\(\int (a+b x^n) (c+d x^n)^3 \, dx\) [285]

Optimal result

Integrand size = 17, antiderivative size = 99 \[ \int \left (a+b x^n\right ) \left (c+d x^n\right )^3 \, dx=a c^3 x+\frac {c^2 (b c+3 a d) x^{1+n}}{1+n}+\frac {3 c d (b c+a d) x^{1+2 n}}{1+2 n}+\frac {d^2 (3 b c+a d) x^{1+3 n}}{1+3 n}+\frac {b d^3 x^{1+4 n}}{1+4 n} \]

[Out]

a*c^3*x+c^2*(3*a*d+b*c)*x^(1+n)/(1+n)+3*c*d*(a*d+b*c)*x^(1+2*n)/(1+2*n)+d^2*(a*d+3*b*c)*x^(1+3*n)/(1+3*n)+b*d^
3*x^(1+4*n)/(1+4*n)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {380} \[ \int \left (a+b x^n\right ) \left (c+d x^n\right )^3 \, dx=\frac {c^2 x^{n+1} (3 a d+b c)}{n+1}+\frac {d^2 x^{3 n+1} (a d+3 b c)}{3 n+1}+\frac {3 c d x^{2 n+1} (a d+b c)}{2 n+1}+a c^3 x+\frac {b d^3 x^{4 n+1}}{4 n+1} \]

[In]

Int[(a + b*x^n)*(c + d*x^n)^3,x]

[Out]

a*c^3*x + (c^2*(b*c + 3*a*d)*x^(1 + n))/(1 + n) + (3*c*d*(b*c + a*d)*x^(1 + 2*n))/(1 + 2*n) + (d^2*(3*b*c + a*
d)*x^(1 + 3*n))/(1 + 3*n) + (b*d^3*x^(1 + 4*n))/(1 + 4*n)

Rule 380

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a c^3+c^2 (b c+3 a d) x^n+3 c d (b c+a d) x^{2 n}+d^2 (3 b c+a d) x^{3 n}+b d^3 x^{4 n}\right ) \, dx \\ & = a c^3 x+\frac {c^2 (b c+3 a d) x^{1+n}}{1+n}+\frac {3 c d (b c+a d) x^{1+2 n}}{1+2 n}+\frac {d^2 (3 b c+a d) x^{1+3 n}}{1+3 n}+\frac {b d^3 x^{1+4 n}}{1+4 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.91 \[ \int \left (a+b x^n\right ) \left (c+d x^n\right )^3 \, dx=\frac {b x \left (c+d x^n\right )^4-(b c-a d (1+4 n)) x \left (c^3+\frac {3 c^2 d x^n}{1+n}+\frac {3 c d^2 x^{2 n}}{1+2 n}+\frac {d^3 x^{3 n}}{1+3 n}\right )}{d+4 d n} \]

[In]

Integrate[(a + b*x^n)*(c + d*x^n)^3,x]

[Out]

(b*x*(c + d*x^n)^4 - (b*c - a*d*(1 + 4*n))*x*(c^3 + (3*c^2*d*x^n)/(1 + n) + (3*c*d^2*x^(2*n))/(1 + 2*n) + (d^3
*x^(3*n))/(1 + 3*n)))/(d + 4*d*n)

Maple [A] (verified)

