3.2756 \(\int (c x)^m (a+b x^3)^2 \, dx\)

Optimal. Leaf size=58 \[ \frac {a^2 (c x)^{m+1}}{c (m+1)}+\frac {2 a b (c x)^{m+4}}{c^4 (m+4)}+\frac {b^2 (c x)^{m+7}}{c^7 (m+7)} \]

[Out]

a^2*(c*x)^(1+m)/c/(1+m)+2*a*b*(c*x)^(4+m)/c^4/(4+m)+b^2*(c*x)^(7+m)/c^7/(7+m)

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Rubi [A]  time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {270} \[ \frac {a^2 (c x)^{m+1}}{c (m+1)}+\frac {2 a b (c x)^{m+4}}{c^4 (m+4)}+\frac {b^2 (c x)^{m+7}}{c^7 (m+7)} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^m*(a + b*x^3)^2,x]

[Out]

(a^2*(c*x)^(1 + m))/(c*(1 + m)) + (2*a*b*(c*x)^(4 + m))/(c^4*(4 + m)) + (b^2*(c*x)^(7 + m))/(c^7*(7 + 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]

Rubi steps

\begin {align*} \int (c x)^m \left (a+b x^3\right )^2 \, dx &=\int \left (a^2 (c x)^m+\frac {2 a b (c x)^{3+m}}{c^3}+\frac {b^2 (c x)^{6+m}}{c^6}\right ) \, dx\\ &=\frac {a^2 (c x)^{1+m}}{c (1+m)}+\frac {2 a b (c x)^{4+m}}{c^4 (4+m)}+\frac {b^2 (c x)^{7+m}}{c^7 (7+m)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 41, normalized size = 0.71 \[ x (c x)^m \left (\frac {a^2}{m+1}+\frac {2 a b x^3}{m+4}+\frac {b^2 x^6}{m+7}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^m*(a + b*x^3)^2,x]

[Out]

x*(c*x)^m*(a^2/(1 + m) + (2*a*b*x^3)/(4 + m) + (b^2*x^6)/(7 + m))

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fricas [A]  time = 0.69, size = 87, normalized size = 1.50 \[ \frac {{\left ({\left (b^{2} m^{2} + 5 \, b^{2} m + 4 \, b^{2}\right )} x^{7} + 2 \, {\left (a b m^{2} + 8 \, a b m + 7 \, a b\right )} x^{4} + {\left (a^{2} m^{2} + 11 \, a^{2} m + 28 \, a^{2}\right )} x\right )} \left (c x\right )^{m}}{m^{3} + 12 \, m^{2} + 39 \, m + 28} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b^2*m^2 + 5*b^2*m + 4*b^2)*x^7 + 2*(a*b*m^2 + 8*a*b*m + 7*a*b)*x^4 + (a^2*m^2 + 11*a^2*m + 28*a^2)*x)*(c*x)^
m/(m^3 + 12*m^2 + 39*m + 28)

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giac [B]  time = 0.19, size = 135, normalized size = 2.33 \[ \frac {\left (c x\right )^{m} b^{2} m^{2} x^{7} + 5 \, \left (c x\right )^{m} b^{2} m x^{7} + 4 \, \left (c x\right )^{m} b^{2} x^{7} + 2 \, \left (c x\right )^{m} a b m^{2} x^{4} + 16 \, \left (c x\right )^{m} a b m x^{4} + 14 \, \left (c x\right )^{m} a b x^{4} + \left (c x\right )^{m} a^{2} m^{2} x + 11 \, \left (c x\right )^{m} a^{2} m x + 28 \, \left (c x\right )^{m} a^{2} x}{m^{3} + 12 \, m^{2} + 39 \, m + 28} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c*x)^m*b^2*m^2*x^7 + 5*(c*x)^m*b^2*m*x^7 + 4*(c*x)^m*b^2*x^7 + 2*(c*x)^m*a*b*m^2*x^4 + 16*(c*x)^m*a*b*m*x^4
+ 14*(c*x)^m*a*b*x^4 + (c*x)^m*a^2*m^2*x + 11*(c*x)^m*a^2*m*x + 28*(c*x)^m*a^2*x)/(m^3 + 12*m^2 + 39*m + 28)

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maple [A]  time = 0.01, size = 94, normalized size = 1.62 \[ \frac {\left (b^{2} m^{2} x^{6}+5 b^{2} m \,x^{6}+4 b^{2} x^{6}+2 a b \,m^{2} x^{3}+16 a b m \,x^{3}+14 a b \,x^{3}+a^{2} m^{2}+11 a^{2} m +28 a^{2}\right ) x \left (c x \right )^{m}}{\left (m +7\right ) \left (m +4\right ) \left (m +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(b^2*m^2*x^6+5*b^2*m*x^6+4*b^2*x^6+2*a*b*m^2*x^3+16*a*b*m*x^3+14*a*b*x^3+a^2*m^2+11*a^2*m+28*a^2)*(c*x)^m/(m
+7)/(m+4)/(m+1)

