3.3.17 \(\int (b x+c x^2) (1+(\frac {b x^2}{2}+\frac {c x^3}{3})^n) \, dx\) [217]

Optimal. Leaf size=44 \[ \frac {b x^2}{2}+\frac {c x^3}{3}+\frac {\left (\frac {b x^2}{2}+\frac {c x^3}{3}\right )^{1+n}}{1+n} \]

[Out]

1/2*b*x^2+1/3*c*x^3+(1/2*b*x^2+1/3*c*x^3)^(1+n)/(1+n)

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Rubi [A]
time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {1605} \begin {gather*} \frac {\left (\frac {b x^2}{2}+\frac {c x^3}{3}\right )^{n+1}}{n+1}+\frac {b x^2}{2}+\frac {c x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)*(1 + ((b*x^2)/2 + (c*x^3)/3)^n),x]

[Out]

(b*x^2)/2 + (c*x^3)/3 + ((b*x^2)/2 + (c*x^3)/3)^(1 + n)/(1 + n)

Rule 1605

Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[Pq, x], r = Expon[Qr, x]}, Dist[Coef
f[Qr, x, r]/(q*Coeff[Pq, x, q]), Subst[Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x,
r]*D[Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] && PolyQ[Qr, x]

Rubi steps

\begin {align*} \int \left (b x+c x^2\right ) \left (1+\left (\frac {b x^2}{2}+\frac {c x^3}{3}\right )^n\right ) \, dx &=\text {Subst}\left (\int \left (1+x^n\right ) \, dx,x,\frac {b x^2}{2}+\frac {c x^3}{3}\right )\\ &=\frac {b x^2}{2}+\frac {c x^3}{3}+\frac {\left (\frac {b x^2}{2}+\frac {c x^3}{3}\right )^{1+n}}{1+n}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 42, normalized size = 0.95 \begin {gather*} \frac {x^2 (3 b+2 c x) \left (1+n+\left (\frac {b x^2}{2}+\frac {c x^3}{3}\right )^n\right )}{6 (1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)*(1 + ((b*x^2)/2 + (c*x^3)/3)^n),x]

[Out]

(x^2*(3*b + 2*c*x)*(1 + n + ((b*x^2)/2 + (c*x^3)/3)^n))/(6*(1 + n))

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Maple [A]
time = 0.20, size = 37, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {b \,x^{2}}{2}+\frac {c \,x^{3}}{3}+\frac {\left (\frac {1}{2} b \,x^{2}+\frac {1}{3} c \,x^{3}\right )^{1+n}}{1+n}\)	\(37\)
default	\(\frac {b \,x^{2}}{2}+\frac {c \,x^{3}}{3}+\frac {\left (\frac {1}{2} b \,x^{2}+\frac {1}{3} c \,x^{3}\right )^{1+n}}{1+n}\)	\(37\)
risch	\(\frac {b \,x^{2}}{2}+\frac {c \,x^{3}}{3}+\frac {x^{2} \left (2 c x +3 b \right ) \left (\frac {1}{3}\right )^{n} \left (\frac {1}{2}\right )^{n} \left (x^{2} \left (2 c x +3 b \right )\right )^{n}}{6 n +6}\)	\(52\)
norman	\(\frac {b \,x^{2}}{2}+\frac {c \,x^{3}}{3}+\frac {b \,x^{2} {\mathrm e}^{n \ln \left (\frac {1}{2} b \,x^{2}+\frac {1}{3} c \,x^{3}\right )}}{2+2 n}+\frac {c \,x^{3} {\mathrm e}^{n \ln \left (\frac {1}{2} b \,x^{2}+\frac {1}{3} c \,x^{3}\right )}}{3+3 n}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)*(1+(1/2*b*x^2+1/3*c*x^3)^n),x,method=_RETURNVERBOSE)

[Out]

1/2*b*x^2+1/3*c*x^3+(1/2*b*x^2+1/3*c*x^3)^(1+n)/(1+n)

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Maxima [A]
time = 0.53, size = 71, normalized size = 1.61 \begin {gather*} \frac {1}{3} \, c x^{3} + \frac {1}{2} \, b x^{2} + \frac {{\left (2 \, c x^{3} + 3 \, b x^{2}\right )} e^{\left (n \log \left (2 \, c x + 3 \, b\right ) + 2 \, n \log \left (x\right )\right )}}{3^{n + 1} 2^{n + 1} n + 3^{n + 1} 2^{n + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)*(1+(1/2*b*x^2+1/3*c*x^3)^n),x, algorithm="maxima")

