\(\int (c+d x)^3 (a+b (c+d x)^4) \, dx\) [2911]

Optimal result

Integrand size = 19, antiderivative size = 23 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right ) \, dx=\frac {\left (a+b (c+d x)^4\right )^2}{8 b d} \]

[Out]

1/8*(a+b*(d*x+c)^4)^2/b/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {379, 14} \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right ) \, dx=\frac {a (c+d x)^4}{4 d}+\frac {b (c+d x)^8}{8 d} \]

[In]

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

[Out]

(a*(c + d*x)^4)/(4*d) + (b*(c + d*x)^8)/(8*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 ) \, dx,x,c+d x\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (a x^3+b x^7\right ) \, dx,x,c+d x\right )}{d} \\ & = \frac {a (c+d x)^4}{4 d}+\frac {b (c+d x)^8}{8 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(80\) vs. \(2(23)=46\).

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.48 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right ) \, dx=\frac {1}{8} x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) \left (2 a+b \left (2 c^4+4 c^3 d x+6 c^2 d^2 x^2+4 c d^3 x^3+d^4 x^4\right )\right ) \]

[In]

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

[Out]

(x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*(2*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)))/8

Maple [A] (verified)

Time = 3.89 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {\left (a +b \left (d x +c \right )^{4}\right )^{2}}{8 b d}\)	\(22\)
norman	\(\frac {d^{7} b \,x^{8}}{8}+c \,d^{6} b \,x^{7}+\frac {7 c^{2} d^{5} b \,x^{6}}{2}+7 c^{3} b \,d^{4} x^{5}+\left (\frac {35}{4} c^{4} b \,d^{3}+\frac {1}{4} a \,d^{3}\right ) x^{4}+\left (7 c^{5} d^{2} b +a c \,d^{2}\right ) x^{3}+\left (\frac {7}{2} c^{6} b d +\frac {3}{2} a \,c^{2} d \right ) x^{2}+\left (b \,c^{7}+c^{3} a \right ) x\)	\(116\)
gosper	\(\frac {x \left (d^{7} b \,x^{7}+8 c \,d^{6} b \,x^{6}+28 c^{2} d^{5} b \,x^{5}+56 c^{3} b \,d^{4} x^{4}+70 x^{3} c^{4} b \,d^{3}+56 b \,c^{5} d^{2} x^{2}+28 x \,c^{6} b d +8 b \,c^{7}+2 x^{3} a \,d^{3}+8 a c \,d^{2} x^{2}+12 x a \,c^{2} d +8 c^{3} a \right )}{8}\)	\(118\)
parallelrisch	\(b \,c^{7} x +a \,c^{3} x +\frac {7}{2} b d \,c^{6} x^{2}+\frac {3}{2} d a \,c^{2} x^{2}+7 b \,d^{2} c^{5} x^{3}+d^{2} x^{3} a c +\frac {35}{4} b \,d^{3} c^{4} x^{4}+\frac {1}{4} d^{3} a \,x^{4}+7 c^{3} b \,d^{4} x^{5}+\frac {7}{2} c^{2} d^{5} b \,x^{6}+c \,d^{6} b \,x^{7}+\frac {1}{8} d^{7} b \,x^{8}\)	\(118\)
risch	\(\frac {d^{7} b \,x^{8}}{8}+c \,d^{6} b \,x^{7}+\frac {7 c^{2} d^{5} b \,x^{6}}{2}+7 c^{3} b \,d^{4} x^{5}+\frac {35 b \,d^{3} c^{4} x^{4}}{4}+7 b \,d^{2} c^{5} x^{3}+\frac {7 b d \,c^{6} x^{2}}{2}+b \,c^{7} x +\frac {d^{3} a \,x^{4}}{4}+\frac {b \,c^{8}}{8 d}+d^{2} x^{3} a c +\frac {3 d a \,c^{2} x^{2}}{2}+a \,c^{3} x +\frac {a \,c^{4}}{4 d}+\frac {a^{2}}{8 b d}\)	\(147\)

[In]

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

[Out]

1/8*(a+b*(d*x+c)^4)^2/b/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (21) = 42\).

Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 4.70 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right ) \, dx=\frac {1}{8} \, b d^{7} x^{8} + b c d^{6} x^{7} + \frac {7}{2} \, b c^{2} d^{5} x^{6} + 7 \, b c^{3} d^{4} x^{5} + \frac {1}{4} \, {\left (35 \, b c^{4} + a\right )} d^{3} x^{4} + {\left (7 \, b c^{5} + a c\right )} d^{2} x^{3} + \frac {1}{2} \, {\left (7 \, b c^{6} + 3 \, a c^{2}\right )} d x^{2} + {\left (b c^{7} + a c^{3}\right )} x \]

[In]

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

[Out]

1/8*b*d^7*x^8 + b*c*d^6*x^7 + 7/2*b*c^2*d^5*x^6 + 7*b*c^3*d^4*x^5 + 1/4*(35*b*c^4 + a)*d^3*x^4 + (7*b*c^5 + a*
c)*d^2*x^3 + 1/2*(7*b*c^6 + 3*a*c^2)*d*x^2 + (b*c^7 + a*c^3)*x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (15) = 30\).

Time = 0.03 (sec) , antiderivative size = 126, normalized size of antiderivative = 5.48 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right ) \, dx=7 b c^{3} d^{4} x^{5} + \frac {7 b c^{2} d^{5} x^{6}}{2} + b c d^{6} x^{7} + \frac {b d^{7} x^{8}}{8} + x^{4} \left (\frac {a d^{3}}{4} + \frac {35 b c^{4} d^{3}}{4}\right ) + x^{3} \left (a c d^{2} + 7 b c^{5} d^{2}\right ) + x^{2} \cdot \left (\frac {3 a c^{2} d}{2} + \frac {7 b c^{6} d}{2}\right ) + x \left (a c^{3} + b c^{7}\right ) \]

[In]

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

[Out]

7*b*c**3*d**4*x**5 + 7*b*c**2*d**5*x**6/2 + b*c*d**6*x**7 + b*d**7*x**8/8 + x**4*(a*d**3/4 + 35*b*c**4*d**3/4)
+ x**3*(a*c*d**2 + 7*b*c**5*d**2) + x**2*(3*a*c**2*d/2 + 7*b*c**6*d/2) + x*(a*c**3 + b*c**7)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right ) \, dx=\frac {{\left ({\left (d x + c\right )}^{4} b + a\right )}^{2}}{8 \, b d} \]

[In]

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

[Out]

1/8*((d*x + c)^4*b + a)^2/(b*d)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right ) \, dx=\frac {{\left (d x + c\right )}^{8} b + 2 \, {\left (d x + c\right )}^{4} a}{8 \, d} \]

[In]

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

[Out]

1/8*((d*x + c)^8*b + 2*(d*x + c)^4*a)/d

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 4.65 \[ \int (c+d x)^3 \left (a+b (c+d x)^4\right ) \, dx=x\,\left (b\,c^7+a\,c^3\right )+\frac {d^3\,x^4\,\left (35\,b\,c^4+a\right )}{4}+\frac {b\,d^7\,x^8}{8}+\frac {c^2\,d\,x^2\,\left (7\,b\,c^4+3\,a\right )}{2}+7\,b\,c^3\,d^4\,x^5+\frac {7\,b\,c^2\,d^5\,x^6}{2}+c\,d^2\,x^3\,\left (7\,b\,c^4+a\right )+b\,c\,d^6\,x^7 \]

[In]

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

[Out]

x*(a*c^3 + b*c^7) + (d^3*x^4*(a + 35*b*c^4))/4 + (b*d^7*x^8)/8 + (c^2*d*x^2*(3*a + 7*b*c^4))/2 + 7*b*c^3*d^4*x
^5 + (7*b*c^2*d^5*x^6)/2 + c*d^2*x^3*(a + 7*b*c^4) + b*c*d^6*x^7