\(\int (a+b x) (c+d x)^3 \, dx\) [1262]

Optimal result

Integrand size = 13, antiderivative size = 38 \[ \int (a+b x) (c+d x)^3 \, dx=-\frac {(b c-a d) (c+d x)^4}{4 d^2}+\frac {b (c+d x)^5}{5 d^2} \]

[Out]

-1/4*(-a*d+b*c)*(d*x+c)^4/d^2+1/5*b*(d*x+c)^5/d^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int (a+b x) (c+d x)^3 \, dx=\frac {b (c+d x)^5}{5 d^2}-\frac {(c+d x)^4 (b c-a d)}{4 d^2} \]

[In]

Int[(a + b*x)*(c + d*x)^3,x]

[Out]

-1/4*((b*c - a*d)*(c + d*x)^4)/d^2 + (b*(c + d*x)^5)/(5*d^2)

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])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(-b c+a d) (c+d x)^3}{d}+\frac {b (c+d x)^4}{d}\right ) \, dx \\ & = -\frac {(b c-a d) (c+d x)^4}{4 d^2}+\frac {b (c+d x)^5}{5 d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.76 \[ \int (a+b x) (c+d x)^3 \, dx=a c^3 x+\frac {1}{2} c^2 (b c+3 a d) x^2+c d (b c+a d) x^3+\frac {1}{4} d^2 (3 b c+a d) x^4+\frac {1}{5} b d^3 x^5 \]

[In]

Integrate[(a + b*x)*(c + d*x)^3,x]

[Out]

a*c^3*x + (c^2*(b*c + 3*a*d)*x^2)/2 + c*d*(b*c + a*d)*x^3 + (d^2*(3*b*c + a*d)*x^4)/4 + (b*d^3*x^5)/5

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(69\) vs. \(2(34)=68\).

Time = 0.37 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.84

method	result	size
norman	\(\frac {b \,d^{3} x^{5}}{5}+\left (\frac {1}{4} a \,d^{3}+\frac {3}{4} b c \,d^{2}\right ) x^{4}+\left (a c \,d^{2}+b \,c^{2} d \right ) x^{3}+\left (\frac {3}{2} a \,c^{2} d +\frac {1}{2} b \,c^{3}\right ) x^{2}+a \,c^{3} x\)	\(70\)
gosper	\(\frac {1}{5} b \,d^{3} x^{5}+\frac {1}{4} x^{4} a \,d^{3}+\frac {3}{4} x^{4} b c \,d^{2}+a c \,d^{2} x^{3}+b \,c^{2} d \,x^{3}+\frac {3}{2} x^{2} a \,c^{2} d +\frac {1}{2} b \,c^{3} x^{2}+a \,c^{3} x\)	\(73\)
default	\(\frac {b \,d^{3} x^{5}}{5}+\frac {\left (a \,d^{3}+3 b c \,d^{2}\right ) x^{4}}{4}+\frac {\left (3 a c \,d^{2}+3 b \,c^{2} d \right ) x^{3}}{3}+\frac {\left (3 a \,c^{2} d +b \,c^{3}\right ) x^{2}}{2}+a \,c^{3} x\)	\(73\)
risch	\(\frac {1}{5} b \,d^{3} x^{5}+\frac {1}{4} x^{4} a \,d^{3}+\frac {3}{4} x^{4} b c \,d^{2}+a c \,d^{2} x^{3}+b \,c^{2} d \,x^{3}+\frac {3}{2} x^{2} a \,c^{2} d +\frac {1}{2} b \,c^{3} x^{2}+a \,c^{3} x\)	\(73\)
parallelrisch	\(\frac {1}{5} b \,d^{3} x^{5}+\frac {1}{4} x^{4} a \,d^{3}+\frac {3}{4} x^{4} b c \,d^{2}+a c \,d^{2} x^{3}+b \,c^{2} d \,x^{3}+\frac {3}{2} x^{2} a \,c^{2} d +\frac {1}{2} b \,c^{3} x^{2}+a \,c^{3} x\)	\(73\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (34) = 68\).

Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.82 \[ \int (a+b x) (c+d x)^3 \, dx=\frac {1}{5} \, b d^{3} x^{5} + a c^{3} x + \frac {1}{4} \, {\left (3 \, b c d^{2} + a d^{3}\right )} x^{4} + {\left (b c^{2} d + a c d^{2}\right )} x^{3} + \frac {1}{2} \, {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (32) = 64\).

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.92 \[ \int (a+b x) (c+d x)^3 \, dx=a c^{3} x + \frac {b d^{3} x^{5}}{5} + x^{4} \left (\frac {a d^{3}}{4} + \frac {3 b c d^{2}}{4}\right ) + x^{3} \left (a c d^{2} + b c^{2} d\right ) + x^{2} \cdot \left (\frac {3 a c^{2} d}{2} + \frac {b c^{3}}{2}\right ) \]

[In]

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

[Out]

a*c**3*x + b*d**3*x**5/5 + x**4*(a*d**3/4 + 3*b*c*d**2/4) + x**3*(a*c*d**2 + b*c**2*d) + x**2*(3*a*c**2*d/2 +
b*c**3/2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (34) = 68\).

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.82 \[ \int (a+b x) (c+d x)^3 \, dx=\frac {1}{5} \, b d^{3} x^{5} + a c^{3} x + \frac {1}{4} \, {\left (3 \, b c d^{2} + a d^{3}\right )} x^{4} + {\left (b c^{2} d + a c d^{2}\right )} x^{3} + \frac {1}{2} \, {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (34) = 68\).

Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.89 \[ \int (a+b x) (c+d x)^3 \, dx=\frac {1}{5} \, b d^{3} x^{5} + \frac {3}{4} \, b c d^{2} x^{4} + \frac {1}{4} \, a d^{3} x^{4} + b c^{2} d x^{3} + a c d^{2} x^{3} + \frac {1}{2} \, b c^{3} x^{2} + \frac {3}{2} \, a c^{2} d x^{2} + a c^{3} x \]

[In]

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

[Out]

1/5*b*d^3*x^5 + 3/4*b*c*d^2*x^4 + 1/4*a*d^3*x^4 + b*c^2*d*x^3 + a*c*d^2*x^3 + 1/2*b*c^3*x^2 + 3/2*a*c^2*d*x^2
+ a*c^3*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.71 \[ \int (a+b x) (c+d x)^3 \, dx=x^2\,\left (\frac {b\,c^3}{2}+\frac {3\,a\,d\,c^2}{2}\right )+x^4\,\left (\frac {a\,d^3}{4}+\frac {3\,b\,c\,d^2}{4}\right )+\frac {b\,d^3\,x^5}{5}+a\,c^3\,x+c\,d\,x^3\,\left (a\,d+b\,c\right ) \]

[In]

int((a + b*x)*(c + d*x)^3,x)

[Out]

x^2*((b*c^3)/2 + (3*a*c^2*d)/2) + x^4*((a*d^3)/4 + (3*b*c*d^2)/4) + (b*d^3*x^5)/5 + a*c^3*x + c*d*x^3*(a*d + b
*c)