Time = 4.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.97

method	result	size
risch	\(a \,c^{3} x +\frac {b \,d^{3} x \,x^{4 n}}{1+4 n}+\frac {c^{2} \left (3 a d +b c \right ) x \,x^{n}}{1+n}+\frac {d^{2} \left (a d +3 b c \right ) x \,x^{3 n}}{1+3 n}+\frac {3 c d \left (a d +b c \right ) x \,x^{2 n}}{1+2 n}\)	\(96\)
norman	\(a \,c^{3} x +\frac {b \,d^{3} x \,{\mathrm e}^{4 n \ln \left (x \right )}}{1+4 n}+\frac {c^{2} \left (3 a d +b c \right ) x \,{\mathrm e}^{n \ln \left (x \right )}}{1+n}+\frac {d^{2} \left (a d +3 b c \right ) x \,{\mathrm e}^{3 n \ln \left (x \right )}}{1+3 n}+\frac {3 c d \left (a d +b c \right ) x \,{\mathrm e}^{2 n \ln \left (x \right )}}{1+2 n}\)	\(104\)
parallelrisch	\(\frac {21 x \,x^{3 n} b c \,d^{2} n +57 x \,x^{2 n} a c \,d^{2} n^{2}+a \,c^{3} x +9 x \,x^{n} b \,c^{3} n +3 x \,x^{n} a \,c^{2} d +78 x \,x^{n} a \,c^{2} d \,n^{2}+27 x \,x^{n} a \,c^{2} d n +24 x \,x^{2 n} a c \,d^{2} n +24 x \,x^{n} b \,c^{3} n^{3}+42 x \,x^{3 n} b c \,d^{2} n^{2}+36 x \,x^{2 n} a c \,d^{2} n^{3}+24 x \,x^{2 n} b \,c^{2} d n +x \,x^{n} b \,c^{3}+10 x a \,c^{3} n +b \,d^{3} x \,x^{4 n}+24 x a \,c^{3} n^{4}+50 x a \,c^{3} n^{3}+35 x a \,c^{3} n^{2}+72 x \,x^{n} a \,c^{2} d \,n^{3}+26 x \,x^{n} b \,c^{3} n^{2}+24 x \,x^{3 n} b c \,d^{2} n^{3}+57 x \,x^{2 n} b \,c^{2} d \,n^{2}+x \,x^{3 n} a \,d^{3}+36 x \,x^{2 n} b \,c^{2} d \,n^{3}+3 x \,x^{2 n} a c \,d^{2}+3 x \,x^{2 n} b \,c^{2} d +6 x \,x^{4 n} b \,d^{3} n^{3}+11 x \,x^{4 n} b \,d^{3} n^{2}+8 x \,x^{3 n} a \,d^{3} n^{3}+6 x \,x^{4 n} b \,d^{3} n +14 x \,x^{3 n} a \,d^{3} n^{2}+7 x \,x^{3 n} a \,d^{3} n +3 x \,x^{3 n} b c \,d^{2}}{\left (1+4 n \right ) \left (1+n \right ) \left (1+3 n \right ) \left (1+2 n \right )}\)	\(455\)

[In]

int((a+b*x^n)*(c+d*x^n)^3,x,method=_RETURNVERBOSE)

[Out]

a*c^3*x+b*d^3/(1+4*n)*x*(x^n)^4+c^2*(3*a*d+b*c)/(1+n)*x*x^n+d^2*(a*d+3*b*c)/(1+3*n)*x*(x^n)^3+3*c*d*(a*d+b*c)/
(1+2*n)*x*(x^n)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (99) = 198\).

Time = 0.25 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.22 \[ \int \left (a+b x^n\right ) \left (c+d x^n\right )^3 \, dx=\frac {{\left (6 \, b d^{3} n^{3} + 11 \, b d^{3} n^{2} + 6 \, b d^{3} n + b d^{3}\right )} x x^{4 \, n} + {\left (3 \, b c d^{2} + a d^{3} + 8 \, {\left (3 \, b c d^{2} + a d^{3}\right )} n^{3} + 14 \, {\left (3 \, b c d^{2} + a d^{3}\right )} n^{2} + 7 \, {\left (3 \, b c d^{2} + a d^{3}\right )} n\right )} x x^{3 \, n} + 3 \, {\left (b c^{2} d + a c d^{2} + 12 \, {\left (b c^{2} d + a c d^{2}\right )} n^{3} + 19 \, {\left (b c^{2} d + a c d^{2}\right )} n^{2} + 8 \, {\left (b c^{2} d + a c d^{2}\right )} n\right )} x x^{2 \, n} + {\left (b c^{3} + 3 \, a c^{2} d + 24 \, {\left (b c^{3} + 3 \, a c^{2} d\right )} n^{3} + 26 \, {\left (b c^{3} + 3 \, a c^{2} d\right )} n^{2} + 9 \, {\left (b c^{3} + 3 \, a c^{2} d\right )} n\right )} x x^{n} + {\left (24 \, a c^{3} n^{4} + 50 \, a c^{3} n^{3} + 35 \, a c^{3} n^{2} + 10 \, a c^{3} n + a c^{3}\right )} x}{24 \, n^{4} + 50 \, n^{3} + 35 \, n^{2} + 10 \, n + 1} \]

