Integrand size = 17, antiderivative size = 93 \[ \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \, dx=\frac {B x \left (d+e x^2\right )^{1+q}}{e (3+2 q)}-\frac {(B d-A e (3+2 q)) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )}{e (3+2 q)} \] Output:
B*x*(e*x^2+d)^(1+q)/e/(3+2*q)-(B*d-A*e*(3+2*q))*x*(e*x^2+d)^q*hypergeom([1 /2, -q],[3/2],-e*x^2/d)/e/(3+2*q)/((1+e*x^2/d)^q)
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.97 \[ \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \, dx=\frac {x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \left (B \left (d+e x^2\right ) \left (1+\frac {e x^2}{d}\right )^q+(-B d+A e (3+2 q)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )\right )}{e (3+2 q)} \] Input:
Integrate[(A + B*x^2)*(d + e*x^2)^q,x]
Output:
(x*(d + e*x^2)^q*(B*(d + e*x^2)*(1 + (e*x^2)/d)^q + (-(B*d) + A*e*(3 + 2*q ))*Hypergeometric2F1[1/2, -q, 3/2, -((e*x^2)/d)]))/(e*(3 + 2*q)*(1 + (e*x^ 2)/d)^q)
Time = 0.21 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {299, 238, 237}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \, dx\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \left (A-\frac {B d}{2 e q+3 e}\right ) \int \left (e x^2+d\right )^qdx+\frac {B x \left (d+e x^2\right )^{q+1}}{e (2 q+3)}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \left (A-\frac {B d}{2 e q+3 e}\right ) \int \left (\frac {e x^2}{d}+1\right )^qdx+\frac {B x \left (d+e x^2\right )^{q+1}}{e (2 q+3)}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \left (A-\frac {B d}{2 e q+3 e}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )+\frac {B x \left (d+e x^2\right )^{q+1}}{e (2 q+3)}\) |
Input:
Int[(A + B*x^2)*(d + e*x^2)^q,x]
Output:
(B*x*(d + e*x^2)^(1 + q))/(e*(3 + 2*q)) + ((A - (B*d)/(3*e + 2*e*q))*x*(d + e*x^2)^q*Hypergeometric2F1[1/2, -q, 3/2, -((e*x^2)/d)])/(1 + (e*x^2)/d)^ q
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] && !IntegerQ[2*p ] && GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) ^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(1 + b*(x^2/a))^p, x], x] / ; FreeQ[{a, b, p}, x] && !IntegerQ[2*p] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
\[\int \left (B \,x^{2}+A \right ) \left (e \,x^{2}+d \right )^{q}d x\]
Input:
int((B*x^2+A)*(e*x^2+d)^q,x)
Output:
int((B*x^2+A)*(e*x^2+d)^q,x)
\[ \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \, dx=\int { {\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{q} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^q,x, algorithm="fricas")
Output:
integral((B*x^2 + A)*(e*x^2 + d)^q, x)
Result contains complex when optimal does not.
