3.174 \(\int (b x^{1+p} (b x+c x^3)^p+2 c x^{3+p} (b x+c x^3)^p) \, dx\)

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

[Out]

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

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Rubi [C]  time = 0.101833, antiderivative size = 116, normalized size of antiderivative = 4.3, number of steps used = 7, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2032, 365, 364} \[ \frac{b x^{p+2} \left (b x+c x^3\right )^p \left (\frac{c x^2}{b}+1\right )^{-p} \, _2F_1\left (-p,p+1;p+2;-\frac{c x^2}{b}\right )}{2 (p+1)}+\frac{c x^{p+4} \left (b x+c x^3\right )^p \left (\frac{c x^2}{b}+1\right )^{-p} \, _2F_1\left (-p,p+2;p+3;-\frac{c x^2}{b}\right )}{p+2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^(2 + p)*(b*x + c*x^3)^p*Hypergeometric2F1[-p, 1 + p, 2 + p, -((c*x^2)/b)])/(2*(1 + p)*(1 + (c*x^2)/b)^p)
+ (c*x^(4 + p)*(b*x + c*x^3)^p*Hypergeometric2F1[-p, 2 + p, 3 + p, -((c*x^2)/b)])/((2 + p)*(1 + (c*x^2)/b)^p)

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \left (b x^{1+p} \left (b x+c x^3\right )^p+2 c x^{3+p} \left (b x+c x^3\right )^p\right ) \, dx &=b \int x^{1+p} \left (b x+c x^3\right )^p \, dx+(2 c) \int x^{3+p} \left (b x+c x^3\right )^p \, dx\\ &=\left (b x^{-p} \left (b+c x^2\right )^{-p} \left (b x+c x^3\right )^p\right ) \int x^{1+2 p} \left (b+c x^2\right )^p \, dx+\left (2 c x^{-p} \left (b+c x^2\right )^{-p} \left (b x+c x^3\right )^p\right ) \int x^{3+2 p} \left (b+c x^2\right )^p \, dx\\ &=\left (b x^{-p} \left (1+\frac{c x^2}{b}\right )^{-p} \left (b x+c x^3\right )^p\right ) \int x^{1+2 p} \left (1+\frac{c x^2}{b}\right )^p \, dx+\left (2 c x^{-p} \left (1+\frac{c x^2}{b}\right )^{-p} \left (b x+c x^3\right )^p\right ) \int x^{3+2 p} \left (1+\frac{c x^2}{b}\right )^p \, dx\\ &=\frac{b x^{2+p} \left (1+\frac{c x^2}{b}\right )^{-p} \left (b x+c x^3\right )^p \, _2F_1\left (-p,1+p;2+p;-\frac{c x^2}{b}\right )}{2 (1+p)}+\frac{c x^{4+p} \left (1+\frac{c x^2}{b}\right )^{-p} \left (b x+c x^3\right )^p \, _2F_1\left (-p,2+p;3+p;-\frac{c x^2}{b}\right )}{2+p}\\ \end{align*}

Mathematica [C]  time = 0.0266186, size = 97, normalized size = 3.59 \[ \frac{x^{p+2} \left (x \left (b+c x^2\right )\right )^p \left (\frac{c x^2}{b}+1\right )^{-p} \left (2 c (p+1) x^2 \, _2F_1\left (-p,p+2;p+3;-\frac{c x^2}{b}\right )+b (p+2) \, _2F_1\left (-p,p+1;p+2;-\frac{c x^2}{b}\right )\right )}{2 (p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(2 + p)*(x*(b + c*x^2))^p*(b*(2 + p)*Hypergeometric2F1[-p, 1 + p, 2 + p, -((c*x^2)/b)] + 2*c*(1 + p)*x^2*Hy
pergeometric2F1[-p, 2 + p, 3 + p, -((c*x^2)/b)]))/(2*(1 + p)*(2 + p)*(1 + (c*x^2)/b)^p)

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Maple [C]  time = 0.114, size = 142, normalized size = 5.3 \begin{align*}{\frac{x \left ( c{x}^{2}+b \right ){x}^{1+p}}{2+2\,p}{{\rm e}^{-{\frac{p \left ( i\pi \, \left ({\it csgn} \left ( ix \left ( c{x}^{2}+b \right ) \right ) \right ) ^{3}-i\pi \, \left ({\it csgn} \left ( ix \left ( c{x}^{2}+b \right ) \right ) \right ) ^{2}{\it csgn} \left ( ix \right ) -i\pi \, \left ({\it csgn} \left ( ix \left ( c{x}^{2}+b \right ) \right ) \right ) ^{2}{\it csgn} \left ( i \left ( c{x}^{2}+b \right ) \right ) +i\pi \,{\it csgn} \left ( ix \left ( c{x}^{2}+b \right ) \right ){\it csgn} \left ( ix \right ){\it csgn} \left ( i \left ( c{x}^{2}+b \right ) \right ) -2\,\ln \left ( x \right ) -2\,\ln \left ( c{x}^{2}+b \right ) \right ) }{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*x^2+b)*x*x^(1+p)/(1+p)*exp(-1/2*p*(I*Pi*csgn(I*x*(c*x^2+b))^3-I*Pi*csgn(I*x*(c*x^2+b))^2*csgn(I*x)-I*Pi
*csgn(I*x*(c*x^2+b))^2*csgn(I*(c*x^2+b))+I*Pi*csgn(I*x*(c*x^2+b))*csgn(I*x)*csgn(I*(c*x^2+b))-2*ln(x)-2*ln(c*x
^2+b)))

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Maxima [A]  time = 1.21373, size = 47, normalized size = 1.74 \begin{align*} \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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(b*x^(1+p)*(c*x^3+b*x)^p+2*c*x^(3+p)*(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 = 1.31456, size = 74, normalized size = 2.74 \begin{align*} \frac{{\left (c x^{2} + b\right )}{\left (c x^{3} + b x\right )}^{p} x^{p + 3}}{2 \,{\left (p + 1\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int 2 \,{\left (c x^{3} + b x\right )}^{p} c x^{p + 3} +{\left (c x^{3} + b x\right )}^{p} b x^{p + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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