[In]

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

[Out]

((6*b*d^3*n^3 + 11*b*d^3*n^2 + 6*b*d^3*n + b*d^3)*x*x^(4*n) + (3*b*c*d^2 + a*d^3 + 8*(3*b*c*d^2 + a*d^3)*n^3 +
14*(3*b*c*d^2 + a*d^3)*n^2 + 7*(3*b*c*d^2 + a*d^3)*n)*x*x^(3*n) + 3*(b*c^2*d + a*c*d^2 + 12*(b*c^2*d + a*c*d^
2)*n^3 + 19*(b*c^2*d + a*c*d^2)*n^2 + 8*(b*c^2*d + a*c*d^2)*n)*x*x^(2*n) + (b*c^3 + 3*a*c^2*d + 24*(b*c^3 + 3*
a*c^2*d)*n^3 + 26*(b*c^3 + 3*a*c^2*d)*n^2 + 9*(b*c^3 + 3*a*c^2*d)*n)*x*x^n + (24*a*c^3*n^4 + 50*a*c^3*n^3 + 35
*a*c^3*n^2 + 10*a*c^3*n + a*c^3)*x)/(24*n^4 + 50*n^3 + 35*n^2 + 10*n + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1540 vs. \(2 (90) = 180\).

Time = 0.49 (sec) , antiderivative size = 1540, normalized size of antiderivative = 15.56 \[ \int \left (a+b x^n\right ) \left (c+d x^n\right )^3 \, dx=\text {Too large to display} \]

[In]

integrate((a+b*x**n)*(c+d*x**n)**3,x)

[Out]

