\(\int (d+e x) (a+c x^2)^2 \, dx\) [463]

Optimal result

Integrand size = 15, antiderivative size = 45 \[ \int (d+e x) \left (a+c x^2\right )^2 \, dx=a^2 d x+\frac {2}{3} a c d x^3+\frac {1}{5} c^2 d x^5+\frac {e \left (a+c x^2\right )^3}{6 c} \]

[Out]

a^2*d*x+2/3*a*c*d*x^3+1/5*c^2*d*x^5+1/6*e*(c*x^2+a)^3/c

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {655, 200} \[ \int (d+e x) \left (a+c x^2\right )^2 \, dx=a^2 d x+\frac {2}{3} a c d x^3+\frac {e \left (a+c x^2\right )^3}{6 c}+\frac {1}{5} c^2 d x^5 \]

[In]

Int[(d + e*x)*(a + c*x^2)^2,x]

[Out]

a^2*d*x + (2*a*c*d*x^3)/3 + (c^2*d*x^5)/5 + (e*(a + c*x^2)^3)/(6*c)

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {e \left (a+c x^2\right )^3}{6 c}+d \int \left (a+c x^2\right )^2 \, dx \\ & = \frac {e \left (a+c x^2\right )^3}{6 c}+d \int \left (a^2+2 a c x^2+c^2 x^4\right ) \, dx \\ & = a^2 d x+\frac {2}{3} a c d x^3+\frac {1}{5} c^2 d x^5+\frac {e \left (a+c x^2\right )^3}{6 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.33 \[ \int (d+e x) \left (a+c x^2\right )^2 \, dx=a^2 d x+\frac {1}{2} a^2 e x^2+\frac {2}{3} a c d x^3+\frac {1}{2} a c e x^4+\frac {1}{5} c^2 d x^5+\frac {1}{6} c^2 e x^6 \]

[In]

Integrate[(d + e*x)*(a + c*x^2)^2,x]

[Out]

a^2*d*x + (a^2*e*x^2)/2 + (2*a*c*d*x^3)/3 + (a*c*e*x^4)/2 + (c^2*d*x^5)/5 + (c^2*e*x^6)/6

Maple [A] (verified)

Time = 2.41 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.13

method	result	size
gosper	\(\frac {1}{6} c^{2} e \,x^{6}+\frac {1}{5} c^{2} d \,x^{5}+\frac {1}{2} a c e \,x^{4}+\frac {2}{3} a c d \,x^{3}+\frac {1}{2} a^{2} e \,x^{2}+a^{2} d x\)	\(51\)
default	\(\frac {1}{6} c^{2} e \,x^{6}+\frac {1}{5} c^{2} d \,x^{5}+\frac {1}{2} a c e \,x^{4}+\frac {2}{3} a c d \,x^{3}+\frac {1}{2} a^{2} e \,x^{2}+a^{2} d x\)	\(51\)
norman	\(\frac {1}{6} c^{2} e \,x^{6}+\frac {1}{5} c^{2} d \,x^{5}+\frac {1}{2} a c e \,x^{4}+\frac {2}{3} a c d \,x^{3}+\frac {1}{2} a^{2} e \,x^{2}+a^{2} d x\)	\(51\)
risch	\(\frac {1}{6} c^{2} e \,x^{6}+\frac {1}{5} c^{2} d \,x^{5}+\frac {1}{2} a c e \,x^{4}+\frac {2}{3} a c d \,x^{3}+\frac {1}{2} a^{2} e \,x^{2}+a^{2} d x\)	\(51\)
parallelrisch	\(\frac {1}{6} c^{2} e \,x^{6}+\frac {1}{5} c^{2} d \,x^{5}+\frac {1}{2} a c e \,x^{4}+\frac {2}{3} a c d \,x^{3}+\frac {1}{2} a^{2} e \,x^{2}+a^{2} d x\)	\(51\)

[In]

int((e*x+d)*(c*x^2+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/6*c^2*e*x^6+1/5*c^2*d*x^5+1/2*a*c*e*x^4+2/3*a*c*d*x^3+1/2*a^2*e*x^2+a^2*d*x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11 \[ \int (d+e x) \left (a+c x^2\right )^2 \, dx=\frac {1}{6} \, c^{2} e x^{6} + \frac {1}{5} \, c^{2} d x^{5} + \frac {1}{2} \, a c e x^{4} + \frac {2}{3} \, a c d x^{3} + \frac {1}{2} \, a^{2} e x^{2} + a^{2} d x \]

[In]

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

[Out]

1/6*c^2*e*x^6 + 1/5*c^2*d*x^5 + 1/2*a*c*e*x^4 + 2/3*a*c*d*x^3 + 1/2*a^2*e*x^2 + a^2*d*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.29 \[ \int (d+e x) \left (a+c x^2\right )^2 \, dx=a^{2} d x + \frac {a^{2} e x^{2}}{2} + \frac {2 a c d x^{3}}{3} + \frac {a c e x^{4}}{2} + \frac {c^{2} d x^{5}}{5} + \frac {c^{2} e x^{6}}{6} \]

[In]

integrate((e*x+d)*(c*x**2+a)**2,x)

[Out]

a**2*d*x + a**2*e*x**2/2 + 2*a*c*d*x**3/3 + a*c*e*x**4/2 + c**2*d*x**5/5 + c**2*e*x**6/6

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11 \[ \int (d+e x) \left (a+c x^2\right )^2 \, dx=\frac {1}{6} \, c^{2} e x^{6} + \frac {1}{5} \, c^{2} d x^{5} + \frac {1}{2} \, a c e x^{4} + \frac {2}{3} \, a c d x^{3} + \frac {1}{2} \, a^{2} e x^{2} + a^{2} d x \]

[In]

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

[Out]

1/6*c^2*e*x^6 + 1/5*c^2*d*x^5 + 1/2*a*c*e*x^4 + 2/3*a*c*d*x^3 + 1/2*a^2*e*x^2 + a^2*d*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11 \[ \int (d+e x) \left (a+c x^2\right )^2 \, dx=\frac {1}{6} \, c^{2} e x^{6} + \frac {1}{5} \, c^{2} d x^{5} + \frac {1}{2} \, a c e x^{4} + \frac {2}{3} \, a c d x^{3} + \frac {1}{2} \, a^{2} e x^{2} + a^{2} d x \]

[In]

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

[Out]

1/6*c^2*e*x^6 + 1/5*c^2*d*x^5 + 1/2*a*c*e*x^4 + 2/3*a*c*d*x^3 + 1/2*a^2*e*x^2 + a^2*d*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11 \[ \int (d+e x) \left (a+c x^2\right )^2 \, dx=\frac {e\,a^2\,x^2}{2}+d\,a^2\,x+\frac {e\,a\,c\,x^4}{2}+\frac {2\,d\,a\,c\,x^3}{3}+\frac {e\,c^2\,x^6}{6}+\frac {d\,c^2\,x^5}{5} \]

[In]

int((a + c*x^2)^2*(d + e*x),x)

[Out]

(a^2*e*x^2)/2 + (c^2*d*x^5)/5 + (c^2*e*x^6)/6 + a^2*d*x + (2*a*c*d*x^3)/3 + (a*c*e*x^4)/2