Integrand size = 24, antiderivative size = 321 \[ \int (e x)^{-2 p} (c+d x)^4 \left (a+b x^2\right )^p \, dx=\frac {d^2 \left (30 b c^2-a d^2 (3-2 p)\right ) (e x)^{1-2 p} \left (a+b x^2\right )^{1+p}}{15 b^2 e}+\frac {c d^3 (e x)^{2-2 p} \left (a+b x^2\right )^{1+p}}{b e^2}+\frac {d^4 (e x)^{3-2 p} \left (a+b x^2\right )^{1+p}}{5 b e^3}+\frac {\left (15 b^2 c^4-30 a b c^2 d^2 (1-2 p)+a^2 d^4 \left (3-8 p+4 p^2\right )\right ) (e x)^{1-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1-2 p),-p,\frac {1}{2} (3-2 p),-\frac {b x^2}{a}\right )}{15 b^2 e (1-2 p)}-\frac {c d \left (a d^2-\frac {2 b c^2}{1-p}\right ) (e x)^{2-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {b x^2}{a}\right )}{b e^2} \] Output:
1/15*d^2*(30*b*c^2-a*d^2*(3-2*p))*(e*x)^(1-2*p)*(b*x^2+a)^(p+1)/b^2/e+c*d^ 3*(e*x)^(2-2*p)*(b*x^2+a)^(p+1)/b/e^2+1/5*d^4*(e*x)^(3-2*p)*(b*x^2+a)^(p+1 )/b/e^3+1/15*(15*b^2*c^4-30*a*b*c^2*d^2*(1-2*p)+a^2*d^4*(4*p^2-8*p+3))*(e* x)^(1-2*p)*(b*x^2+a)^p*hypergeom([-p, 1/2-p],[3/2-p],-b*x^2/a)/b^2/e/(1-2* p)/((1+b*x^2/a)^p)-c*d*(a*d^2-2*b*c^2/(1-p))*(e*x)^(2-2*p)*(b*x^2+a)^p*hyp ergeom([-p, 1-p],[2-p],-b*x^2/a)/b/e^2/((1+b*x^2/a)^p)
Time = 0.27 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.69 \[ \int (e x)^{-2 p} (c+d x)^4 \left (a+b x^2\right )^p \, dx=x (e x)^{-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (\frac {c^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-p,-p,\frac {3}{2}-p,-\frac {b x^2}{a}\right )}{1-2 p}+d x \left (-\frac {2 c^3 \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {b x^2}{a}\right )}{-1+p}+d x \left (\frac {6 c^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-p,-p,\frac {5}{2}-p,-\frac {b x^2}{a}\right )}{3-2 p}+d x \left (-\frac {2 c \operatorname {Hypergeometric2F1}\left (2-p,-p,3-p,-\frac {b x^2}{a}\right )}{-2+p}+\frac {d x \operatorname {Hypergeometric2F1}\left (\frac {5}{2}-p,-p,\frac {7}{2}-p,-\frac {b x^2}{a}\right )}{5-2 p}\right )\right )\right )\right ) \] Input:
Integrate[((c + d*x)^4*(a + b*x^2)^p)/(e*x)^(2*p),x]
Output:
(x*(a + b*x^2)^p*((c^4*Hypergeometric2F1[1/2 - p, -p, 3/2 - p, -((b*x^2)/a )])/(1 - 2*p) + d*x*((-2*c^3*Hypergeometric2F1[1 - p, -p, 2 - p, -((b*x^2) /a)])/(-1 + p) + d*x*((6*c^2*Hypergeometric2F1[3/2 - p, -p, 5/2 - p, -((b* x^2)/a)])/(3 - 2*p) + d*x*((-2*c*Hypergeometric2F1[2 - p, -p, 3 - p, -((b* x^2)/a)])/(-2 + p) + (d*x*Hypergeometric2F1[5/2 - p, -p, 7/2 - p, -((b*x^2 )/a)])/(5 - 2*p))))))/((e*x)^(2*p)*(1 + (b*x^2)/a)^p)
Time = 0.73 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {559, 2340, 27, 2340, 27, 557, 279, 278}
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.
