3.342 \(\int (a+b x^2)^p (c+d x^2)^3 \, dx\)

Optimal. Leaf size=296 \[ \frac {d x \left (a+b x^2\right )^{p+1} \left (15 a^2 d^2-8 a b c d (p+6)+b^2 c^2 \left (4 p^2+28 p+57\right )\right )}{b^3 (2 p+3) (2 p+5) (2 p+7)}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (15 a^3 d^3-9 a^2 b c d^2 (2 p+7)+3 a b^2 c^2 d \left (4 p^2+24 p+35\right )-b^3 c^3 \left (8 p^3+60 p^2+142 p+105\right )\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b^3 (2 p+3) (2 p+5) (2 p+7)}-\frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1} (5 a d-b c (2 p+11))}{b^2 (2 p+5) (2 p+7)}+\frac {d x \left (c+d x^2\right )^2 \left (a+b x^2\right )^{p+1}}{b (2 p+7)} \]

[Out]

d*(15*a^2*d^2-8*a*b*c*d*(6+p)+b^2*c^2*(4*p^2+28*p+57))*x*(b*x^2+a)^(1+p)/b^3/(8*p^3+60*p^2+142*p+105)-d*(5*a*d
-b*c*(11+2*p))*x*(b*x^2+a)^(1+p)*(d*x^2+c)/b^2/(4*p^2+24*p+35)+d*x*(b*x^2+a)^(1+p)*(d*x^2+c)^2/b/(7+2*p)-(15*a
^3*d^3-9*a^2*b*c*d^2*(7+2*p)+3*a*b^2*c^2*d*(4*p^2+24*p+35)-b^3*c^3*(8*p^3+60*p^2+142*p+105))*x*(b*x^2+a)^p*hyp
ergeom([1/2, -p],[3/2],-b*x^2/a)/b^3/(8*p^3+60*p^2+142*p+105)/((1+b*x^2/a)^p)

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Rubi [A]  time = 0.28, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {416, 528, 388, 246, 245} \[ -\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (-9 a^2 b c d^2 (2 p+7)+15 a^3 d^3+3 a b^2 c^2 d \left (4 p^2+24 p+35\right )-b^3 c^3 \left (8 p^3+60 p^2+142 p+105\right )\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b^3 (2 p+3) (2 p+5) (2 p+7)}+\frac {d x \left (a+b x^2\right )^{p+1} \left (15 a^2 d^2-8 a b c d (p+6)+b^2 c^2 \left (4 p^2+28 p+57\right )\right )}{b^3 (2 p+3) (2 p+5) (2 p+7)}-\frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1} (5 a d-b c (2 p+11))}{b^2 (2 p+5) (2 p+7)}+\frac {d x \left (c+d x^2\right )^2 \left (a+b x^2\right )^{p+1}}{b (2 p+7)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p*(c + d*x^2)^3,x]

[Out]

(d*(15*a^2*d^2 - 8*a*b*c*d*(6 + p) + b^2*c^2*(57 + 28*p + 4*p^2))*x*(a + b*x^2)^(1 + p))/(b^3*(3 + 2*p)*(5 + 2
*p)*(7 + 2*p)) - (d*(5*a*d - b*c*(11 + 2*p))*x*(a + b*x^2)^(1 + p)*(c + d*x^2))/(b^2*(5 + 2*p)*(7 + 2*p)) + (d
*x*(a + b*x^2)^(1 + p)*(c + d*x^2)^2)/(b*(7 + 2*p)) - ((15*a^3*d^3 - 9*a^2*b*c*d^2*(7 + 2*p) + 3*a*b^2*c^2*d*(
35 + 24*p + 4*p^2) - b^3*c^3*(105 + 142*p + 60*p^2 + 8*p^3))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -
((b*x^2)/a)])/(b^3*(3 + 2*p)*(5 + 2*p)*(7 + 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]

