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

Optimal. Leaf size=176 \[ \frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 d^2-2 a b c d (2 p+5)+b^2 c^2 \left (4 p^2+16 p+15\right )\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b^2 (2 p+3) (2 p+5)}-\frac {d x \left (a+b x^2\right )^{p+1} (3 a d-b c (2 p+7))}{b^2 (2 p+3) (2 p+5)}+\frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]

[Out]

-d*(3*a*d-b*c*(7+2*p))*x*(b*x^2+a)^(1+p)/b^2/(4*p^2+16*p+15)+d*x*(b*x^2+a)^(1+p)*(d*x^2+c)/b/(5+2*p)+(3*a^2*d^
2-2*a*b*c*d*(5+2*p)+b^2*c^2*(4*p^2+16*p+15))*x*(b*x^2+a)^p*hypergeom([1/2, -p],[3/2],-b*x^2/a)/b^2/(4*p^2+16*p
+15)/((1+b*x^2/a)^p)

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Rubi [A]  time = 0.12, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {416, 388, 246, 245} \[ \frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 d^2-2 a b c d (2 p+5)+b^2 c^2 \left (4 p^2+16 p+15\right )\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b^2 (2 p+3) (2 p+5)}-\frac {d x \left (a+b x^2\right )^{p+1} (3 a d-b c (2 p+7))}{b^2 (2 p+3) (2 p+5)}+\frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p*(c + d*x^2)^2,x]

[Out]

-((d*(3*a*d - b*c*(7 + 2*p))*x*(a + b*x^2)^(1 + p))/(b^2*(3 + 2*p)*(5 + 2*p))) + (d*x*(a + b*x^2)^(1 + p)*(c +
d*x^2))/(b*(5 + 2*p)) + ((3*a^2*d^2 - 2*a*b*c*d*(5 + 2*p) + b^2*c^2*(15 + 16*p + 4*p^2))*x*(a + b*x^2)^p*Hype
rgeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(b^2*(3 + 2*p)*(5 + 2*p)*(1 + (b*x^2)/a)^p)

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])

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 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 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps

\begin {align*} \int \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx &=\frac {d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b (5+2 p)}+\frac {\int \left (a+b x^2\right )^p \left (-c (a d-b c (5+2 p))-d (3 a d-b c (7+2 p)) x^2\right ) \, dx}{b (5+2 p)}\\ &=-\frac {d (3 a d-b c (7+2 p)) x \left (a+b x^2\right )^{1+p}}{b^2 (3+2 p) (5+2 p)}+\frac {d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b (5+2 p)}+\frac {\left (3 a^2 d^2-2 a b c d (5+2 p)+b^2 c^2 \left (15+16 p+4 p^2\right )\right ) \int \left (a+b x^2\right )^p \, dx}{b^2 (3+2 p) (5+2 p)}\\ &=-\frac {d (3 a d-b c (7+2 p)) x \left (a+b x^2\right )^{1+p}}{b^2 (3+2 p) (5+2 p)}+\frac {d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b (5+2 p)}+\frac {\left (\left (3 a^2 d^2-2 a b c d (5+2 p)+b^2 c^2 \left (15+16 p+4 p^2\right )\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}{b^2 (3+2 p) (5+2 p)}\\ &=-\frac {d (3 a d-b c (7+2 p)) x \left (a+b x^2\right )^{1+p}}{b^2 (3+2 p) (5+2 p)}+\frac {d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b (5+2 p)}+\frac {\left (3 a^2 d^2-2 a b c d (5+2 p)+b^2 c^2 \left (15+16 p+4 p^2\right )\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 )}{b^2 (3+2 p) (5+2 p)}\\ \end {align*}

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Mathematica [A]  time = 5.05, size = 106, normalized size = 0.60 \[ \frac {1}{15} x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (15 c^2 \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )+d x^2 \left (10 c \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )+3 d x^2 \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^2)^p*(15*c^2*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + d*x^2*(10*c*Hypergeometric2F1[3/2, -p
, 5/2, -((b*x^2)/a)] + 3*d*x^2*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])))/(15*(1 + (b*x^2)/a)^p)

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fricas [F]  time = 1.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d^{2} x^{4} + 2 \, c d x^{2} + c^{2}\right )} {\left (b x^{2} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d^2*x^4 + 2*c*d*x^2 + c^2)*(b*x^2 + a)^p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \left (d \,x^{2}+c \right )^{2} \left (b \,x^{2}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p*(c + d*x^2)^2,x)

[Out]

int((a + b*x^2)^p*(c + d*x^2)^2, x)

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sympy [C]  time = 20.40, size = 88, normalized size = 0.50 \[ a^{p} c^{2} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {2 a^{p} c d x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {a^{p} d^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*c**2*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + 2*a**p*c*d*x**3*hyper((3/2, -p), (5/2,), b*x*
*2*exp_polar(I*pi)/a)/3 + a**p*d**2*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5

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