3.2912 \(\int (c+d x)^3 (a+b (c+d x)^4)^2 \, dx\)

Optimal. Leaf size=23 \[ \frac{\left (a+b (c+d x)^4\right )^3}{12 b d} \]

[Out]

(a + b*(c + d*x)^4)^3/(12*b*d)

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Rubi [A]  time = 0.0931545, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {372, 261} \[ \frac{\left (a+b (c+d x)^4\right )^3}{12 b d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*(a + b*(c + d*x)^4)^2,x]

[Out]

(a + b*(c + d*x)^4)^3/(12*b*d)

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 261

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]

Rubi steps

\begin{align*} \int (c+d x)^3 \left (a+b (c+d x)^4\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int x^3 \left (a+b x^4\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{\left (a+b (c+d x)^4\right )^3}{12 b d}\\ \end{align*}

Mathematica [B]  time = 0.0357989, size = 172, normalized size = 7.48 \[ \frac{1}{12} x \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right ) \left (3 a^2+3 a b \left (6 c^2 d^2 x^2+4 c^3 d x+2 c^4+4 c d^3 x^3+d^4 x^4\right )+b^2 \left (34 c^6 d^2 x^2+60 c^5 d^3 x^3+71 c^4 d^4 x^4+56 c^3 d^5 x^5+28 c^2 d^6 x^6+12 c^7 d x+3 c^8+8 c d^7 x^7+d^8 x^8\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3*(a + b*(c + d*x)^4)^2,x]

[Out]

(x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*(3*a^2 + 3*a*b*(2*c^4 + 4*c^3*d*x + 6*c^2*d^2*x^2 + 4*c*d^3*x^3
+ d^4*x^4) + b^2*(3*c^8 + 12*c^7*d*x + 34*c^6*d^2*x^2 + 60*c^5*d^3*x^3 + 71*c^4*d^4*x^4 + 56*c^3*d^5*x^5 + 28
*c^2*d^6*x^6 + 8*c*d^7*x^7 + d^8*x^8)))/12

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Maple [B]  time = 0.002, size = 622, normalized size = 27. \begin{align*}{\frac{{d}^{11}{b}^{2}{x}^{12}}{12}}+c{d}^{10}{b}^{2}{x}^{11}+{\frac{11\,{c}^{2}{d}^{9}{b}^{2}{x}^{10}}{2}}+{\frac{55\,{c}^{3}{b}^{2}{d}^{8}{x}^{9}}{3}}+{\frac{ \left ( 260\,{c}^{4}{b}^{2}{d}^{7}+{d}^{3} \left ( 2\, \left ( b{c}^{4}+a \right ) b{d}^{4}+68\,{c}^{4}{d}^{4}{b}^{2} \right ) \right ){x}^{8}}{8}}+{\frac{ \left ( 196\,{c}^{5}{d}^{6}{b}^{2}+3\,c{d}^{2} \left ( 2\, \left ( b{c}^{4}+a \right ) b{d}^{4}+68\,{c}^{4}{d}^{4}{b}^{2} \right ) +{d}^{3} \left ( 8\, \left ( b{c}^{4}+a \right ) bc{d}^{3}+48\,{c}^{5}{d}^{3}{b}^{2} \right ) \right ){x}^{7}}{7}}+{\frac{ \left ( 56\,{c}^{6}{d}^{5}{b}^{2}+3\,{c}^{2}d \left ( 2\, \left ( b{c}^{4}+a \right ) b{d}^{4}+68\,{c}^{4}{d}^{4}{b}^{2} \right ) +3\,c{d}^{2} \left ( 8\, \left ( b{c}^{4}+a \right ) bc{d}^{3}+48\,{c}^{5}{d}^{3}{b}^{2} \right ) +{d}^{3} \left ( 12\, \left ( b{c}^{4}+a \right ){c}^{2}{d}^{2}b+16\,{c}^{6}{d}^{2}{b}^{2} \right ) \right ){x}^{6}}{6}}+{\frac{ \left ({c}^{3} \left ( 2\, \left ( b{c}^{4}+a \right ) b{d}^{4}+68\,{c}^{4}{d}^{4}{b}^{2} \right ) +3\,{c}^{2}d \left ( 8\, \left ( b{c}^{4}+a \right ) bc{d}^{3}+48\,{c}^{5}{d}^{3}{b}^{2} \right ) +3\,c{d}^{2} \left ( 12\, \left ( b{c}^{4}+a \right ){c}^{2}{d}^{2}b+16\,{c}^{6}{d}^{2}{b}^{2} \right ) +8\,{d}^{4} \left ( b{c}^{4}+a \right ){c}^{3}b \right ){x}^{5}}{5}}+{\frac{ \left ({c}^{3} \left ( 8\, \left ( b{c}^{4}+a \right ) bc{d}^{3}+48\,{c}^{5}{d}^{3}{b}^{2} \right ) +3\,{c}^{2}d \left ( 12\, \left ( b{c}^{4}+a \right ){c}^{2}{d}^{2}b+16\,{c}^{6}{d}^{2}{b}^{2} \right ) +24\,{c}^{4}{d}^{3} \left ( b{c}^{4}+a \right ) b+{d}^{3} \left ( b{c}^{4}+a \right ) ^{2} \right ){x}^{4}}{4}}+{\frac{ \left ({c}^{3} \left ( 12\, \left ( b{c}^{4}+a \right ){c}^{2}{d}^{2}b+16\,{c}^{6}{d}^{2}{b}^{2} \right ) +24\,{c}^{5}{d}^{2} \left ( b{c}^{4}+a \right ) b+3\,c{d}^{2} \left ( b{c}^{4}+a \right ) ^{2} \right ){x}^{3}}{3}}+{\frac{ \left ( 8\,{c}^{6} \left ( b{c}^{4}+a \right ) db+3\,{c}^{2}d \left ( b{c}^{4}+a \right ) ^{2} \right ){x}^{2}}{2}}+{c}^{3} \left ( b{c}^{4}+a \right ) ^{2}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*(a+b*(d*x+c)^4)^2,x)

