3.3.63 \(\int (d+e x)^3 (d^2-e^2 x^2)^p \, dx\) [263]

3.3.63.1 Optimal result

Integrand size = 22, antiderivative size = 73 \[ \int (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=-\frac {2^{3+p} d^2 \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-3-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{e (1+p)} \]

output

-2^(3+p)*d^2*(1+e*x/d)^(-1-p)*(-e^2*x^2+d^2)^(p+1)*hypergeom([p+1, -3-p],[ 
2+p],1/2*(-e*x+d)/d)/e/(p+1)
 

3.3.63.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(155\) vs. \(2(73)=146\).

Time = 0.29 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.12 \[ \int (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\frac {1}{2} \left (d^2-e^2 x^2\right )^p \left (\frac {\left (-d^2+e^2 x^2\right ) \left (d^2 (7+3 p)+e^2 (1+p) x^2\right )}{e (1+p) (2+p)}+2 d^3 x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )+2 d e^2 x^3 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )\right ) \]

input

Integrate[(d + e*x)^3*(d^2 - e^2*x^2)^p,x]
 

output

((d^2 - e^2*x^2)^p*(((-d^2 + e^2*x^2)*(d^2*(7 + 3*p) + e^2*(1 + p)*x^2))/( 
e*(1 + p)*(2 + p)) + (2*d^3*x*Hypergeometric2F1[1/2, -p, 3/2, (e^2*x^2)/d^ 
2])/(1 - (e^2*x^2)/d^2)^p + (2*d*e^2*x^3*Hypergeometric2F1[3/2, -p, 5/2, ( 
e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^2)^p))/2
 

3.3.63.3 Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {473, 79}

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.

\(\displaystyle \int (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx\)

\(\Big \downarrow \) 473

\(\displaystyle d^2 (d-e x)^{-p-1} \left (\frac {e x}{d}+1\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \int (d-e x)^p \left (\frac {e x}{d}+1\right )^{p+3}dx\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {d^2 2^{p+3} \left (\frac {e x}{d}+1\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p-3,p+1,p+2,\frac {d-e x}{2 d}\right )}{e (p+1)}\)

input

Int[(d + e*x)^3*(d^2 - e^2*x^2)^p,x]
 

output

-((2^(3 + p)*d^2*(1 + (e*x)/d)^(-1 - p)*(d^2 - e^2*x^2)^(1 + p)*Hypergeome 
tric2F1[-3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/(e*(1 + p)))
 

3.3.63.3.1 Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 473

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 
1)))   Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, 
c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) &&  !Gt 
Q[a, 0] &&  !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
 

3.3.63.4 Maple [F]

\[\int \left (e x +d \right )^{3} \left (-e^{2} x^{2}+d^{2}\right )^{p}d x\]

input

int((e*x+d)^3*(-e^2*x^2+d^2)^p,x)
 

output

int((e*x+d)^3*(-e^2*x^2+d^2)^p,x)
 

3.3.63.5 Fricas [F]

\[ \int (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} \,d x } \]

input

integrate((e*x+d)^3*(-e^2*x^2+d^2)^p,x, algorithm="fricas")
 

output

integral((e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*(-e^2*x^2 + d^2)^p, x)
 

3.3.63.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (58) = 116\).

Time = 1.85 (sec) , antiderivative size = 476, normalized size of antiderivative = 6.52 \[ \int (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=d^{3} d^{2 p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + 3 d^{2} e \left (\begin {cases} \frac {x^{2} \left (d^{2}\right )^{p}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\begin {cases} \frac {\left (d^{2} - e^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (d^{2} - e^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 e^{2}} & \text {otherwise} \end {cases}\right ) + d d^{2 p} e^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + e^{3} \left (\begin {cases} \frac {x^{4} \left (d^{2}\right )^{p}}{4} & \text {for}\: e = 0 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{2 e^{4}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{2 e^{4}} - \frac {x^{2}}{2 e^{2}} & \text {for}\: p = -1 \\- \frac {d^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} - \frac {d^{2} e^{2} p x^{2} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} p x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} & \text {otherwise} \end {cases}\right ) \]

input

integrate((e*x+d)**3*(-e**2*x**2+d**2)**p,x)
 

output

d**3*d**(2*p)*x*hyper((1/2, -p), (3/2,), e**2*x**2*exp_polar(2*I*pi)/d**2) 
 + 3*d**2*e*Piecewise((x**2*(d**2)**p/2, Eq(e**2, 0)), (-Piecewise(((d**2 
- e**2*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(d**2 - e**2*x**2), True))/ 
(2*e**2), True)) + d*d**(2*p)*e**2*x**3*hyper((3/2, -p), (5/2,), e**2*x**2 
*exp_polar(2*I*pi)/d**2) + e**3*Piecewise((x**4*(d**2)**p/4, Eq(e, 0)), (- 
d**2*log(-d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2) - d**2*log(d/e + x)/(-2*d* 
*2*e**4 + 2*e**6*x**2) - d**2/(-2*d**2*e**4 + 2*e**6*x**2) + e**2*x**2*log 
(-d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2) + e**2*x**2*log(d/e + x)/(-2*d**2* 
e**4 + 2*e**6*x**2), Eq(p, -2)), (-d**2*log(-d/e + x)/(2*e**4) - d**2*log( 
d/e + x)/(2*e**4) - x**2/(2*e**2), Eq(p, -1)), (-d**4*(d**2 - e**2*x**2)** 
p/(2*e**4*p**2 + 6*e**4*p + 4*e**4) - d**2*e**2*p*x**2*(d**2 - e**2*x**2)* 
*p/(2*e**4*p**2 + 6*e**4*p + 4*e**4) + e**4*p*x**4*(d**2 - e**2*x**2)**p/( 
2*e**4*p**2 + 6*e**4*p + 4*e**4) + e**4*x**4*(d**2 - e**2*x**2)**p/(2*e**4 
*p**2 + 6*e**4*p + 4*e**4), True))
 

3.3.63.7 Maxima [F]

\[ \int (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} \,d x } \]

input

integrate((e*x+d)^3*(-e^2*x^2+d^2)^p,x, algorithm="maxima")
 

output

integrate((e*x + d)^3*(-e^2*x^2 + d^2)^p, x)
 

3.3.63.8 Giac [F]

\[ \int (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} \,d x } \]

input

integrate((e*x+d)^3*(-e^2*x^2+d^2)^p,x, algorithm="giac")
 

output

integrate((e*x + d)^3*(-e^2*x^2 + d^2)^p, x)
 

3.3.63.9 Mupad [F(-1)]

Timed out. \[ \int (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int {\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^3 \,d x \]

input

int((d^2 - e^2*x^2)^p*(d + e*x)^3,x)
 

output

int((d^2 - e^2*x^2)^p*(d + e*x)^3, x)