\(\int (c x)^m (a+b x^n)^2 \, dx\) [2755]

Optimal result

Integrand size = 15, antiderivative size = 64 \[ \int (c x)^m \left (a+b x^n\right )^2 \, dx=\frac {2 a b x^{1+n} (c x)^m}{1+m+n}+\frac {b^2 x^{1+2 n} (c x)^m}{1+m+2 n}+\frac {a^2 (c x)^{1+m}}{c (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {276, 20, 30} \[ \int (c x)^m \left (a+b x^n\right )^2 \, dx=\frac {a^2 (c x)^{m+1}}{c (m+1)}+\frac {2 a b x^{n+1} (c x)^m}{m+n+1}+\frac {b^2 x^{2 n+1} (c x)^m}{m+2 n+1} \]

[In]

Int[(c*x)^m*(a + b*x^n)^2,x]

[Out]

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

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[b^IntPart[n]*((b*v)^FracPart[n]/(a^IntPart[n]
*(a*v)^FracPart[n])), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 276

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*} \text {integral}& = \int \left (a^2 (c x)^m+2 a b x^n (c x)^m+b^2 x^{2 n} (c x)^m\right ) \, dx \\ & = \frac {a^2 (c x)^{1+m}}{c (1+m)}+(2 a b) \int x^n (c x)^m \, dx+b^2 \int x^{2 n} (c x)^m \, dx \\ & = \frac {a^2 (c x)^{1+m}}{c (1+m)}+\left (2 a b x^{-m} (c x)^m\right ) \int x^{m+n} \, dx+\left (b^2 x^{-m} (c x)^m\right ) \int x^{m+2 n} \, dx \\ & = \frac {2 a b x^{1+n} (c x)^m}{1+m+n}+\frac {b^2 x^{1+2 n} (c x)^m}{1+m+2 n}+\frac {a^2 (c x)^{1+m}}{c (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.73 \[ \int (c x)^m \left (a+b x^n\right )^2 \, dx=x (c x)^m \left (\frac {a^2}{1+m}+\frac {2 a b x^n}{1+m+n}+\frac {b^2 x^{2 n}}{1+m+2 n}\right ) \]

[In]

Integrate[(c*x)^m*(a + b*x^n)^2,x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 3.82 (sec) , antiderivative size = 201, normalized size of antiderivative = 3.14

method	result	size
risch	\(\frac {x \left (b^{2} m^{2} x^{2 n}+b^{2} m n \,x^{2 n}+2 a b \,m^{2} x^{n}+4 a b m n \,x^{n}+2 m \,b^{2} x^{2 n}+b^{2} n \,x^{2 n}+a^{2} m^{2}+3 a^{2} m n +2 a^{2} n^{2}+4 m a b \,x^{n}+4 a b n \,x^{n}+b^{2} x^{2 n}+2 a^{2} m +3 a^{2} n +2 a b \,x^{n}+a^{2}\right ) x^{m} c^{m} {\mathrm e}^{\frac {i \operatorname {csgn}\left (i c x \right ) \pi m \left (\operatorname {csgn}\left (i c x \right )-\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i c x \right )+\operatorname {csgn}\left (i c \right )\right )}{2}}}{\left (1+m \right ) \left (1+m +n \right ) \left (1+m +2 n \right )}\)	\(201\)
parallelrisch	\(\frac {x \,x^{2 n} \left (c x \right )^{m} b^{2} m n +2 x \,x^{n} \left (c x \right )^{m} a b \,m^{2}+4 x \,x^{n} \left (c x \right )^{m} a b m +4 x \,x^{n} \left (c x \right )^{m} a b n +4 x \,x^{n} \left (c x \right )^{m} a b m n +x \,x^{2 n} \left (c x \right )^{m} b^{2}+x \left (c x \right )^{m} a^{2} m^{2}+2 x \left (c x \right )^{m} a^{2} n^{2}+2 x \left (c x \right )^{m} a^{2} m +3 x \left (c x \right )^{m} a^{2} n +x \left (c x \right )^{m} a^{2}+x \,x^{2 n} \left (c x \right )^{m} b^{2} m^{2}+2 x \,x^{2 n} \left (c x \right )^{m} b^{2} m +x \,x^{2 n} \left (c x \right )^{m} b^{2} n +3 x \left (c x \right )^{m} a^{2} m n +2 x \,x^{n} \left (c x \right )^{m} a b}{\left (1+m \right ) \left (1+m +n \right ) \left (1+m +2 n \right )}\)	\(251\)