[Out]

1/12*d^11*b^2*x^12+c*d^10*b^2*x^11+11/2*c^2*d^9*b^2*x^10+55/3*c^3*b^2*d^8*x^9+1/8*(260*c^4*b^2*d^7+d^3*(2*(b*c
^4+a)*b*d^4+68*c^4*d^4*b^2))*x^8+1/7*(196*c^5*d^6*b^2+3*c*d^2*(2*(b*c^4+a)*b*d^4+68*c^4*d^4*b^2)+d^3*(8*(b*c^4
+a)*b*c*d^3+48*c^5*d^3*b^2))*x^7+1/6*(56*c^6*d^5*b^2+3*c^2*d*(2*(b*c^4+a)*b*d^4+68*c^4*d^4*b^2)+3*c*d^2*(8*(b*
c^4+a)*b*c*d^3+48*c^5*d^3*b^2)+d^3*(12*(b*c^4+a)*c^2*d^2*b+16*c^6*d^2*b^2))*x^6+1/5*(c^3*(2*(b*c^4+a)*b*d^4+68
*c^4*d^4*b^2)+3*c^2*d*(8*(b*c^4+a)*b*c*d^3+48*c^5*d^3*b^2)+3*c*d^2*(12*(b*c^4+a)*c^2*d^2*b+16*c^6*d^2*b^2)+8*d
^4*(b*c^4+a)*c^3*b)*x^5+1/4*(c^3*(8*(b*c^4+a)*b*c*d^3+48*c^5*d^3*b^2)+3*c^2*d*(12*(b*c^4+a)*c^2*d^2*b+16*c^6*d
^2*b^2)+24*c^4*d^3*(b*c^4+a)*b+d^3*(b*c^4+a)^2)*x^4+1/3*(c^3*(12*(b*c^4+a)*c^2*d^2*b+16*c^6*d^2*b^2)+24*c^5*d^
2*(b*c^4+a)*b+3*c*d^2*(b*c^4+a)^2)*x^3+1/2*(8*c^6*(b*c^4+a)*d*b+3*c^2*d*(b*c^4+a)^2)*x^2+c^3*(b*c^4+a)^2*x

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Maxima [A]  time = 0.979569, size = 28, normalized size = 1.22 \begin{align*} \frac{{\left ({\left (d x + c\right )}^{4} b + a\right )}^{3}}{12 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*(a+b*(d*x+c)^4)^2,x, algorithm="maxima")

[Out]

1/12*((d*x + c)^4*b + a)^3/(b*d)

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Fricas [B]  time = 1.11316, size = 639, normalized size = 27.78 \begin{align*} \frac{1}{12} x^{12} d^{11} b^{2} + x^{11} d^{10} c b^{2} + \frac{11}{2} x^{10} d^{9} c^{2} b^{2} + \frac{55}{3} x^{9} d^{8} c^{3} b^{2} + \frac{165}{4} x^{8} d^{7} c^{4} b^{2} + 66 x^{7} d^{6} c^{5} b^{2} + 77 x^{6} d^{5} c^{6} b^{2} + 66 x^{5} d^{4} c^{7} b^{2} + \frac{165}{4} x^{4} d^{3} c^{8} b^{2} + \frac{1}{4} x^{8} d^{7} b a + \frac{55}{3} x^{3} d^{2} c^{9} b^{2} + 2 x^{7} d^{6} c b a + \frac{11}{2} x^{2} d c^{10} b^{2} + 7 x^{6} d^{5} c^{2} b a + x c^{11} b^{2} + 14 x^{5} d^{4} c^{3} b a + \frac{35}{2} x^{4} d^{3} c^{4} b a + 14 x^{3} d^{2} c^{5} b a + 7 x^{2} d c^{6} b a + 2 x c^{7} b a + \frac{1}{4} x^{4} d^{3} a^{2} + x^{3} d^{2} c a^{2} + \frac{3}{2} x^{2} d c^{2} a^{2} + x c^{3} a^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*(a+b*(d*x+c)^4)^2,x, algorithm="fricas")

