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]

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

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Rubi [C]  time = 0.10, antiderivative size = 116, normalized size of antiderivative = 4.30, 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 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])

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 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]

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*}

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Mathematica [C]  time = 0.03, 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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fricas [A]  time = 0.75, size = 33, normalized size = 1.22 \[ \frac {{\left (c x^{2} + b\right )} {\left (c x^{3} + b x\right )}^{p} x^{p + 3}}{2 \, {\left (p + 1\right )} x} \]

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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giac [B]  time = 0.49, size = 54, normalized size = 2.00 \[ \frac {c x^{3} e^{\left (p \log \left (c x^{2} + b\right ) + 2 \, p \log \relax (x) + \log \relax (x)\right )} + b x e^{\left (p \log \left (c x^{2} + b\right ) + 2 \, p \log \relax (x) + \log \relax (x)\right )}}{2 \, {\left (p + 1\right )}} \]

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]

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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maple [C]  time = 0.26, size = 142, normalized size = 5.26 \[ \frac {\left (c \,x^{2}+b \right ) x \,x^{p +1} {\mathrm e}^{\frac {\left (-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (c \,x^{2}+b \right )\right ) \mathrm {csgn}\left (i \left (c \,x^{2}+b \right ) x \right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (c \,x^{2}+b \right ) x \right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (c \,x^{2}+b \right )\right ) \mathrm {csgn}\left (i \left (c \,x^{2}+b \right ) x \right )^{2}-i \pi \mathrm {csgn}\left (i \left (c \,x^{2}+b \right ) x \right )^{3}+2 \ln \relax (x )+2 \ln \left (c \,x^{2}+b \right )\right ) p}{2}}}{2+2 p} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 1.07, size = 35, normalized size = 1.30 \[ \frac {{\left (c x^{4} + b x^{2}\right )} e^{\left (p \log \left (c x^{2} + b\right ) + 2 \, p \log \relax (x)\right )}}{2 \, {\left (p + 1\right )}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int b\,x^{p+1}\,{\left (c\,x^3+b\,x\right )}^p+2\,c\,x^{p+3}\,{\left (c\,x^3+b\,x\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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