\(\displaystyle \int (c+d x)^4 (e x)^{-2 p} \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 559 |
\(\displaystyle \frac {\int (e x)^{-2 p} \left (b x^2+a\right )^p \left (5 b c^4+20 b d x c^3+20 b d^3 x^3 c+d^2 \left (30 b c^2-a d^2 (3-2 p)\right ) x^2\right )dx}{5 b}+\frac {d^4 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\) |
\(\Big \downarrow \) 2340 |
\(\displaystyle \frac {\frac {\int 4 (e x)^{-2 p} \left (b x^2+a\right )^p \left (5 b^2 c^4+10 b d \left (2 b c^2-a d^2 (1-p)\right ) x c+b d^2 \left (30 b c^2-a d^2 (3-2 p)\right ) x^2\right )dx}{4 b}+\frac {5 c d^3 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{e^2}}{5 b}+\frac {d^4 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int (e x)^{-2 p} \left (b x^2+a\right )^p \left (5 b^2 c^4+10 b d \left (2 b c^2-a d^2 (1-p)\right ) x c+b d^2 \left (30 b c^2-a d^2 (3-2 p)\right ) x^2\right )dx}{b}+\frac {5 c d^3 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{e^2}}{5 b}+\frac {d^4 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\) |
\(\Big \downarrow \) 2340 |
\(\displaystyle \frac {\frac {\frac {\int b (e x)^{-2 p} \left (15 b^2 c^4-30 a b d^2 (1-2 p) c^2+30 b d \left (2 b c^2-a d^2 (1-p)\right ) x c+a^2 d^4 \left (4 p^2-8 p+3\right )\right ) \left (b x^2+a\right )^pdx}{3 b}+\frac {d^2 (e x)^{1-2 p} \left (a+b x^2\right )^{p+1} \left (30 b c^2-a d^2 (3-2 p)\right )}{3 e}}{b}+\frac {5 c d^3 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{e^2}}{5 b}+\frac {d^4 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {1}{3} \int (e x)^{-2 p} \left (15 b^2 c^4-30 a b d^2 (1-2 p) c^2+30 b d \left (2 b c^2-a d^2 (1-p)\right ) x c+a^2 d^4 \left (4 p^2-8 p+3\right )\right ) \left (b x^2+a\right )^pdx+\frac {d^2 (e x)^{1-2 p} \left (a+b x^2\right )^{p+1} \left (30 b c^2-a d^2 (3-2 p)\right )}{3 e}}{b}+\frac {5 c d^3 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{e^2}}{5 b}+\frac {d^4 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\) |
\(\Big \downarrow \) 557 |
\(\displaystyle \frac {\frac {\frac {1}{3} \left (\left (a^2 d^4 \left (4 p^2-8 p+3\right )-30 a b c^2 d^2 (1-2 p)+15 b^2 c^4\right ) \int (e x)^{-2 p} \left (b x^2+a\right )^pdx+\frac {30 b c d \left (2 b c^2-a d^2 (1-p)\right ) \int (e x)^{1-2 p} \left (b x^2+a\right )^pdx}{e}\right )+\frac {d^2 (e x)^{1-2 p} \left (a+b x^2\right )^{p+1} \left (30 b c^2-a d^2 (3-2 p)\right )}{3 e}}{b}+\frac {5 c d^3 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{e^2}}{5 b}+\frac {d^4 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\frac {\frac {1}{3} \left (\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a^2 d^4 \left (4 p^2-8 p+3\right )-30 a b c^2 d^2 (1-2 p)+15 b^2 c^4\right ) \int (e x)^{-2 p} \left (\frac {b x^2}{a}+1\right )^pdx+\frac {30 b c d \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 b c^2-a d^2 (1-p)\right ) \int (e x)^{1-2 p} \left (\frac {b x^2}{a}+1\right )^pdx}{e}\right )+\frac {d^2 (e x)^{1-2 