3.2847 \(\int (c+d x)^5 (a+b (c+d x)^2)^p \, dx\)

Optimal. Leaf size=93 \[ \frac{a^2 \left (a+b (c+d x)^2\right )^{p+1}}{2 b^3 d (p+1)}-\frac{a \left (a+b (c+d x)^2\right )^{p+2}}{b^3 d (p+2)}+\frac{\left (a+b (c+d x)^2\right )^{p+3}}{2 b^3 d (p+3)} \]

[Out]

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

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Rubi [A]  time = 0.0950768, antiderivative size = 93, 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} \[ \frac{a^2 \left (a+b (c+d x)^2\right )^{p+1}}{2 b^3 d (p+1)}-\frac{a \left (a+b (c+d x)^2\right )^{p+2}}{b^3 d (p+2)}+\frac{\left (a+b (c+d x)^2\right )^{p+3}}{2 b^3 d (p+3)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^5*(a + b*(c + d*x)^2)^p,x]

[Out]

(a^2*(a + b*(c + d*x)^2)^(1 + p))/(2*b^3*d*(1 + p)) - (a*(a + b*(c + d*x)^2)^(2 + p))/(b^3*d*(2 + p)) + (a + b
*(c + d*x)^2)^(3 + p)/(2*b^3*d*(3 + 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)^5 \left (a+b (c+d x)^2\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int x^5 \left (a+b x^2\right )^p \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int x^2 (a+b x)^p \, dx,x,(c+d x)^2\right )}{2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2 (a+b x)^p}{b^2}-\frac{2 a (a+b x)^{1+p}}{b^2}+\frac{(a+b x)^{2+p}}{b^2}\right ) \, dx,x,(c+d x)^2\right )}{2 d}\\ &=\frac{a^2 \left (a+b (c+d x)^2\right )^{1+p}}{2 b^3 d (1+p)}-\frac{a \left (a+b (c+d x)^2\right )^{2+p}}{b^3 d (2+p)}+\frac{\left (a+b (c+d x)^2\right )^{3+p}}{2 b^3 d (3+p)}\\ \end{align*}

Mathematica [A]  time = 0.0768374, size = 73, normalized size = 0.78 \[ \frac{\left (a+b (c+d x)^2\right )^{p+1} \left (\frac{a^2}{p+1}-\frac{2 a \left (a+b (c+d x)^2\right )}{p+2}+\frac{\left (a+b (c+d x)^2\right )^2}{p+3}\right )}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^5*(a + b*(c + d*x)^2)^p,x]

[Out]

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

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Maple [B]  time = 0.015, size = 289, normalized size = 3.1 \begin{align*}{\frac{ \left ( b{d}^{2}{x}^{2}+2\,bcdx+b{c}^{2}+a \right ) ^{1+p} \left ({b}^{2}{d}^{4}{p}^{2}{x}^{4}+4\,{b}^{2}c{d}^{3}{p}^{2}{x}^{3}+3\,{b}^{2}{d}^{4}p{x}^{4}+6\,{b}^{2}{c}^{2}{d}^{2}{p}^{2}{x}^{2}+12\,{b}^{2}c{d}^{3}p{x}^{3}+2\,{d}^{4}{x}^{4}{b}^{2}+4\,{b}^{2}{c}^{3}d{p}^{2}x+18\,{b}^{2}{c}^{2}{d}^{2}p{x}^{2}+8\,c{d}^{3}{x}^{3}{b}^{2}+{b}^{2}{c}^{4}{p}^{2}+12\,{b}^{2}{c}^{3}dpx+12\,{b}^{2}{c}^{2}{d}^{2}{x}^{2}-2\,ab{d}^{2}p{x}^{2}+3\,{b}^{2}{c}^{4}p+8\,{b}^{2}{c}^{3}dx-4\,abcdpx-2\,ab{d}^{2}{x}^{2}+2\,{b}^{2}{c}^{4}-2\,ab{c}^{2}p-4\,abcdx-2\,ab{c}^{2}+2\,{a}^{2} \right ) }{2\,{b}^{3}d \left ({p}^{3}+6\,{p}^{2}+11\,p+6 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^5*(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^2*d^4*p^2*x^4+4*b^2*c*d^3*p^2*x^3+3*b^2*d^4*p*x^4+6*b^2*c^2*d^2*p^2
*x^2+12*b^2*c*d^3*p*x^3+2*b^2*d^4*x^4+4*b^2*c^3*d*p^2*x+18*b^2*c^2*d^2*p*x^2+8*b^2*c*d^3*x^3+b^2*c^4*p^2+12*b^
2*c^3*d*p*x+12*b^2*c^2*d^2*x^2-2*a*b*d^2*p*x^2+3*b^2*c^4*p+8*b^2*c^3*d*x-4*a*b*c*d*p*x-2*a*b*d^2*x^2+2*b^2*c^4
-2*a*b*c^2*p-4*a*b*c*d*x-2*a*b*c^2+2*a^2)/b^3/d/(p^3+6*p^2+11*p+6)