Piecewise((a*c**3*x + 3*a*c**2*d*log(x) - 3*a*c*d**2/x - a*d**3/(2*x**2) + b*c**3*log(x) - 3*b*c**2*d/x - 3*b*
c*d**2/(2*x**2) - b*d**3/(3*x**3), Eq(n, -1)), (a*c**3*x + 6*a*c**2*d*sqrt(x) + 3*a*c*d**2*log(x) - 2*a*d**3/s
qrt(x) + 2*b*c**3*sqrt(x) + 3*b*c**2*d*log(x) - 6*b*c*d**2/sqrt(x) - b*d**3/x, Eq(n, -1/2)), (a*c**3*x + 9*a*c
**2*d*x**(2/3)/2 + 9*a*c*d**2*x**(1/3) + a*d**3*log(x) + 3*b*c**3*x**(2/3)/2 + 9*b*c**2*d*x**(1/3) + 3*b*c*d**
2*log(x) - 3*b*d**3/x**(1/3), Eq(n, -1/3)), (a*c**3*x + 4*a*c**2*d*x**(3/4) + 6*a*c*d**2*sqrt(x) + 4*a*d**3*x*
*(1/4) + 4*b*c**3*x**(3/4)/3 + 6*b*c**2*d*sqrt(x) + 12*b*c*d**2*x**(1/4) + b*d**3*log(x), Eq(n, -1/4)), (24*a*
c**3*n**4*x/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 50*a*c**3*n**3*x/(24*n**4 + 50*n**3 + 35*n**2 + 10*n +
1) + 35*a*c**3*n**2*x/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 10*a*c**3*n*x/(24*n**4 + 50*n**3 + 35*n**2 +
10*n + 1) + a*c**3*x/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 72*a*c**2*d*n**3*x*x**n/(24*n**4 + 50*n**3 + 3
5*n**2 + 10*n + 1) + 78*a*c**2*d*n**2*x*x**n/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 27*a*c**2*d*n*x*x**n/(
24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 3*a*c**2*d*x*x**n/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 36*a*c*
d**2*n**3*x*x**(2*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 57*a*c*d**2*n**2*x*x**(2*n)/(24*n**4 + 50*n**3
+ 35*n**2 + 10*n + 1) + 24*a*c*d**2*n*x*x**(2*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 3*a*c*d**2*x*x**(
2*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 8*a*d**3*n**3*x*x**(3*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n +
1) + 14*a*d**3*n**2*x*x**(3*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 7*a*d**3*n*x*x**(3*n)/(24*n**4 + 50
*n**3 + 35*n**2 + 10*n + 1) + a*d**3*x*x**(3*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 24*b*c**3*n**3*x*x*
*n/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 26*b*c**3*n**2*x*x**n/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) +
9*b*c**3*n*x*x**n/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + b*c**3*x*x**n/(24*n**4 + 50*n**3 + 35*n**2 + 10*
n + 1) + 36*b*c**2*d*n**3*x*x**(2*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 57*b*c**2*d*n**2*x*x**(2*n)/(2
4*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 24*b*c**2*d*n*x*x**(2*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 3
*b*c**2*d*x*x**(2*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 24*b*c*d**2*n**3*x*x**(3*n)/(24*n**4 + 50*n**3
+ 35*n**2 + 10*n + 1) + 42*b*c*d**2*n**2*x*x**(3*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 21*b*c*d**2*n*
x*x**(3*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 3*b*c*d**2*x*x**(3*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*
n + 1) + 6*b*d**3*n**3*x*x**(4*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 11*b*d**3*n**2*x*x**(4*n)/(24*n**
4 + 50*n**3 + 35*n**2 + 10*n + 1) + 6*b*d**3*n*x*x**(4*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + b*d**3*x*
x**(4*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.41 \[ \int \left (a+b x^n\right ) \left (c+d x^n\right )^3 \, dx=a c^{3} x + \frac {b d^{3} x^{4 \, n + 1}}{4 \, n + 1} + \frac {3 \, b c d^{2} x^{3 \, n + 1}}{3 \, n + 1} + \frac {a d^{3} x^{3 \, n + 1}}{3 \, n + 1} + \frac {3 \, b c^{2} d x^{2 \, n + 1}}{2 \, n + 1} + \frac {3 \, a c d^{2} x^{2 \, n + 1}}{2 \, n + 1} + \frac {b c^{3} x^{n + 1}}{n + 1} + \frac {3 \, a c^{2} d x^{n + 1}}{n + 1} \]

[In]

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

[Out]

a*c^3*x + b*d^3*x^(4*n + 1)/(4*n + 1) + 3*b*c*d^2*x^(3*n + 1)/(3*n + 1) + a*d^3*x^(3*n + 1)/(3*n + 1) + 3*b*c^
2*d*x^(2*n + 1)/(2*n + 1) + 3*a*c*d^2*x^(2*n + 1)/(2*n + 1) + b*c^3*x^(n + 1)/(n + 1) + 3*a*c^2*d*x^(n + 1)/(n
+ 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 450 vs. \(2 (99) = 198\).

