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

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Rubi [A]  time = 0.05, antiderivative size = 41, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.11, size = 119, normalized size = 2.90 \[ \frac {f \left (d+f x^2\right )^p \left (\frac {f x^2}{d}+1\right )^{-p} \left ((2 p+3) \left (3 b \left (d+f x^2\right ) \left (\frac {f x^2}{d}+1\right )^p+2 c f (p+1) x^3 \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {f x^2}{d}\right )\right )+6 c d (p+1) x \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {f x^2}{d}\right )\right )}{3 (p+1)} \]

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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fricas [A]  time = 0.58, size = 75, normalized size = 1.83 \[ \frac {{\left (2 \, b d f p + 2 \, {\left (c f^{2} p + c f^{2}\right )} x^{3} + 3 \, b d f + {\left (2 \, b f^{2} p + 3 \, b f^{2}\right )} x^{2} + 2 \, {\left (c d f p + c d f\right )} x\right )} {\left (f x^{2} + d\right )}^{p}}{p + 1} \]

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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giac [B]  time = 0.18, size = 141, normalized size = 3.44 \[ \frac {2 \, {\left (f x^{2} + d\right )}^{p} c f^{2} p x^{3} + 2 \, {\left (f x^{2} + d\right )}^{p} b f^{2} p x^{2} + 2 \, {\left (f x^{2} + d\right )}^{p} c f^{2} x^{3} + 2 \, {\left (f x^{2} + d\right )}^{p} c d f p x + 3 \, {\left (f x^{2} + d\right )}^{p} b f^{2} x^{2} + 2 \, {\left (f x^{2} + d\right )}^{p} b d f p + 2 \, {\left (f x^{2} + d\right )}^{p} c d f x + 3 \, {\left (f x^{2} + d\right )}^{p} b d f}{p + 1} \]

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)

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maple [A]  time = 0.00, size = 36, normalized size = 0.88 \[ \frac {\left (2 p c x +2 p b +2 c x +3 b \right ) f \left (f \,x^{2}+d \right )^{p +1}}{p +1} \]

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 = 0.58, size = 59, normalized size = 1.44 \[ \frac {{\left (2 \, c f^{2} {\left (p + 1\right )} x^{3} + b f^{2} {\left (2 \, p + 3\right )} x^{2} + 2 \, c d f {\left (p + 1\right )} x + b d f {\left (2 \, p + 3\right )}\right )} {\left (f x^{2} + d\right )}^{p}}{p + 1} \]

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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mupad [B]  time = 4.25, size = 58, normalized size = 1.41 \[ {\left (f\,x^2+d\right )}^p\,\left (2\,c\,f^2\,x^3+2\,c\,d\,f\,x+\frac {b\,f^2\,x^2\,\left (2\,p+3\right )}{p+1}+\frac {b\,d\,f\,\left (2\,p+3\right )}{p+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 13.25, size = 221, normalized size = 5.39 \[ \begin {cases} \frac {2 b d f p \left (d + f x^{2}\right )^{p}}{p + 1} + \frac {3 b d f \left (d + f x^{2}\right )^{p}}{p + 1} + \frac {2 b f^{2} p x^{2} \left (d + f x^{2}\right )^{p}}{p + 1} + \frac {3 b f^{2} x^{2} \left (d + f x^{2}\right )^{p}}{p + 1} + \frac {2 c d f p x \left (d + f x^{2}\right )^{p}}{p + 1} + \frac {2 c d f x \left (d + f x^{2}\right )^{p}}{p + 1} + \frac {2 c f^{2} p x^{3} \left (d + f x^{2}\right )^{p}}{p + 1} + \frac {2 c f^{2} x^{3} \left (d + f x^{2}\right )^{p}}{p + 1} & \text {for}\: p \neq -1 \\b f \log {\left (- i \sqrt {d} \sqrt {\frac {1}{f}} + x \right )} + b f \log {\left (i \sqrt {d} \sqrt {\frac {1}{f}} + x \right )} + 2 c f x & \text {otherwise} \end {cases} \]

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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