3.2849 \(\int (c+d x)^3 (a+b (c+d x)^2)^p \, dx\)

Optimal. Leaf size=62 \[ \frac{\left (a+b (c+d x)^2\right )^{p+2}}{2 b^2 d (p+2)}-\frac{a \left (a+b (c+d x)^2\right )^{p+1}}{2 b^2 d (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0590412, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {372, 266, 43} \[ \frac{\left (a+b (c+d x)^2\right )^{p+2}}{2 b^2 d (p+2)}-\frac{a \left (a+b (c+d x)^2\right )^{p+1}}{2 b^2 d (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 372

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]

Rule 266

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 43

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])

Rubi steps

\begin{align*} \int (c+d x)^3 \left (a+b (c+d x)^2\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int x^3 \left (a+b x^2\right )^p \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int x (a+b x)^p \, dx,x,(c+d x)^2\right )}{2 d}\\ &=\frac{\operatorname{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]  time = 0.0416965, size = 51, normalized size = 0.82 \[ \frac{\left (a+b (c+d x)^2\right )^{p+1} \left (b (p+1) (c+d x)^2-a\right )}{2 b^2 d (p+1) (p+2)} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.006, size = 91, normalized size = 1.5 \begin{align*} -{\frac{ \left ( b{d}^{2}{x}^{2}+2\,bcdx+b{c}^{2}+a \right ) ^{1+p} \left ( -b{d}^{2}p{x}^{2}-2\,bcdpx-b{d}^{2}{x}^{2}-b{c}^{2}p-2\,bcdx-b{c}^{2}+a \right ) }{2\,{b}^{2}d \left ({p}^{2}+3\,p+2 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.39967, size = 189, normalized size = 3.05 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 1.6184, size = 369, normalized size = 5.95 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.12388, size = 662, normalized size = 10.68 \begin{align*} \frac{{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} b^{2} d^{4} p x^{4} + 4 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} b^{2} c d^{3} p x^{3} +{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} b^{2} d^{4} x^{4} + 6 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} b^{2} c^{2} d^{2} p x^{2} + 4 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} b^{2} c d^{3} x^{3} + 4 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} b^{2} c^{3} d p x + 6 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} b^{2} c^{2} d^{2} x^{2} +{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} b^{2} c^{4} p + 4 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} b^{2} c^{3} d x +{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} a b d^{2} p x^{2} +{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} b^{2} c^{4} + 2 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} a b c d p x +{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} a b c^{2} p -{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p} a^{2}}{2 \,{\left (b^{2} d p^{2} + 3 \, b^{2} d p + 2 \, b^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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