[In]

int((c*x)^m*(a+b*x^n)^2,x,method=_RETURNVERBOSE)

[Out]

x*(b^2*m^2*(x^n)^2+b^2*m*n*(x^n)^2+2*a*b*m^2*x^n+4*a*b*m*n*x^n+2*m*b^2*(x^n)^2+b^2*n*(x^n)^2+a^2*m^2+3*a^2*m*n
+2*a^2*n^2+4*m*a*b*x^n+4*a*b*n*x^n+b^2*(x^n)^2+2*a^2*m+3*a^2*n+2*a*b*x^n+a^2)/(1+m)/(1+m+n)/(1+m+2*n)*x^m*c^m*
exp(1/2*I*csgn(I*c*x)*Pi*m*(csgn(I*c*x)-csgn(I*x))*(-csgn(I*c*x)+csgn(I*c)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (64) = 128\).

Time = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.69 \[ \int (c x)^m \left (a+b x^n\right )^2 \, dx=\frac {{\left (b^{2} m^{2} + 2 \, b^{2} m + b^{2} + {\left (b^{2} m + b^{2}\right )} n\right )} x x^{2 \, n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 2 \, {\left (a b m^{2} + 2 \, a b m + a b + 2 \, {\left (a b m + a b\right )} n\right )} x x^{n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + {\left (a^{2} m^{2} + 2 \, a^{2} n^{2} + 2 \, a^{2} m + a^{2} + 3 \, {\left (a^{2} m + a^{2}\right )} n\right )} x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )}}{m^{3} + 2 \, {\left (m + 1\right )} n^{2} + 3 \, m^{2} + 3 \, {\left (m^{2} + 2 \, m + 1\right )} n + 3 \, m + 1} \]

[In]

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

[Out]

((b^2*m^2 + 2*b^2*m + b^2 + (b^2*m + b^2)*n)*x*x^(2*n)*e^(m*log(c) + m*log(x)) + 2*(a*b*m^2 + 2*a*b*m + a*b +
2*(a*b*m + a*b)*n)*x*x^n*e^(m*log(c) + m*log(x)) + (a^2*m^2 + 2*a^2*n^2 + 2*a^2*m + a^2 + 3*(a^2*m + a^2)*n)*x
*e^(m*log(c) + m*log(x)))/(m^3 + 2*(m + 1)*n^2 + 3*m^2 + 3*(m^2 + 2*m + 1)*n + 3*m + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1148 vs. \(2 (56) = 112\).

