + 143 m + 315} + \frac {35 c^{2} c^{m} x^{9} x^{m}}{m^{3} + 21 m^{2} + 143 m + 315} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**m*(c*x**4+b*x**2)**2,x)

[Out]

Piecewise(((-b**2/(4*x**4) - b*c/x**2 + c**2*log(x))/c**9, Eq(m, -9)), ((-b**2/(2*x**2) + 2*b*c*log(x) + c**2*
x**2/2)/c**7, Eq(m, -7)), ((b**2*log(x) + b*c*x**2 + c**2*x**4/4)/c**5, Eq(m, -5)), (b**2*c**m*m**2*x**5*x**m/
(m**3 + 21*m**2 + 143*m + 315) + 16*b**2*c**m*m*x**5*x**m/(m**3 + 21*m**2 + 143*m + 315) + 63*b**2*c**m*x**5*x
**m/(m**3 + 21*m**2 + 143*m + 315) + 2*b*c*c**m*m**2*x**7*x**m/(m**3 + 21*m**2 + 143*m + 315) + 28*b*c*c**m*m*
x**7*x**m/(m**3 + 21*m**2 + 143*m + 315) + 90*b*c*c**m*x**7*x**m/(m**3 + 21*m**2 + 143*m + 315) + c**2*c**m*m*
*2*x**9*x**m/(m**3 + 21*m**2 + 143*m + 315) + 12*c**2*c**m*m*x**9*x**m/(m**3 + 21*m**2 + 143*m + 315) + 35*c**
2*c**m*x**9*x**m/(m**3 + 21*m**2 + 143*m + 315), True))

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