\(\int (c+d x)^3 (a+b (c+d x)^4)^p \, dx\) [2910]

Optimal result

Integrand size = 21, antiderivative size = 30 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right )^p \, dx=\frac {\left (a+b (c+d x)^4\right )^{1+p}}{4 b d (1+p)} \]

[Out]

1/4*(a+b*(d*x+c)^4)^(p+1)/b/d/(p+1)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {379, 267} \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right )^p \, dx=\frac {\left (a+b (c+d x)^4\right )^{p+1}}{4 b d (p+1)} \]

[In]

Int[(c + d*x)^3*(a + b*(c + d*x)^4)^p,x]

[Out]

(a + b*(c + d*x)^4)^(1 + p)/(4*b*d*(1 + p))

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 379

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right )^p \, dx=\frac {\left (a+b (c+d x)^4\right )^{1+p}}{4 b d (1+p)} \]

[In]

Integrate[(c + d*x)^3*(a + b*(c + d*x)^4)^p,x]

[Out]

(a + b*(c + d*x)^4)^(1 + p)/(4*b*d*(1 + p))

Maple [A] (verified)

Time = 4.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {\left (a +b \left (d x +c \right )^{4}\right )^{1+p}}{4 b d \left (1+p \right )}\)	\(29\)
default	\(\frac {\left (a +b \left (d x +c \right )^{4}\right )^{1+p}}{4 b d \left (1+p \right )}\)	\(29\)
gosper	\(\frac {\left (b \,d^{4} x^{4}+4 b c \,d^{3} x^{3}+6 b \,c^{2} d^{2} x^{2}+4 b \,c^{3} d x +b \,c^{4}+a \right )^{1+p}}{4 b d \left (1+p \right )}\)	\(63\)
risch	\(\frac {\left (b \,d^{4} x^{4}+4 b c \,d^{3} x^{3}+6 b \,c^{2} d^{2} x^{2}+4 b \,c^{3} d x +b \,c^{4}+a \right ) \left (a +b \left (d x +c \right )^{4}\right )^{p}}{4 b d \left (1+p \right )}\)	\(72\)
parallelrisch	\(\frac {d^{7} \left (a +b \left (d x +c \right )^{4}\right )^{p} x^{4} b +4 c \,d^{6} \left (a +b \left (d x +c \right )^{4}\right )^{p} x^{3} b +6 c^{2} d^{5} \left (a +b \left (d x +c \right )^{4}\right )^{p} x^{2} b +4 c^{3} \left (a +b \left (d x +c \right )^{4}\right )^{p} x b \,d^{4}+\left (a +b \left (d x +c \right )^{4}\right )^{p} b \,c^{4} d^{3}+\left (a +b \left (d x +c \right )^{4}\right )^{p} a \,d^{3}}{4 b \,d^{4} \left (1+p \right )}\)	\(146\)
norman	\(\frac {c^{3} x \,{\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{4}\right )}}{1+p}+\frac {c \,d^{2} x^{3} {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{4}\right )}}{1+p}+\frac {d^{3} x^{4} {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{4}\right )}}{4+4 p}+\frac {3 c^{2} d \,x^{2} {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{4}\right )}}{2 \left (1+p \right )}+\frac {\left (b \,c^{4}+a \right ) {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{4}\right )}}{4 b d \left (1+p \right )}\)	\(147\)

[In]

int((d*x+c)^3*(a+b*(d*x+c)^4)^p,x,method=_RETURNVERBOSE)

[Out]

1/4*(a+b*(d*x+c)^4)^(1+p)/b/d/(1+p)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (28) = 56\).

Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.47 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right )^p \, dx=\frac {{\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a\right )} {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a\right )}^{p}}{4 \, {\left (b d p + b d\right )}} \]

[In]

integrate((d*x+c)^3*(a+b*(d*x+c)^4)^p,x, algorithm="fricas")

[Out]

1/4*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + a)*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c
^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + a)^p/(b*d*p + b*d)

Sympy [F(-1)]

Timed out. \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right )^p \, dx=\text {Timed out} \]

[In]

integrate((d*x+c)**3*(a+b*(d*x+c)**4)**p,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right )^p \, dx=\frac {{\left ({\left (d x + c\right )}^{4} b + a\right )}^{p + 1}}{4 \, b d {\left (p + 1\right )}} \]

[In]

integrate((d*x+c)^3*(a+b*(d*x+c)^4)^p,x, algorithm="maxima")

[Out]

1/4*((d*x + c)^4*b + a)^(p + 1)/(b*d*(p + 1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (28) = 56\).

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.07 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right )^p \, dx=\frac {{\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a\right )}^{p + 1}}{4 \, b d {\left (p + 1\right )}} \]

[In]

integrate((d*x+c)^3*(a+b*(d*x+c)^4)^p,x, algorithm="giac")

[Out]

1/4*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + a)^(p + 1)/(b*d*(p + 1))

Mupad [B] (verification not implemented)

Time = 6.01 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.93 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right )^p \, dx={\left (a+b\,{\left (c+d\,x\right )}^4\right )}^p\,\left (\frac {d^3\,x^4}{4\,\left (p+1\right )}+\frac {c^3\,x}{p+1}+\frac {b\,c^4+a}{4\,b\,d\,\left (p+1\right )}+\frac {3\,c^2\,d\,x^2}{2\,\left (p+1\right )}+\frac {c\,d^2\,x^3}{p+1}\right ) \]

[In]

int((a + b*(c + d*x)^4)^p*(c + d*x)^3,x)

[Out]

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