3.275 \(\int (d+f x^2)^p (2 c d f+2 b f^2 (3+2 p) x+2 c f^2 (3+2 p) x^2) \, dx\)

Optimal. Leaf size=41 \[ \frac{b f (2 p+3) \left (d+f x^2\right )^{p+1}}{p+1}+2 c f x \left (d+f x^2\right )^{p+1} \]

[Out]

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

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Rubi [A]  time = 0.0515075, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {1815, 12, 261} \[ \frac{b f (2 p+3) \left (d+f x^2\right )^{p+1}}{p+1}+2 c f x \left (d+f x^2\right )^{p+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

Mathematica [C]  time = 0.0946046, size = 119, normalized size = 2.9 \[ \frac{f \left (d+f x^2\right )^p \left (\frac{f x^2}{d}+1\right )^{-p} \left ((2 p+3) \left (3 b \left (d+f x^2\right ) \left (\frac{f x^2}{d}+1\right )^p+2 c f (p+1) x^3 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{f x^2}{d}\right )\right )+6 c d (p+1) x \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{f x^2}{d}\right )\right )}{3 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(d + f*x^2)^p*(6*c*d*(1 + p)*x*Hypergeometric2F1[1/2, -p, 3/2, -((f*x^2)/d)] + (3 + 2*p)*(3*b*(d + f*x^2)*(
1 + (f*x^2)/d)^p + 2*c*f*(1 + p)*x^3*Hypergeometric2F1[3/2, -p, 5/2, -((f*x^2)/d)])))/(3*(1 + p)*(1 + (f*x^2)/
d)^p)

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Maple [A]  time = 0.044, size = 36, normalized size = 0.9 \begin{align*}{\frac{f \left ( f{x}^{2}+d \right ) ^{1+p} \left ( 2\,cxp+2\,bp+2\,cx+3\,b \right ) }{1+p}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.11117, size = 80, normalized size = 1.95 \begin{align*} \frac{{\left (2 \, c f^{2}{\left (p + 1\right )} x^{3} + b f^{2}{\left (2 \, p + 3\right )} x^{2} + 2 \, c d f{\left (p + 1\right )} x + b d f{\left (2 \, p + 3\right )}\right )}{\left (f x^{2} + d\right )}^{p}}{p + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.39543, size = 166, normalized size = 4.05 \begin{align*} \frac{{\left (2 \, b d f p + 2 \,{\left (c f^{2} p + c f^{2}\right )} x^{3} + 3 \, b d f +{\left (2 \, b f^{2} p + 3 \, b f^{2}\right )} x^{2} + 2 \,{\left (c d f p + c d f\right )} x\right )}{\left (f x^{2} + d\right )}^{p}}{p + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 9.56548, size = 221, normalized size = 5.39 \begin{align*} \begin{cases} \frac{2 b d f p \left (d + f x^{2}\right )^{p}}{p + 1} + \frac{3 b d f \left (d + f x^{2}\right )^{p}}{p + 1} + \frac{2 b f^{2} p x^{2} \left (d + f x^{2}\right )^{p}}{p + 1} + \frac{3 b f^{2} x^{2} \left (d + f x^{2}\right )^{p}}{p + 1} + \frac{2 c d f p x \left (d + f x^{2}\right )^{p}}{p + 1} + \frac{2 c d f x \left (d + f x^{2}\right )^{p}}{p + 1} + \frac{2 c f^{2} p x^{3} \left (d + f x^{2}\right )^{p}}{p + 1} + \frac{2 c f^{2} x^{3} \left (d + f x^{2}\right )^{p}}{p + 1} & \text{for}\: p \neq -1 \\b f \log{\left (- i \sqrt{d} \sqrt{\frac{1}{f}} + x \right )} + b f \log{\left (i \sqrt{d} \sqrt{\frac{1}{f}} + x \right )} + 2 c f x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*b*d*f*p*(d + f*x**2)**p/(p + 1) + 3*b*d*f*(d + f*x**2)**p/(p + 1) + 2*b*f**2*p*x**2*(d + f*x**2)*
*p/(p + 1) + 3*b*f**2*x**2*(d + f*x**2)**p/(p + 1) + 2*c*d*f*p*x*(d + f*x**2)**p/(p + 1) + 2*c*d*f*x*(d + f*x*
*2)**p/(p + 1) + 2*c*f**2*p*x**3*(d + f*x**2)**p/(p + 1) + 2*c*f**2*x**3*(d + f*x**2)**p/(p + 1), Ne(p, -1)),
(b*f*log(-I*sqrt(d)*sqrt(1/f) + x) + b*f*log(I*sqrt(d)*sqrt(1/f) + x) + 2*c*f*x, True))

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Giac [B]  time = 1.22614, size = 190, normalized size = 4.63 \begin{align*} \frac{2 \,{\left (f x^{2} + d\right )}^{p} c f^{2} p x^{3} + 2 \,{\left (f x^{2} + d\right )}^{p} b f^{2} p x^{2} + 2 \,{\left (f x^{2} + d\right )}^{p} c f^{2} x^{3} + 2 \,{\left (f x^{2} + d\right )}^{p} c d f p x + 3 \,{\left (f x^{2} + d\right )}^{p} b f^{2} x^{2} + 2 \,{\left (f x^{2} + d\right )}^{p} b d f p + 2 \,{\left (f x^{2} + d\right )}^{p} c d f x + 3 \,{\left (f x^{2} + d\right )}^{p} b d f}{p + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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