Time = 0.29 (sec) , antiderivative size = 450, normalized size of antiderivative = 4.55 \[ \int \left (a+b x^n\right ) \left (c+d x^n\right )^3 \, dx=\frac {24 \, a c^{3} n^{4} x + 6 \, b d^{3} n^{3} x x^{4 \, n} + 24 \, b c d^{2} n^{3} x x^{3 \, n} + 8 \, a d^{3} n^{3} x x^{3 \, n} + 36 \, b c^{2} d n^{3} x x^{2 \, n} + 36 \, a c d^{2} n^{3} x x^{2 \, n} + 24 \, b c^{3} n^{3} x x^{n} + 72 \, a c^{2} d n^{3} x x^{n} + 50 \, a c^{3} n^{3} x + 11 \, b d^{3} n^{2} x x^{4 \, n} + 42 \, b c d^{2} n^{2} x x^{3 \, n} + 14 \, a d^{3} n^{2} x x^{3 \, n} + 57 \, b c^{2} d n^{2} x x^{2 \, n} + 57 \, a c d^{2} n^{2} x x^{2 \, n} + 26 \, b c^{3} n^{2} x x^{n} + 78 \, a c^{2} d n^{2} x x^{n} + 35 \, a c^{3} n^{2} x + 6 \, b d^{3} n x x^{4 \, n} + 21 \, b c d^{2} n x x^{3 \, n} + 7 \, a d^{3} n x x^{3 \, n} + 24 \, b c^{2} d n x x^{2 \, n} + 24 \, a c d^{2} n x x^{2 \, n} + 9 \, b c^{3} n x x^{n} + 27 \, a c^{2} d n x x^{n} + 10 \, a c^{3} n x + b d^{3} x x^{4 \, n} + 3 \, b c d^{2} x x^{3 \, n} + a d^{3} x x^{3 \, n} + 3 \, b c^{2} d x x^{2 \, n} + 3 \, a c d^{2} x x^{2 \, n} + b c^{3} x x^{n} + 3 \, a c^{2} d x x^{n} + a c^{3} x}{24 \, n^{4} + 50 \, n^{3} + 35 \, n^{2} + 10 \, n + 1} \]

[In]

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

[Out]

(24*a*c^3*n^4*x + 6*b*d^3*n^3*x*x^(4*n) + 24*b*c*d^2*n^3*x*x^(3*n) + 8*a*d^3*n^3*x*x^(3*n) + 36*b*c^2*d*n^3*x*
x^(2*n) + 36*a*c*d^2*n^3*x*x^(2*n) + 24*b*c^3*n^3*x*x^n + 72*a*c^2*d*n^3*x*x^n + 50*a*c^3*n^3*x + 11*b*d^3*n^2
*x*x^(4*n) + 42*b*c*d^2*n^2*x*x^(3*n) + 14*a*d^3*n^2*x*x^(3*n) + 57*b*c^2*d*n^2*x*x^(2*n) + 57*a*c*d^2*n^2*x*x
^(2*n) + 26*b*c^3*n^2*x*x^n + 78*a*c^2*d*n^2*x*x^n + 35*a*c^3*n^2*x + 6*b*d^3*n*x*x^(4*n) + 21*b*c*d^2*n*x*x^(
3*n) + 7*a*d^3*n*x*x^(3*n) + 24*b*c^2*d*n*x*x^(2*n) + 24*a*c*d^2*n*x*x^(2*n) + 9*b*c^3*n*x*x^n + 27*a*c^2*d*n*
x*x^n + 10*a*c^3*n*x + b*d^3*x*x^(4*n) + 3*b*c*d^2*x*x^(3*n) + a*d^3*x*x^(3*n) + 3*b*c^2*d*x*x^(2*n) + 3*a*c*d
^2*x*x^(2*n) + b*c^3*x*x^n + 3*a*c^2*d*x*x^n + a*c^3*x)/(24*n^4 + 50*n^3 + 35*n^2 + 10*n + 1)

Mupad [B] (verification not implemented)

Time = 5.65 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^n\right ) \left (c+d x^n\right )^3 \, dx=a\,c^3\,x+\frac {x\,x^n\,\left (b\,c^3+3\,a\,d\,c^2\right )}{n+1}+\frac {x\,x^{3\,n}\,\left (a\,d^3+3\,b\,c\,d^2\right )}{3\,n+1}+\frac {b\,d^3\,x\,x^{4\,n}}{4\,n+1}+\frac {3\,c\,d\,x\,x^{2\,n}\,\left (a\,d+b\,c\right )}{2\,n+1} \]

[In]

int((a + b*x^n)*(c + d*x^n)^3,x)

[Out]

a*c^3*x + (x*x^n*(b*c^3 + 3*a*c^2*d))/(n + 1) + (x*x^(3*n)*(a*d^3 + 3*b*c*d^2))/(3*n + 1) + (b*d^3*x*x^(4*n))/
(4*n + 1) + (3*c*d*x*x^(2*n)*(a*d + b*c))/(2*n + 1)