Rule 528

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

Rubi steps

\begin {align*} \int \left (a+b x^2\right )^p \left (c+d x^2\right )^3 \, dx &=\frac {d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^2}{b (7+2 p)}+\frac {\int \left (a+b x^2\right )^p \left (c+d x^2\right ) \left (-c (a d-b c (7+2 p))-d (5 a d-b c (11+2 p)) x^2\right ) \, dx}{b (7+2 p)}\\ &=-\frac {d (5 a d-b c (11+2 p)) x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b^2 (5+2 p) (7+2 p)}+\frac {d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^2}{b (7+2 p)}+\frac {\int \left (a+b x^2\right )^p \left (c \left (5 a^2 d^2-4 a b c d (4+p)+b^2 c^2 \left (35+24 p+4 p^2\right )\right )+d \left (15 a^2 d^2-8 a b c d (6+p)+b^2 c^2 \left (57+28 p+4 p^2\right )\right ) x^2\right ) \, dx}{b^2 (5+2 p) (7+2 p)}\\ &=\frac {d \left (15 a^2 d^2-8 a b c d (6+p)+b^2 c^2 \left (57+28 p+4 p^2\right )\right ) x \left (a+b x^2\right )^{1+p}}{b^3 (3+2 p) (5+2 p) (7+2 p)}-\frac {d (5 a d-b c (11+2 p)) x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b^2 (5+2 p) (7+2 p)}+\frac {d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^2}{b (7+2 p)}-\frac {\left (15 a^3 d^3-9 a^2 b c d^2 (7+2 p)+3 a b^2 c^2 d \left (35+24 p+4 p^2\right )-b^3 c^3 \left (105+142 p+60 p^2+8 p^3\right )\right ) \int \left (a+b x^2\right )^p \, dx}{b^3 (3+2 p) (5+2 p) (7+2 p)}\\ &=\frac {d \left (15 a^2 d^2-8 a b c d (6+p)+b^2 c^2 \left (57+28 p+4 p^2\right )\right ) x \left (a+b x^2\right )^{1+p}}{b^3 (3+2 p) (5+2 p) (7+2 p)}-\frac {d (5 a d-b c (11+2 p)) x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b^2 (5+2 p) (7+2 p)}+\frac {d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^2}{b (7+2 p)}-\frac {\left (\left (15 a^3 d^3-9 a^2 b c d^2 (7+2 p)+3 a b^2 c^2 d \left (35+24 p+4 p^2\right )-b^3 c^3 \left (105+142 p+60 p^2+8 p^3\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^3 (3+2 p) (5+2 p) (7+2 p)}\\ &=\frac {d \left (15 a^2 d^2-8 a b c d (6+p)+b^2 c^2 \left (57+28 p+4 p^2\right )\right ) x \left (a+b x^2\right )^{1+p}}{b^3 (3+2 p) (5+2 p) (7+2 p)}-\frac {d (5 a d-b c (11+2 p)) x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b^2 (5+2 p) (7+2 p)}+\frac {d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^2}{b (7+2 p)}-\frac {\left (15 a^3 d^3-9 a^2 b c d^2 (7+2 p)+3 a b^2 c^2 d \left (35+24 p+4 p^2\right )-b^3 c^3 \left (105+142 p+60 p^2+8 p^3\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^3 (3+2 p) (5+2 p) (7+2 p)}\\ \end {align*}

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Mathematica [A]  time = 5.07, size = 136, normalized size = 0.46 \[ \frac {1}{35} x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (35 c^3 \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )+d x^2 \left (35 c^2 \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )+d x^2 \left (21 c \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right )+5 d x^2 \, _2F_1\left (\frac {7}{2},-p;\frac {9}{2};-\frac {b x^2}{a}\right )\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^2)^p*(35*c^3*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + d*x^2*(35*c^2*Hypergeometric2F1[3/2,
-p, 5/2, -((b*x^2)/a)] + d*x^2*(21*c*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)] + 5*d*x^2*Hypergeometric2F1
[7/2, -p, 9/2, -((b*x^2)/a)]))))/(35*(1 + (b*x^2)/a)^p)

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fricas [F]  time = 1.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}\right )} {\left (b x^{2} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d*x^2 + c^3)*(b*x^2 + a)^p, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x^{2} + c\right )}^{3} {\left (b x^{2} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \left (d \,x^{2}+c \right )^{3} \left (b \,x^{2}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x^{2} + c\right )}^{3} {\left (b x^{2} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (b\,x^2+a\right )}^p\,{\left (d\,x^2+c\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p*(c + d*x^2)^3,x)

[Out]

int((a + b*x^2)^p*(c + d*x^2)^3, x)

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sympy [C]  time = 39.51, size = 121, normalized size = 0.41 \[ a^{p} c^{3} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + a^{p} c^{2} d x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {3 a^{p} c d^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + \frac {a^{p} d^{3} x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{2}, - p \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*c**3*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**p*c**2*d*x**3*hyper((3/2, -p), (5/2,), b*x
**2*exp_polar(I*pi)/a) + 3*a**p*c*d**2*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + a**p*d**3*x
**7*hyper((7/2, -p), (9/2,), b*x**2*exp_polar(I*pi)/a)/7

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