\(\int (c+d x)^3 (a+b (c+d x)^2)^p \, dx\) [2849]

Optimal result

Integrand size = 21, antiderivative size = 62 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=-\frac {a \left (a+b (c+d x)^2\right )^{1+p}}{2 b^2 d (1+p)}+\frac {\left (a+b (c+d x)^2\right )^{2+p}}{2 b^2 d (2+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {379, 272, 45} \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\frac {\left (a+b (c+d x)^2\right )^{p+2}}{2 b^2 d (p+2)}-\frac {a \left (a+b (c+d x)^2\right )^{p+1}}{2 b^2 d (p+1)} \]

[In]

Int[(c + d*x)^3*(a + b*(c + d*x)^2)^p,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.82 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\frac {\left (a+b (c+d x)^2\right )^{1+p} \left (-a+b (1+p) (c+d x)^2\right )}{2 b^2 d (1+p) (2+p)} \]

[In]

Integrate[(c + d*x)^3*(a + b*(c + d*x)^2)^p,x]

[Out]

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

Maple [A] (verified)

Time = 4.00 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.47

method	result	size
gosper	\(-\frac {\left (b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a \right )^{1+p} \left (-b \,d^{2} p \,x^{2}-2 b c d p x -b \,d^{2} x^{2}-b \,c^{2} p -2 b c d x -b \,c^{2}+a \right )}{2 b^{2} d \left (p^{2}+3 p +2\right )}\)	\(91\)
risch	\(-\frac {\left (-b^{2} d^{4} p \,x^{4}-4 b^{2} c \,d^{3} p \,x^{3}-d^{4} x^{4} b^{2}-6 b^{2} c^{2} d^{2} p \,x^{2}-4 c \,d^{3} x^{3} b^{2}-4 b^{2} c^{3} d p x -6 b^{2} c^{2} d^{2} x^{2}-a b \,d^{2} p \,x^{2}-b^{2} c^{4} p -4 b^{2} d x \,c^{3}-2 a b c d p x -b^{2} c^{4}-a b \,c^{2} p +a^{2}\right ) \left (a +b \left (d x +c \right )^{2}\right )^{p}}{2 \left (1+p \right ) \left (2+p \right ) b^{2} d}\)	\(178\)
norman	\(\frac {c \left (2 b \,c^{2} p +2 b \,c^{2}+a p \right ) x \,{\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{2}\right )}}{b \left (p^{2}+3 p +2\right )}+\frac {d^{3} x^{4} {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{2}\right )}}{4+2 p}+\frac {2 c \,d^{2} x^{3} {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{2}\right )}}{2+p}-\frac {\left (-b^{2} c^{4} p -b^{2} c^{4}-a b \,c^{2} p +a^{2}\right ) {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{2}\right )}}{2 b^{2} \left (p^{2}+3 p +2\right ) d}+\frac {d \left (6 b \,c^{2} p +6 b \,c^{2}+a p \right ) x^{2} {\mathrm e}^{p \ln \left (a +b \left (d x +c \right )^{2}\right )}}{2 b \left (p^{2}+3 p +2\right )}\)	\(220\)
parallelrisch	\(\frac {x^{4} \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} d^{5} p +d^{5} \left (a +b \left (d x +c \right )^{2}\right )^{p} x^{4} b^{2}+4 x^{3} \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c \,d^{4} p +4 c \,d^{4} \left (a +b \left (d x +c \right )^{2}\right )^{p} x^{3} b^{2}+6 x^{2} \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{2} d^{3} p +6 x^{2} \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{2} d^{3}+4 x \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{3} d^{2} p +x^{2} \left (a +b \left (d x +c \right )^{2}\right )^{p} a b \,d^{3} p +4 x \left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{3} d^{2}+\left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{4} d p +2 x \left (a +b \left (d x +c \right )^{2}\right )^{p} a b c \,d^{2} p +\left (a +b \left (d x +c \right )^{2}\right )^{p} b^{2} c^{4} d +\left (a +b \left (d x +c \right )^{2}\right )^{p} a b \,c^{2} d p -\left (a +b \left (d x +c \right )^{2}\right )^{p} a^{2} d}{2 \left (2+p \right ) \left (1+p \right ) b^{2} d^{2}}\)	\(353\)

[In]

int((d*x+c)^3*(a+b*(d*x+c)^2)^p,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (58) = 116\).

Time = 0.24 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.95 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\frac {{\left (b^{2} c^{4} + {\left (b^{2} d^{4} p + b^{2} d^{4}\right )} x^{4} + 4 \, {\left (b^{2} c d^{3} p + b^{2} c d^{3}\right )} x^{3} + {\left (6 \, b^{2} c^{2} d^{2} + {\left (6 \, b^{2} c^{2} + a b\right )} d^{2} p\right )} x^{2} - a^{2} + {\left (b^{2} c^{4} + a b c^{2}\right )} p + 2 \, {\left (2 \, b^{2} c^{3} d + {\left (2 \, b^{2} c^{3} + a b c\right )} d p\right )} x\right )} {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p}}{2 \, {\left (b^{2} d p^{2} + 3 \, b^{2} d p + 2 \, b^{2} d\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (58) = 116\).

Time = 0.24 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.26 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\frac {{\left (b^{2} d^{4} {\left (p + 1\right )} x^{4} + 4 \, b^{2} c d^{3} {\left (p + 1\right )} x^{3} + b^{2} c^{4} {\left (p + 1\right )} + a b c^{2} p + {\left (6 \, b^{2} c^{2} d^{2} {\left (p + 1\right )} + a b d^{2} p\right )} x^{2} - a^{2} + 2 \, {\left (2 \, b^{2} c^{3} d {\left (p + 1\right )} + a b c d p\right )} x\right )} {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p}}{2 \, {\left (p^{2} + 3 \, p + 2\right )} b^{2} d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.60 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx=\frac {{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{2} {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p}}{2 \, b^{2} d {\left (p + 2\right )}} - \frac {{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p + 1} a}{2 \, b^{2} d {\left (p + 1\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.86 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.71 \[ \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx={\left (a+b\,{\left (c+d\,x\right )}^2\right )}^p\,\left (\frac {d^3\,x^4\,\left (p+1\right )}{2\,\left (p^2+3\,p+2\right )}+\frac {\left (b\,c^2+a\right )\,\left (b\,c^2-a+b\,c^2\,p\right )}{2\,b^2\,d\,\left (p^2+3\,p+2\right )}+\frac {c\,x\,\left (a\,p+2\,b\,c^2+2\,b\,c^2\,p\right )}{b\,\left (p^2+3\,p+2\right )}+\frac {2\,c\,d^2\,x^3\,\left (p+1\right )}{p^2+3\,p+2}+\frac {d\,x^2\,\left (a\,p+6\,b\,c^2+6\,b\,c^2\,p\right )}{2\,b\,\left (p^2+3\,p+2\right )}\right ) \]

[In]

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

[Out]

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