3.4.44 \(\int (a+b x^2)^p (c+d x^2) \, dx\) [344]

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

[Out]

d*x*(b*x^2+a)^(1+p)/b/(3+2*p)-(a*d-b*c*(3+2*p))*x*(b*x^2+a)^p*hypergeom([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 = 0.03, 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 = {396, 252, 251} \begin {gather*} 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)} \end {gather*}

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 251

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])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 396

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]

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*}

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Mathematica [A]
time = 0.07, size = 90, normalized size = 0.97 \begin {gather*} \frac {x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (d \left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p+(-a d+b c (3+2 p)) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )\right )}{b (3+2 p)} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \left (b \,x^{2}+a \right )^{p} \left (d \,x^{2}+c \right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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] Result contains complex when optimal does not.
time = 4.68, size = 53, normalized size = 0.57 \begin {gather*} 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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,x^2+a\right )}^p\,\left (d\,x^2+c\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p*(c + d*x^2),x)

[Out]

int((a + b*x^2)^p*(c + d*x^2), x)

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