p} \left (a+b x^2\right )^{p+1} \left (30 b c^2-a d^2 (3-2 p)\right )}{3 e}}{b}+\frac {5 c d^3 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{e^2}}{5 b}+\frac {d^4 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {(e x)^{1-2 p} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a^2 d^4 \left (4 p^2-8 p+3\right )-30 a b c^2 d^2 (1-2 p)+15 b^2 c^4\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1-2 p),-p,\frac {1}{2} (3-2 p),-\frac {b x^2}{a}\right )}{e (1-2 p)}+\frac {15 b c d (e x)^{2-2 p} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 b c^2-a d^2 (1-p)\right ) \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {b x^2}{a}\right )}{e^2 (1-p)}\right )+\frac {d^2 (e x)^{1-2 p} \left (a+b x^2\right )^{p+1} \left (30 b c^2-a d^2 (3-2 p)\right )}{3 e}}{b}+\frac {5 c d^3 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{e^2}}{5 b}+\frac {d^4 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\) |
Input:
Int[((c + d*x)^4*(a + b*x^2)^p)/(e*x)^(2*p),x]
Output:
(d^4*(e*x)^(3 - 2*p)*(a + b*x^2)^(1 + p))/(5*b*e^3) + ((5*c*d^3*(e*x)^(2 - 2*p)*(a + b*x^2)^(1 + p))/e^2 + ((d^2*(30*b*c^2 - a*d^2*(3 - 2*p))*(e*x)^ (1 - 2*p)*(a + b*x^2)^(1 + p))/(3*e) + (((15*b^2*c^4 - 30*a*b*c^2*d^2*(1 - 2*p) + a^2*d^4*(3 - 8*p + 4*p^2))*(e*x)^(1 - 2*p)*(a + b*x^2)^p*Hypergeom etric2F1[(1 - 2*p)/2, -p, (3 - 2*p)/2, -((b*x^2)/a)])/(e*(1 - 2*p)*(1 + (b *x^2)/a)^p) + (15*b*c*d*(2*b*c^2 - a*d^2*(1 - p))*(e*x)^(2 - 2*p)*(a + b*x ^2)^p*Hypergeometric2F1[1 - p, -p, 2 - p, -((b*x^2)/a)])/(e^2*(1 - p)*(1 + (b*x^2)/a)^p))/3)/b)/(5*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym bol] :> Simp[c Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e Int[(e*x)^( m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d^n*(e*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*e^(n - 1)*( m + n + 2*p + 1))), x] + Simp[1/(b*(m + n + 2*p + 1)) Int[(e*x)^m*(a + b* x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1 )*x^n - a*d^n*(m + n - 1)*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && IGtQ[n, 1] && !IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 )*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m + q + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) *Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ [Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
\[\int \left (d x +c \right )^{4} \left (b \,x^{2}+a \right )^{p} \left (e x \right )^{-2 p}d x\]
Input:
int((d*x+c)^4*(b*x^2+a)^p/((e*x)^(2*p)),x)
Output:
int((d*x+c)^4*(b*x^2+a)^p/((e*x)^(2*p)),x)
\[ \int (e x)^{-2 p} (c+d x)^4 \left (a+b x^2\right )^p \, dx=\int { \frac {{\left (d x + c\right )}^{4} {\left (b x^{2} + a\right )}^{p}}{\left (e x\right )^{2 \, p}} \,d x } \] Input:
integrate((d*x+c)^4*(b*x^2+a)^p/((e*x)^(2*p)),x, algorithm="fricas")
Output:
integral((d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4)*(b*x^2 + a)^p/(e*x)^(2*p), x)
Result contains complex when optimal does not.