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maxima [A]  time = 0.53, size = 56, normalized size = 0.97 \[ \frac {b^{2} c^{m} x^{7} x^{m}}{m + 7} + \frac {2 \, a b c^{m} x^{4} x^{m}}{m + 4} + \frac {\left (c x\right )^{m + 1} a^{2}}{c {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*c^m*x^7*x^m/(m + 7) + 2*a*b*c^m*x^4*x^m/(m + 4) + (c*x)^(m + 1)*a^2/(c*(m + 1))

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mupad [B]  time = 1.22, size = 95, normalized size = 1.64 \[ {\left (c\,x\right )}^m\,\left (\frac {a^2\,x\,\left (m^2+11\,m+28\right )}{m^3+12\,m^2+39\,m+28}+\frac {b^2\,x^7\,\left (m^2+5\,m+4\right )}{m^3+12\,m^2+39\,m+28}+\frac {2\,a\,b\,x^4\,\left (m^2+8\,m+7\right )}{m^3+12\,m^2+39\,m+28}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^m*(a + b*x^3)^2,x)

[Out]

(c*x)^m*((a^2*x*(11*m + m^2 + 28))/(39*m + 12*m^2 + m^3 + 28) + (b^2*x^7*(5*m + m^2 + 4))/(39*m + 12*m^2 + m^3
+ 28) + (2*a*b*x^4*(8*m + m^2 + 7))/(39*m + 12*m^2 + m^3 + 28))

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sympy [A]  time = 1.39, size = 352, normalized size = 6.07 \[ \begin {cases} \frac {- \frac {a^{2}}{6 x^{6}} - \frac {2 a b}{3 x^{3}} + b^{2} \log {\relax (x )}}{c^{7}} & \text {for}\: m = -7 \\\frac {- \frac {a^{2}}{3 x^{3}} + 2 a b \log {\relax (x )} + \frac {b^{2} x^{3}}{3}}{c^{4}} & \text {for}\: m = -4 \\\frac {a^{2} \log {\relax (x )} + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{6}}{6}}{c} & \text {for}\: m = -1 \\\frac {a^{2} c^{m} m^{2} x x^{m}}{m^{3} + 12 m^{2} + 39 m + 28} + \frac {11 a^{2} c^{m} m x x^{m}}{m^{3} + 12 m^{2} + 39 m + 28} + \frac {28 a^{2} c^{m} x x^{m}}{m^{3} + 12 m^{2} + 39 m + 28} + \frac {2 a b c^{m} m^{2} x^{4} x^{m}}{m^{3} + 12 m^{2} + 39 m + 28} + \frac {16 a b c^{m} m x^{4} x^{m}}{m^{3} + 12 m^{2} + 39 m + 28} + \frac {14 a b c^{m} x^{4} x^{m}}{m^{3} + 12 m^{2} + 39 m + 28} + \frac {b^{2} c^{m} m^{2} x^{7} x^{m}}{m^{3} + 12 m^{2} + 39 m + 28} + \frac {5 b^{2} c^{m} m x^{7} x^{m}}{m^{3} + 12 m^{2} + 39 m + 28} + \frac {4 b^{2} c^{m} x^{7} x^{m}}{m^{3} + 12 m^{2} + 39 m + 28} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-a**2/(6*x**6) - 2*a*b/(3*x**3) + b**2*log(x))/c**7, Eq(m, -7)), ((-a**2/(3*x**3) + 2*a*b*log(x) +
b**2*x**3/3)/c**4, Eq(m, -4)), ((a**2*log(x) + 2*a*b*x**3/3 + b**2*x**6/6)/c, Eq(m, -1)), (a**2*c**m*m**2*x*x
**m/(m**3 + 12*m**2 + 39*m + 28) + 11*a**2*c**m*m*x*x**m/(m**3 + 12*m**2 + 39*m + 28) + 28*a**2*c**m*x*x**m/(m
**3 + 12*m**2 + 39*m + 28) + 2*a*b*c**m*m**2*x**4*x**m/(m**3 + 12*m**2 + 39*m + 28) + 16*a*b*c**m*m*x**4*x**m/
(m**3 + 12*m**2 + 39*m + 28) + 14*a*b*c**m*x**4*x**m/(m**3 + 12*m**2 + 39*m + 28) + b**2*c**m*m**2*x**7*x**m/(
m**3 + 12*m**2 + 39*m + 28) + 5*b**2*c**m*m*x**7*x**m/(m**3 + 12*m**2 + 39*m + 28) + 4*b**2*c**m*x**7*x**m/(m*
*3 + 12*m**2 + 39*m + 28), True))

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