[Out]

1/3*c*x^3 + 1/2*b*x^2 + (2*c*x^3 + 3*b*x^2)*e^(n*log(2*c*x + 3*b) + 2*n*log(x))/(3^(n + 1)*2^(n + 1)*n + 3^(n
+ 1)*2^(n + 1))

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Fricas [A]
time = 0.39, size = 57, normalized size = 1.30 \begin {gather*} \frac {2 \, {\left (c n + c\right )} x^{3} + 3 \, {\left (b n + b\right )} x^{2} + {\left (2 \, c x^{3} + 3 \, b x^{2}\right )} {\left (\frac {1}{3} \, c x^{3} + \frac {1}{2} \, b x^{2}\right )}^{n}}{6 \, {\left (n + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)*(1+(1/2*b*x^2+1/3*c*x^3)^n),x, algorithm="fricas")

[Out]

1/6*(2*(c*n + c)*x^3 + 3*(b*n + b)*x^2 + (2*c*x^3 + 3*b*x^2)*(1/3*c*x^3 + 1/2*b*x^2)^n)/(n + 1)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (32) = 64\).
time = 96.62, size = 189, normalized size = 4.30 \begin {gather*} \begin {cases} \frac {3 \cdot 6^{n} b n x^{2}}{6 \cdot 6^{n} n + 6 \cdot 6^{n}} + \frac {3 \cdot 6^{n} b x^{2}}{6 \cdot 6^{n} n + 6 \cdot 6^{n}} + \frac {2 \cdot 6^{n} c n x^{3}}{6 \cdot 6^{n} n + 6 \cdot 6^{n}} + \frac {2 \cdot 6^{n} c x^{3}}{6 \cdot 6^{n} n + 6 \cdot 6^{n}} + \frac {3 b x^{2} \left (3 b x^{2} + 2 c x^{3}\right )^{n}}{6 \cdot 6^{n} n + 6 \cdot 6^{n}} + \frac {2 c x^{3} \left (3 b x^{2} + 2 c x^{3}\right )^{n}}{6 \cdot 6^{n} n + 6 \cdot 6^{n}} & \text {for}\: n \neq -1 \\\frac {b x^{2}}{2} + \frac {c x^{3}}{3} + 2 \log {\left (x \right )} + \log {\left (\frac {3 b}{2 c} + x \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)*(1+(1/2*b*x**2+1/3*c*x**3)**n),x)

[Out]

Piecewise((3*6**n*b*n*x**2/(6*6**n*n + 6*6**n) + 3*6**n*b*x**2/(6*6**n*n + 6*6**n) + 2*6**n*c*n*x**3/(6*6**n*n
+ 6*6**n) + 2*6**n*c*x**3/(6*6**n*n + 6*6**n) + 3*b*x**2*(3*b*x**2 + 2*c*x**3)**n/(6*6**n*n + 6*6**n) + 2*c*x
**3*(3*b*x**2 + 2*c*x**3)**n/(6*6**n*n + 6*6**n), Ne(n, -1)), (b*x**2/2 + c*x**3/3 + 2*log(x) + log(3*b/(2*c)
+ x), True))

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Giac [A]
time = 3.70, size = 36, normalized size = 0.82 \begin {gather*} \frac {1}{3} \, c x^{3} + \frac {1}{2} \, b x^{2} + \frac {{\left (\frac {1}{3} \, c x^{3} + \frac {1}{2} \, b x^{2}\right )}^{n + 1}}{n + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)*(1+(1/2*b*x^2+1/3*c*x^3)^n),x, algorithm="giac")

[Out]

1/3*c*x^3 + 1/2*b*x^2 + (1/3*c*x^3 + 1/2*b*x^2)^(n + 1)/(n + 1)

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Mupad [B]
time = 2.21, size = 37, normalized size = 0.84 \begin {gather*} \frac {x^2\,\left (3\,b+2\,c\,x\right )\,\left (n+{\left (\frac {c\,x^3}{3}+\frac {b\,x^2}{2}\right )}^n+1\right )}{6\,\left (n+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)*(((b*x^2)/2 + (c*x^3)/3)^n + 1),x)

[Out]

(x^2*(3*b + 2*c*x)*(n + ((b*x^2)/2 + (c*x^3)/3)^n + 1))/(6*(n + 1))

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