Time = 119.08 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.80 \[ \int (e x)^{-2 p} (c+d x)^4 \left (a+b x^2\right )^p \, dx=\frac {2 a^{p} c^{3} d e^{- 2 p} x^{2 - 2 p} \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\Gamma \left (2 - p\right )} + \frac {3 a^{p} c^{2} d^{2} e^{- 2 p} x^{3 - 2 p} \Gamma \left (\frac {3}{2} - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {3}{2} - p \\ \frac {5}{2} - p \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\Gamma \left (\frac {5}{2} - p\right )} + \frac {2 a^{p} c d^{3} e^{- 2 p} x^{4 - 2 p} \Gamma \left (2 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 2 - p \\ 3 - p \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\Gamma \left (3 - p\right )} + \frac {a^{p} d^{4} e^{- 2 p} x^{5 - 2 p} \Gamma \left (\frac {5}{2} - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {5}{2} - p \\ \frac {7}{2} - p \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {7}{2} - p\right )} + \sqrt {b} b^{p - \frac {1}{2}} c^{4} e^{- 2 p} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )} \] Input:
integrate((d*x+c)**4*(b*x**2+a)**p/((e*x)**(2*p)),x)
Output:
2*a**p*c**3*d*x**(2 - 2*p)*gamma(1 - p)*hyper((-p, 1 - p), (2 - p,), b*x** 2*exp_polar(I*pi)/a)/(e**(2*p)*gamma(2 - p)) + 3*a**p*c**2*d**2*x**(3 - 2* p)*gamma(3/2 - p)*hyper((-p, 3/2 - p), (5/2 - p,), b*x**2*exp_polar(I*pi)/ a)/(e**(2*p)*gamma(5/2 - p)) + 2*a**p*c*d**3*x**(4 - 2*p)*gamma(2 - p)*hyp er((-p, 2 - p), (3 - p,), b*x**2*exp_polar(I*pi)/a)/(e**(2*p)*gamma(3 - p) ) + a**p*d**4*x**(5 - 2*p)*gamma(5/2 - p)*hyper((-p, 5/2 - p), (7/2 - p,), b*x**2*exp_polar(I*pi)/a)/(2*e**(2*p)*gamma(7/2 - p)) + sqrt(b)*b**(p - 1 /2)*c**4*x*hyper((-1/2, -p), (1/2,), a*exp_polar(I*pi)/(b*x**2))/e**(2*p)
\[ \int (e x)^{-2 p} (c+d x)^4 \left (a+b x^2\right )^p \, dx=\int { \frac {{\left (d x + c\right )}^{4} {\left (b x^{2} + a\right )}^{p}}{\left (e x\right )^{2 \, p}} \,d x } \] Input:
integrate((d*x+c)^4*(b*x^2+a)^p/((e*x)^(2*p)),x, algorithm="maxima")
Output:
integrate((d*x + c)^4*(b*x^2 + a)^p/(e*x)^(2*p), x)
\[ \int (e x)^{-2 p} (c+d x)^4 \left (a+b x^2\right )^p \, dx=\int { \frac {{\left (d x + c\right )}^{4} {\left (b x^{2} + a\right )}^{p}}{\left (e x\right )^{2 \, p}} \,d x } \] Input:
integrate((d*x+c)^4*(b*x^2+a)^p/((e*x)^(2*p)),x, algorithm="giac")
Output:
integrate((d*x + c)^4*(b*x^2 + a)^p/(e*x)^(2*p), x)
Timed out. \[ \int (e x)^{-2 p} (c+d x)^4 \left (a+b x^2\right )^p \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,{\left (c+d\,x\right )}^4}{{\left (e\,x\right )}^{2\,p}} \,d x \] Input:
int(((a + b*x^2)^p*(c + d*x)^4)/(e*x)^(2*p),x)
Output:
int(((a + b*x^2)^p*(c + d*x)^4)/(e*x)^(2*p), x)