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Maxima [B]  time = 1.46512, size = 405, normalized size = 4.35 \begin{align*} \frac{{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} d^{6} x^{6} + 6 \,{\left (p^{2} + 3 \, p + 2\right )} b^{3} c d^{5} x^{5} +{\left (p^{2} + 3 \, p + 2\right )} b^{3} c^{6} +{\left (p^{2} + p\right )} a b^{2} c^{4} - 2 \, a^{2} b c^{2} p +{\left (15 \,{\left (p^{2} + 3 \, p + 2\right )} b^{3} c^{2} d^{4} +{\left (p^{2} + p\right )} a b^{2} d^{4}\right )} x^{4} + 4 \,{\left (5 \,{\left (p^{2} + 3 \, p + 2\right )} b^{3} c^{3} d^{3} +{\left (p^{2} + p\right )} a b^{2} c d^{3}\right )} x^{3} + 2 \, a^{3} +{\left (15 \,{\left (p^{2} + 3 \, p + 2\right )} b^{3} c^{4} d^{2} + 6 \,{\left (p^{2} + p\right )} a b^{2} c^{2} d^{2} - 2 \, a^{2} b d^{2} p\right )} x^{2} + 2 \,{\left (3 \,{\left (p^{2} + 3 \, p + 2\right )} b^{3} c^{5} d + 2 \,{\left (p^{2} + p\right )} a b^{2} c^{3} d - 2 \, a^{2} 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^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^5*(a+b*(d*x+c)^2)^p,x, algorithm="maxima")

[Out]

1/2*((p^2 + 3*p + 2)*b^3*d^6*x^6 + 6*(p^2 + 3*p + 2)*b^3*c*d^5*x^5 + (p^2 + 3*p + 2)*b^3*c^6 + (p^2 + p)*a*b^2
*c^4 - 2*a^2*b*c^2*p + (15*(p^2 + 3*p + 2)*b^3*c^2*d^4 + (p^2 + p)*a*b^2*d^4)*x^4 + 4*(5*(p^2 + 3*p + 2)*b^3*c
^3*d^3 + (p^2 + p)*a*b^2*c*d^3)*x^3 + 2*a^3 + (15*(p^2 + 3*p + 2)*b^3*c^4*d^2 + 6*(p^2 + p)*a*b^2*c^2*d^2 - 2*
a^2*b*d^2*p)*x^2 + 2*(3*(p^2 + 3*p + 2)*b^3*c^5*d + 2*(p^2 + p)*a*b^2*c^3*d - 2*a^2*b*c*d*p)*x)*(b*d^2*x^2 + 2
*b*c*d*x + b*c^2 + a)^p/((p^3 + 6*p^2 + 11*p + 6)*b^3*d)

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Fricas [B]  time = 1.6406, size = 887, normalized size = 9.54 \begin{align*} \frac{{\left (2 \, b^{3} c^{6} +{\left (b^{3} d^{6} p^{2} + 3 \, b^{3} d^{6} p + 2 \, b^{3} d^{6}\right )} x^{6} + 6 \,{\left (b^{3} c d^{5} p^{2} + 3 \, b^{3} c d^{5} p + 2 \, b^{3} c d^{5}\right )} x^{5} +{\left (30 \, b^{3} c^{2} d^{4} +{\left (15 \, b^{3} c^{2} + a b^{2}\right )} d^{4} p^{2} +{\left (45 \, b^{3} c^{2} + a b^{2}\right )} d^{4} p\right )} x^{4} + 4 \,{\left (10 \, b^{3} c^{3} d^{3} +{\left (5 \, b^{3} c^{3} + a b^{2} c\right )} d^{3} p^{2} +{\left (15 \, b^{3} c^{3} + a b^{2} c\right )} d^{3} p\right )} x^{3} + 2 \, a^{3} +{\left (b^{3} c^{6} + a b^{2} c^{4}\right )} p^{2} +{\left (30 \, b^{3} c^{4} d^{2} + 3 \,{\left (5 \, b^{3} c^{4} + 2 \, a b^{2} c^{2}\right )} d^{2} p^{2} +{\left (45 \, b^{3} c^{4} + 6 \, a b^{2} c^{2} - 2 \, a^{2} b\right )} d^{2} p\right )} x^{2} +{\left (3 \, b^{3} c^{6} + a b^{2} c^{4} - 2 \, a^{2} b c^{2}\right )} p + 2 \,{\left (6 \, b^{3} c^{5} d +{\left (3 \, b^{3} c^{5} + 2 \, a b^{2} c^{3}\right )} d p^{2} +{\left (9 \, b^{3} c^{5} + 2 \, a b^{2} c^{3} - 2 \, a^{2} 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^{3} d p^{3} + 6 \, b^{3} d p^{2} + 11 \, b^{3} d p + 6 \, b^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^5*(a+b*(d*x+c)^2)^p,x, algorithm="fricas")

