3.344 \(\int (a+b x^2)^p (c+d x^2) \, dx\)

Optimal. Leaf size=93 \[ \frac{d x \left (a+b x^2\right )^{p+1}}{b (2 p+3)}-\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} (a d-b c (2 p+3)) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b (2 p+3)} \]

[Out]

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

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Rubi [A]  time = 0.0390779, antiderivative size = 85, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {388, 246, 245} \[ x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (c-\frac{a d}{2 b p+3 b}\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+\frac{d x \left (a+b x^2\right )^{p+1}}{b (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p*(c + d*x^2),x]

[Out]

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

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

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

Rubi steps

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

Mathematica [A]  time = 0.029336, size = 90, normalized size = 0.97 \[ \frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left ((b c (2 p+3)-a d) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+d \left (a+b x^2\right ) \left (\frac{b x^2}{a}+1\right )^p\right )}{b (2 p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int \left ( b{x}^{2}+a \right ) ^{p} \left ( d{x}^{2}+c \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x^2 + c)*(b*x^2 + a)^p, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d x^{2} + c\right )}{\left (b x^{2} + a\right )}^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x^2 + c)*(b*x^2 + a)^p, x)

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Sympy [C]  time = 10.7052, size = 53, normalized size = 0.57 \begin{align*} a^{p} c x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} + \frac{a^{p} d x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*c*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**p*d*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_
polar(I*pi)/a)/3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x^2 + c)*(b*x^2 + a)^p, x)