3.2.73 \(\int x^{1+p} (b+2 c x^2) (b x+c x^3)^p \, dx\) [173]

Optimal. Leaf size=27 \[ \frac {x^{1+p} \left (b x+c x^3\right )^{1+p}}{2 (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {1604} \begin {gather*} \frac {x^{p+1} \left (b x+c x^3\right )^{p+1}}{2 (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1604

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*(Rr^(n + 1)/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x,
r])), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 27, normalized size = 1.00 \begin {gather*} \frac {x^{1+p} \left (x \left (b+c x^2\right )\right )^{1+p}}{2 (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + p)*(x*(b + c*x^2))^(1 + p))/(2*(1 + p))

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Maple [A]
time = 0.24, size = 31, normalized size = 1.15

method	result	size
gosper	\(\frac {x^{2+p} \left (c \,x^{2}+b \right ) \left (c \,x^{3}+b x \right )^{p}}{2+2 p}\)	\(31\)
risch	\(\frac {\left (c \,x^{2}+b \right ) x \,x^{1+p} {\mathrm e}^{\frac {p \left (-i \pi \mathrm {csgn}\left (i x \left (c \,x^{2}+b \right )\right )^{3}+i \pi \mathrm {csgn}\left (i x \left (c \,x^{2}+b \right )\right )^{2} \mathrm {csgn}\left (i x \right )+i \pi \mathrm {csgn}\left (i x \left (c \,x^{2}+b \right )\right )^{2} \mathrm {csgn}\left (i \left (c \,x^{2}+b \right )\right )-i \pi \,\mathrm {csgn}\left (i x \left (c \,x^{2}+b \right )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (c \,x^{2}+b \right )\right )+2 \ln \left (c \,x^{2}+b \right )+2 \ln \left (x \right )\right )}{2}}}{2+2 p}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 35, normalized size = 1.30 \begin {gather*} \frac {{\left (c x^{4} + b x^{2}\right )} e^{\left (p \log \left (c x^{2} + b\right ) + 2 \, p \log \left (x\right )\right )}}{2 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*x^4 + b*x^2)*e^(p*log(c*x^2 + b) + 2*p*log(x))/(p + 1)

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Fricas [A]
time = 0.38, size = 32, normalized size = 1.19 \begin {gather*} \frac {{\left (c x^{3} + b x\right )} {\left (c x^{3} + b x\right )}^{p} x^{p + 1}}{2 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).
time = 2.96, size = 54, normalized size = 2.00 \begin {gather*} \frac {c x^{3} e^{\left (p \log \left (c x^{2} + b\right ) + 2 \, p \log \left (x\right ) + \log \left (x\right )\right )} + b x e^{\left (p \log \left (c x^{2} + b\right ) + 2 \, p \log \left (x\right ) + \log \left (x\right )\right )}}{2 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.21, size = 45, normalized size = 1.67 \begin {gather*} {\left (c\,x^3+b\,x\right )}^p\,\left (\frac {b\,x\,x^{p+1}}{2\,p+2}+\frac {c\,x^{p+1}\,x^3}{2\,p+2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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