[Out]

1/2*(2*b^3*c^6 + (b^3*d^6*p^2 + 3*b^3*d^6*p + 2*b^3*d^6)*x^6 + 6*(b^3*c*d^5*p^2 + 3*b^3*c*d^5*p + 2*b^3*c*d^5)
*x^5 + (30*b^3*c^2*d^4 + (15*b^3*c^2 + a*b^2)*d^4*p^2 + (45*b^3*c^2 + a*b^2)*d^4*p)*x^4 + 4*(10*b^3*c^3*d^3 +
(5*b^3*c^3 + a*b^2*c)*d^3*p^2 + (15*b^3*c^3 + a*b^2*c)*d^3*p)*x^3 + 2*a^3 + (b^3*c^6 + a*b^2*c^4)*p^2 + (30*b^
3*c^4*d^2 + 3*(5*b^3*c^4 + 2*a*b^2*c^2)*d^2*p^2 + (45*b^3*c^4 + 6*a*b^2*c^2 - 2*a^2*b)*d^2*p)*x^2 + (3*b^3*c^6
+ a*b^2*c^4 - 2*a^2*b*c^2)*p + 2*(6*b^3*c^5*d + (3*b^3*c^5 + 2*a*b^2*c^3)*d*p^2 + (9*b^3*c^5 + 2*a*b^2*c^3 -
2*a^2*b*c)*d*p)*x)*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p/(b^3*d*p^3 + 6*b^3*d*p^2 + 11*b^3*d*p + 6*b^3*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)**5*(a+b*(d*x+c)**2)**p,x)

[Out]

Timed out

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Giac [B]  time = 1.14976, size = 1742, normalized size = 18.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^5*(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^3*d^6*p^2*x^6 + 6*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c*d^5
*p^2*x^5 + 3*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*d^6*p*x^6 + 15*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^
3*c^2*d^4*p^2*x^4 + 18*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c*d^5*p*x^5 + 2*(b*d^2*x^2 + 2*b*c*d*x + b*c^
2 + a)^p*b^3*d^6*x^6 + 20*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^3*d^3*p^2*x^3 + 45*(b*d^2*x^2 + 2*b*c*d*
x + b*c^2 + a)^p*b^3*c^2*d^4*p*x^4 + 12*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c*d^5*x^5 + 15*(b*d^2*x^2 +
2*b*c*d*x + b*c^2 + a)^p*b^3*c^4*d^2*p^2*x^2 + 60*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^3*d^3*p*x^3 + 30
*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^2*d^4*x^4 + (b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*d^4*p^2*x
^4 + 6*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^5*d*p^2*x + 45*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^
4*d^2*p*x^2 + 40*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^3*d^3*x^3 + 4*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)
^p*a*b^2*c*d^3*p^2*x^3 + (b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*d^4*p*x^4 + (b*d^2*x^2 + 2*b*c*d*x + b*c^
2 + a)^p*b^3*c^6*p^2 + 18*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^5*d*p*x + 30*(b*d^2*x^2 + 2*b*c*d*x + b*
c^2 + a)^p*b^3*c^4*d^2*x^2 + 6*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^2*d^2*p^2*x^2 + 4*(b*d^2*x^2 + 2*
b*c*d*x + b*c^2 + a)^p*a*b^2*c*d^3*p*x^3 + 3*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^6*p + 12*(b*d^2*x^2 +
2*b*c*d*x + b*c^2 + a)^p*b^3*c^5*d*x + 4*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^3*d*p^2*x + 6*(b*d^2*x
^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^2*d^2*p*x^2 + 2*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^6 + (b*d^2*x
^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^4*p^2 + 4*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^3*d*p*x + (b*d^2
*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^4*p - 2*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a^2*b*d^2*p*x^2 - 4*(b*d
^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a^2*b*c*d*p*x - 2*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a^2*b*c^2*p + 2*(b*d
^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a^3)/(b^3*d*p^3 + 6*b^3*d*p^2 + 11*b^3*d*p + 6*b^3*d)