Time = 1.45 (sec) , antiderivative size = 1148, normalized size of antiderivative = 17.94 \[ \int (c x)^m \left (a+b x^n\right )^2 \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise(((a + b)**2*log(x)/c, Eq(m, -1) & Eq(n, 0)), ((a**2*log(x) + 2*a*b*x**n/n + b**2*x**(2*n)/(2*n))/c,
Eq(m, -1)), (a**2*Piecewise((0**(-2*n - 1)*x, Eq(c, 0)), (Piecewise((-1/(2*n*(c*x)**(2*n)), Ne(n, 0)), (log(c*
x), True))/c, True)) + 2*a*b*Piecewise((-x*x**n*(c*x)**(-2*n - 1)/n, Ne(n, 0)), (x*x**n*(c*x)**(-2*n - 1)*log(
x), True)) + b**2*x*x**(2*n)*(c*x)**(-2*n - 1)*log(x), Eq(m, -2*n - 1)), (a**2*Piecewise((0**(-n - 1)*x, Eq(c,
0)), (Piecewise((-1/(n*(c*x)**n), Ne(n, 0)), (log(c*x), True))/c, True)) + 2*a*b*x*x**n*(c*x)**(-n - 1)*log(x
) + b**2*Piecewise((x*x**(2*n)*(c*x)**(-n - 1)/n, Ne(n, 0)), (x*x**(2*n)*(c*x)**(-n - 1)*log(x), True)), Eq(m,
-n - 1)), (a**2*m**2*x*(c*x)**m/(m**3 + 3*m**2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) + 3*a*
*2*m*n*x*(c*x)**m/(m**3 + 3*m**2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) + 2*a**2*m*x*(c*x)**m
/(m**3 + 3*m**2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) + 2*a**2*n**2*x*(c*x)**m/(m**3 + 3*m**
2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) + 3*a**2*n*x*(c*x)**m/(m**3 + 3*m**2*n + 3*m**2 + 2*
m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) + a**2*x*(c*x)**m/(m**3 + 3*m**2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3*m
+ 2*n**2 + 3*n + 1) + 2*a*b*m**2*x*x**n*(c*x)**m/(m**3 + 3*m**2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3*m + 2*n**2 +
3*n + 1) + 4*a*b*m*n*x*x**n*(c*x)**m/(m**3 + 3*m**2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) +
4*a*b*m*x*x**n*(c*x)**m/(m**3 + 3*m**2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) + 4*a*b*n*x*x*
*n*(c*x)**m/(m**3 + 3*m**2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) + 2*a*b*x*x**n*(c*x)**m/(m*
*3 + 3*m**2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) + b**2*m**2*x*x**(2*n)*(c*x)**m/(m**3 + 3*
m**2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) + b**2*m*n*x*x**(2*n)*(c*x)**m/(m**3 + 3*m**2*n +
3*m**2 + 2*m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) + 2*b**2*m*x*x**(2*n)*(c*x)**m/(m**3 + 3*m**2*n + 3*m**2
+ 2*m*n**2 + 6*m*n + 3*m + 2*n**2 + 3*n + 1) + b**2*n*x*x**(2*n)*(c*x)**m/(m**3 + 3*m**2*n + 3*m**2 + 2*m*n**2
+ 6*m*n + 3*m + 2*n**2 + 3*n + 1) + b**2*x*x**(2*n)*(c*x)**m/(m**3 + 3*m**2*n + 3*m**2 + 2*m*n**2 + 6*m*n + 3
*m + 2*n**2 + 3*n + 1), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11 \[ \int (c x)^m \left (a+b x^n\right )^2 \, dx=\frac {b^{2} c^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {2 \, a b c^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {\left (c x\right )^{m + 1} a^{2}}{c {\left (m + 1\right )}} \]

[In]

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

[Out]

b^2*c^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + 2*a*b*c^m*x*e^(m*log(x) + n*log(x))/(m + n + 1) + (c*x)^(m
+ 1)*a^2/(c*(m + 1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 613 vs. \(2 (64) = 128\).

Time = 0.29 (sec) , antiderivative size = 613, normalized size of antiderivative = 9.58 \[ \int (c x)^m \left (a+b x^n\right )^2 \, dx=\frac {b^{2} m^{2} x x^{2 \, n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + b^{2} m n x x^{2 \, n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 2 \, a b m^{2} x x^{n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + b^{2} m^{2} x x^{n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 4 \, a b m n x x^{n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + b^{2} m n x x^{n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + a^{2} m^{2} x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 2 \, a b m^{2} x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + b^{2} m^{2} x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 3 \, a^{2} m n x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 4 \, a b m n x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + b^{2} m n x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 2 \, a^{2} n^{2} x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 2 \, b^{2} m x x^{2 \, n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + b^{2} n x x^{2 \, n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 4 \, a b m x x^{n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 2 \, b^{2} m x x^{n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 4 \, a b n x x^{n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + b^{2} n x x^{n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 2 \, a^{2} m x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 4 \, a b m x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 2 \, b^{2} m x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 3 \, a^{2} n x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 4 \, a b n x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + b^{2} n x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + b^{2} x x^{2 \, n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 2 \, a b x x^{n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + b^{2} x x^{n} e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + a^{2} x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + 2 \, a b x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )} + b^{2} x e^{\left (m \log \left (c\right ) + m \log \left (x\right )\right )}}{m^{3} + 3 \, m^{2} n + 2 \, m n^{2} + 3 \, m^{2} + 6 \, m n + 2 \, n^{2} + 3 \, m + 3 \, n + 1} \]