\[ \int (e x)^{-2 p} (c+d x)^4 \left (a+b x^2\right )^p \, dx=\frac {4 \left (b \,x^{2}+a \right )^{p} a^{2} d^{4} p^{2} x -6 \left (b \,x^{2}+a \right )^{p} a^{2} d^{4} p x +60 \left (b \,x^{2}+a \right )^{p} a b \,c^{2} d^{2} p x +15 \left (b \,x^{2}+a \right )^{p} a b c \,d^{3} p \,x^{2}+2 \left (b \,x^{2}+a \right )^{p} a b \,d^{4} p \,x^{3}+15 \left (b \,x^{2}+a \right )^{p} b^{2} c^{4} x +30 \left (b \,x^{2}+a \right )^{p} b^{2} c^{3} d \,x^{2}+30 \left (b \,x^{2}+a \right )^{p} b^{2} c^{2} d^{2} x^{3}+15 \left (b \,x^{2}+a \right )^{p} b^{2} c \,d^{3} x^{4}+3 \left (b \,x^{2}+a \right )^{p} b^{2} d^{4} x^{5}+8 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{3} d^{4} p^{3}-16 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{3} d^{4} p^{2}+6 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{3} d^{4} p +120 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{2} b \,c^{2} d^{2} p^{2}-60 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{2} b \,c^{2} d^{2} p +30 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a \,b^{2} c^{4} p +30 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p} x}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{2} b c \,d^{3} p^{2}-30 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p} x}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{2} b c \,d^{3} p +60 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p} x}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a \,b^{2} c^{3} d p}{15 x^{2 p} e^{2 p} b^{2}} \] Input:
int((d*x+c)^4*(b*x^2+a)^p/((e*x)^(2*p)),x)
Output:
(4*(a + b*x**2)**p*a**2*d**4*p**2*x - 6*(a + b*x**2)**p*a**2*d**4*p*x + 60 *(a + b*x**2)**p*a*b*c**2*d**2*p*x + 15*(a + b*x**2)**p*a*b*c*d**3*p*x**2 + 2*(a + b*x**2)**p*a*b*d**4*p*x**3 + 15*(a + b*x**2)**p*b**2*c**4*x + 30* (a + b*x**2)**p*b**2*c**3*d*x**2 + 30*(a + b*x**2)**p*b**2*c**2*d**2*x**3 + 15*(a + b*x**2)**p*b**2*c*d**3*x**4 + 3*(a + b*x**2)**p*b**2*d**4*x**5 + 8*x**(2*p)*int((a + b*x**2)**p/(x**(2*p)*a + x**(2*p)*b*x**2),x)*a**3*d** 4*p**3 - 16*x**(2*p)*int((a + b*x**2)**p/(x**(2*p)*a + x**(2*p)*b*x**2),x) *a**3*d**4*p**2 + 6*x**(2*p)*int((a + b*x**2)**p/(x**(2*p)*a + x**(2*p)*b* x**2),x)*a**3*d**4*p + 120*x**(2*p)*int((a + b*x**2)**p/(x**(2*p)*a + x**( 2*p)*b*x**2),x)*a**2*b*c**2*d**2*p**2 - 60*x**(2*p)*int((a + b*x**2)**p/(x **(2*p)*a + x**(2*p)*b*x**2),x)*a**2*b*c**2*d**2*p + 30*x**(2*p)*int((a + b*x**2)**p/(x**(2*p)*a + x**(2*p)*b*x**2),x)*a*b**2*c**4*p + 30*x**(2*p)*i nt(((a + b*x**2)**p*x)/(x**(2*p)*a + x**(2*p)*b*x**2),x)*a**2*b*c*d**3*p** 2 - 30*x**(2*p)*int(((a + b*x**2)**p*x)/(x**(2*p)*a + x**(2*p)*b*x**2),x)* a**2*b*c*d**3*p + 60*x**(2*p)*int(((a + b*x**2)**p*x)/(x**(2*p)*a + x**(2* p)*b*x**2),x)*a*b**2*c**3*d*p)/(15*x**(2*p)*e**(2*p)*b**2)