Time = 4.53 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.57 \[ \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \, dx=A d^{q} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - q \\ \frac {3}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )} + \frac {B d^{q} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - q \\ \frac {5}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{3} \] Input:
integrate((B*x**2+A)*(e*x**2+d)**q,x)
Output:
A*d**q*x*hyper((1/2, -q), (3/2,), e*x**2*exp_polar(I*pi)/d) + B*d**q*x**3* hyper((3/2, -q), (5/2,), e*x**2*exp_polar(I*pi)/d)/3
\[ \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \, dx=\int { {\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{q} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^q,x, algorithm="maxima")
Output:
integrate((B*x^2 + A)*(e*x^2 + d)^q, x)
\[ \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \, dx=\int { {\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{q} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^q,x, algorithm="giac")
Output:
integrate((B*x^2 + A)*(e*x^2 + d)^q, x)
Timed out. \[ \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \, dx=\int \left (B\,x^2+A\right )\,{\left (e\,x^2+d\right )}^q \,d x \] Input:
int((A + B*x^2)*(d + e*x^2)^q,x)
Output:
int((A + B*x^2)*(d + e*x^2)^q, x)
\[ \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \, dx=\frac {2 \left (e \,x^{2}+d \right )^{q} a e q x +3 \left (e \,x^{2}+d \right )^{q} a e x +2 \left (e \,x^{2}+d \right )^{q} b d q x +2 \left (e \,x^{2}+d \right )^{q} b e q \,x^{3}+\left (e \,x^{2}+d \right )^{q} b e \,x^{3}+16 \left (\int \frac {\left (e \,x^{2}+d \right )^{q}}{4 e \,q^{2} x^{2}+8 e q \,x^{2}+4 d \,q^{2}+3 e \,x^{2}+8 d q +3 d}d x \right ) a d e \,q^{4}+56 \left (\int \frac {\left (e \,x^{2}+d \right )^{q}}{4 e \,q^{2} x^{2}+8 e q \,x^{2}+4 d \,q^{2}+3 e \,x^{2}+8 d q +3 d}d x \right ) a d e \,q^{3}+60 \left (\int \frac {\left (e \,x^{2}+d \right )^{q}}{4 e \,q^{2} x^{2}+8 e q \,x^{2}+4 d \,q^{2}+3 e \,x^{2}+8 d q +3 d}d x \right ) a d e \,q^{2}+18 \left (\int \frac {\left (e \,x^{2}+d \right )^{q}}{4 e \,q^{2} x^{2}+8 e q \,x^{2}+4 d \,q^{2}+3 e \,x^{2}+8 d q +3 d}d x \right ) a d e q -8 \left (\int \frac {\left (e \,x^{2}+d \right )^{q}}{4 e \,q^{2} x^{2}+8 e q \,x^{2}+4 d \,q^{2}+3 e \,x^{2}+8 d q +3 d}d x \right ) b \,d^{2} q^{3}-16 \left (\int \frac {\left (e \,x^{2}+d \right )^{q}}{4 e \,q^{2} x^{2}+8 e q \,x^{2}+4 d \,q^{2}+3 e \,x^{2}+8 d q +3 d}d x \right ) b \,d^{2} q^{2}-6 \left (\int \frac {\left (e \,x^{2}+d \right )^{q}}{4 e \,q^{2} x^{2}+8 e q \,x^{2}+4 d \,q^{2}+3 e \,x^{2}+8 d q +3 d}d x \right ) b \,d^{2} q}{e \left (4 q^{2}+8 q +3\right )} \] Input:
int((B*x^2+A)*(e*x^2+d)^q,x)
Output:
(2*(d + e*x**2)**q*a*e*q*x + 3*(d + e*x**2)**q*a*e*x + 2*(d + e*x**2)**q*b *d*q*x + 2*(d + e*x**2)**q*b*e*q*x**3 + (d + e*x**2)**q*b*e*x**3 + 16*int( (d + e*x**2)**q/(4*d*q**2 + 8*d*q + 3*d + 4*e*q**2*x**2 + 8*e*q*x**2 + 3*e *x**2),x)*a*d*e*q**4 + 56*int((d + e*x**2)**q/(4*d*q**2 + 8*d*q + 3*d + 4* e*q**2*x**2 + 8*e*q*x**2 + 3*e*x**2),x)*a*d*e*q**3 + 60*int((d + e*x**2)** q/(4*d*q**2 + 8*d*q + 3*d + 4*e*q**2*x**2 + 8*e*q*x**2 + 3*e*x**2),x)*a*d* e*q**2 + 18*int((d + e*x**2)**q/(4*d*q**2 + 8*d*q + 3*d + 4*e*q**2*x**2 + 8*e*q*x**2 + 3*e*x**2),x)*a*d*e*q - 8*int((d + e*x**2)**q/(4*d*q**2 + 8*d* q + 3*d + 4*e*q**2*x**2 + 8*e*q*x**2 + 3*e*x**2),x)*b*d**2*q**3 - 16*int(( d + e*x**2)**q/(4*d*q**2 + 8*d*q + 3*d + 4*e*q**2*x**2 + 8*e*q*x**2 + 3*e* x**2),x)*b*d**2*q**2 - 6*int((d + e*x**2)**q/(4*d*q**2 + 8*d*q + 3*d + 4*e *q**2*x**2 + 8*e*q*x**2 + 3*e*x**2),x)*b*d**2*q)/(e*(4*q**2 + 8*q + 3))