[In]

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

[Out]

(b^2*m^2*x*x^(2*n)*e^(m*log(c) + m*log(x)) + b^2*m*n*x*x^(2*n)*e^(m*log(c) + m*log(x)) + 2*a*b*m^2*x*x^n*e^(m*
log(c) + m*log(x)) + b^2*m^2*x*x^n*e^(m*log(c) + m*log(x)) + 4*a*b*m*n*x*x^n*e^(m*log(c) + m*log(x)) + b^2*m*n
*x*x^n*e^(m*log(c) + m*log(x)) + a^2*m^2*x*e^(m*log(c) + m*log(x)) + 2*a*b*m^2*x*e^(m*log(c) + m*log(x)) + b^2
*m^2*x*e^(m*log(c) + m*log(x)) + 3*a^2*m*n*x*e^(m*log(c) + m*log(x)) + 4*a*b*m*n*x*e^(m*log(c) + m*log(x)) + b
^2*m*n*x*e^(m*log(c) + m*log(x)) + 2*a^2*n^2*x*e^(m*log(c) + m*log(x)) + 2*b^2*m*x*x^(2*n)*e^(m*log(c) + m*log
(x)) + b^2*n*x*x^(2*n)*e^(m*log(c) + m*log(x)) + 4*a*b*m*x*x^n*e^(m*log(c) + m*log(x)) + 2*b^2*m*x*x^n*e^(m*lo
g(c) + m*log(x)) + 4*a*b*n*x*x^n*e^(m*log(c) + m*log(x)) + b^2*n*x*x^n*e^(m*log(c) + m*log(x)) + 2*a^2*m*x*e^(
m*log(c) + m*log(x)) + 4*a*b*m*x*e^(m*log(c) + m*log(x)) + 2*b^2*m*x*e^(m*log(c) + m*log(x)) + 3*a^2*n*x*e^(m*
log(c) + m*log(x)) + 4*a*b*n*x*e^(m*log(c) + m*log(x)) + b^2*n*x*e^(m*log(c) + m*log(x)) + b^2*x*x^(2*n)*e^(m*
log(c) + m*log(x)) + 2*a*b*x*x^n*e^(m*log(c) + m*log(x)) + b^2*x*x^n*e^(m*log(c) + m*log(x)) + a^2*x*e^(m*log(
c) + m*log(x)) + 2*a*b*x*e^(m*log(c) + m*log(x)) + b^2*x*e^(m*log(c) + m*log(x)))/(m^3 + 3*m^2*n + 2*m*n^2 + 3
*m^2 + 6*m*n + 2*n^2 + 3*m + 3*n + 1)

Mupad [B] (verification not implemented)

Time = 5.71 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.39 \[ \int (c x)^m \left (a+b x^n\right )^2 \, dx={\left (c\,x\right )}^m\,\left (\frac {a^2\,x}{m+1}+\frac {b^2\,x\,x^{2\,n}\,\left (m+n+1\right )}{m^2+3\,m\,n+2\,m+2\,n^2+3\,n+1}+\frac {2\,a\,b\,x\,x^n\,\left (m+2\,n+1\right )}{m^2+3\,m\,n+2\,m+2\,n^2+3\,n+1}\right ) \]

[In]

int((c*x)^m*(a + b*x^n)^2,x)

[Out]

(c*x)^m*((a^2*x)/(m + 1) + (b^2*x*x^(2*n)*(m + n + 1))/(2*m + 3*n + 3*m*n + m^2 + 2*n^2 + 1) + (2*a*b*x*x^n*(m
+ 2*n + 1))/(2*m + 3*n + 3*m*n + m^2 + 2*n^2 + 1))