[Out]

1/12*x^12*d^11*b^2 + x^11*d^10*c*b^2 + 11/2*x^10*d^9*c^2*b^2 + 55/3*x^9*d^8*c^3*b^2 + 165/4*x^8*d^7*c^4*b^2 +
66*x^7*d^6*c^5*b^2 + 77*x^6*d^5*c^6*b^2 + 66*x^5*d^4*c^7*b^2 + 165/4*x^4*d^3*c^8*b^2 + 1/4*x^8*d^7*b*a + 55/3*
x^3*d^2*c^9*b^2 + 2*x^7*d^6*c*b*a + 11/2*x^2*d*c^10*b^2 + 7*x^6*d^5*c^2*b*a + x*c^11*b^2 + 14*x^5*d^4*c^3*b*a
+ 35/2*x^4*d^3*c^4*b*a + 14*x^3*d^2*c^5*b*a + 7*x^2*d*c^6*b*a + 2*x*c^7*b*a + 1/4*x^4*d^3*a^2 + x^3*d^2*c*a^2
+ 3/2*x^2*d*c^2*a^2 + x*c^3*a^2

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Sympy [B]  time = 0.125388, size = 299, normalized size = 13. \begin{align*} \frac{55 b^{2} c^{3} d^{8} x^{9}}{3} + \frac{11 b^{2} c^{2} d^{9} x^{10}}{2} + b^{2} c d^{10} x^{11} + \frac{b^{2} d^{11} x^{12}}{12} + x^{8} \left (\frac{a b d^{7}}{4} + \frac{165 b^{2} c^{4} d^{7}}{4}\right ) + x^{7} \left (2 a b c d^{6} + 66 b^{2} c^{5} d^{6}\right ) + x^{6} \left (7 a b c^{2} d^{5} + 77 b^{2} c^{6} d^{5}\right ) + x^{5} \left (14 a b c^{3} d^{4} + 66 b^{2} c^{7} d^{4}\right ) + x^{4} \left (\frac{a^{2} d^{3}}{4} + \frac{35 a b c^{4} d^{3}}{2} + \frac{165 b^{2} c^{8} d^{3}}{4}\right ) + x^{3} \left (a^{2} c d^{2} + 14 a b c^{5} d^{2} + \frac{55 b^{2} c^{9} d^{2}}{3}\right ) + x^{2} \left (\frac{3 a^{2} c^{2} d}{2} + 7 a b c^{6} d + \frac{11 b^{2} c^{10} d}{2}\right ) + x \left (a^{2} c^{3} + 2 a b c^{7} + b^{2} c^{11}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*(a+b*(d*x+c)**4)**2,x)

[Out]

55*b**2*c**3*d**8*x**9/3 + 11*b**2*c**2*d**9*x**10/2 + b**2*c*d**10*x**11 + b**2*d**11*x**12/12 + x**8*(a*b*d*
*7/4 + 165*b**2*c**4*d**7/4) + x**7*(2*a*b*c*d**6 + 66*b**2*c**5*d**6) + x**6*(7*a*b*c**2*d**5 + 77*b**2*c**6*
d**5) + x**5*(14*a*b*c**3*d**4 + 66*b**2*c**7*d**4) + x**4*(a**2*d**3/4 + 35*a*b*c**4*d**3/2 + 165*b**2*c**8*d
**3/4) + x**3*(a**2*c*d**2 + 14*a*b*c**5*d**2 + 55*b**2*c**9*d**2/3) + x**2*(3*a**2*c**2*d/2 + 7*a*b*c**6*d +
11*b**2*c**10*d/2) + x*(a**2*c**3 + 2*a*b*c**7 + b**2*c**11)

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Giac [A]  time = 1.18806, size = 28, normalized size = 1.22 \begin{align*} \frac{{\left ({\left (d x + c\right )}^{4} b + a\right )}^{3}}{12 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*(a+b*(d*x+c)^4)^2,x, algorithm="giac")

[Out]

1/12*((d*x + c)^4*b + a)^3/(b*d)