Integrand size = 26, antiderivative size = 251 \[ \int (e x)^{1-2 p} (c+d x)^3 \left (a+b x^2\right )^p \, dx=\frac {3 c d^2 (e x)^{2-2 p} \left (a+b x^2\right )^{1+p}}{4 b e}+\frac {d^3 (e x)^{3-2 p} \left (a+b x^2\right )^{1+p}}{5 b e^2}-\frac {d \left (\frac {a d^2}{b}-\frac {15 c^2}{3-2 p}\right ) (e x)^{3-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (3-2 p),-p,\frac {1}{2} (5-2 p),-\frac {b x^2}{a}\right )}{5 e^2}+\frac {c \left (2 b c^2-3 a d^2 (1-p)\right ) (e x)^{2-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {b x^2}{a}\right )}{4 b e (1-p)} \] Output:
3/4*c*d^2*(e*x)^(2-2*p)*(b*x^2+a)^(p+1)/b/e+1/5*d^3*(e*x)^(3-2*p)*(b*x^2+a )^(p+1)/b/e^2-1/5*d*(a*d^2/b-15*c^2/(3-2*p))*(e*x)^(3-2*p)*(b*x^2+a)^p*hyp ergeom([-p, 3/2-p],[5/2-p],-b*x^2/a)/e^2/((1+b*x^2/a)^p)+1/4*c*(2*b*c^2-3* a*d^2*(1-p))*(e*x)^(2-2*p)*(b*x^2+a)^p*hypergeom([-p, 1-p],[2-p],-b*x^2/a) /b/e/(1-p)/((1+b*x^2/a)^p)
Time = 0.15 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.75 \[ \int (e x)^{1-2 p} (c+d x)^3 \left (a+b x^2\right )^p \, dx=\frac {1}{2} e x^2 (e x)^{-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (\frac {c^3 \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {b x^2}{a}\right )}{1-p}+d x \left (\frac {6 c^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-p,-p,\frac {5}{2}-p,-\frac {b x^2}{a}\right )}{3-2 p}+d x \left (-\frac {3 c \operatorname {Hypergeometric2F1}\left (2-p,-p,3-p,-\frac {b x^2}{a}\right )}{-2+p}+\frac {2 d x \operatorname {Hypergeometric2F1}\left (\frac {5}{2}-p,-p,\frac {7}{2}-p,-\frac {b x^2}{a}\right )}{5-2 p}\right )\right )\right ) \] Input:
Integrate[(e*x)^(1 - 2*p)*(c + d*x)^3*(a + b*x^2)^p,x]
Output:
(e*x^2*(a + b*x^2)^p*((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*((-3*c*Hypergeometric2F1[2 - p, -p, 3 - p, -((b*x^2 )/a)])/(-2 + p) + (2*d*x*Hypergeometric2F1[5/2 - p, -p, 7/2 - p, -((b*x^2) /a)])/(5 - 2*p)))))/(2*(e*x)^(2*p)*(1 + (b*x^2)/a)^p)
Time = 0.51 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {559, 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.
\(\displaystyle \int (c+d x)^3 (e x)^{1-2 p} \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 559 |
\(\displaystyle \frac {\int (e x)^{1-2 p} \left (b x^2+a\right )^p \left (5 b c^3+15 b d^2 x^2 c+d \left (15 b c^2-a d^2 (3-2 p)\right ) x\right )dx}{5 b}+\frac {d^3 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^2}\) |
\(\Big \downarrow \) 2340 |
\(\displaystyle \frac {\frac {\int 2 b (e x)^{1-2 p} \left (5 c \left (2 b c^2-3 a d^2 (1-p)\right )+2 d \left (15 b c^2-a d^2 (3-2 p)\right ) x\right ) \left (b x^2+a\right )^pdx}{4 b}+\frac {15 c d^2 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{4 e}}{5 b}+\frac {d^3 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{2} \int (e x)^{1-2 p} \left (5 c \left (2 b c^2-3 a d^2 (1-p)\right )+2 d \left (15 b c^2-a d^2 (3-2 p)\right ) x\right ) \left (b x^2+a\right )^pdx+\frac {15 c d^2 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{4 e}}{5 b}+\frac {d^3 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^2}\) |
\(\Big \downarrow \) 557 |
\(\displaystyle \frac {\frac {1}{2} \left (5 c \left (2 b c^2-3 a d^2 (1-p)\right ) \int (e x)^{1-2 p} \left (b x^2+a\right )^pdx+\frac {2 d \left (15 b c^2-a d^2 (3-2 p)\right ) \int (e x)^{2-2 p} \left (b x^2+a\right )^pdx}{e}\right )+\frac {15 c d^2 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{4 e}}{5 b}+\frac {d^3 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^2}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\frac {1}{2} \left (5 c \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 b c^2-3 a d^2 (1-p)\right ) \int (e x)^{1-2 p} \left (\frac {b x^2}{a}+1\right )^pdx+\frac {2 d \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (15 b c^2-a d^2 (3-2 p)\right ) \int (e x)^{2-2 p} \left (\frac {b x^2}{a}+1\right )^pdx}{e}\right )+\frac {15 c d^2 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{4 e}}{5 b}+\frac {d^3 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^2}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 d (e x)^{3-2 p} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (15 b c^2-a d^2 (3-2 p)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (3-2 p),-p,\frac {1}{2} (5-2 p),-\frac {b x^2}{a}\right )}{e^2 (3-2 p)}+\frac {5 c (e x)^{2-2 p} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 b c^2-3 a d^2 (1-p)\right ) \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {b x^2}{a}\right )}{2 e (1-p)}\right )+\frac {15 c d^2 (e x)^{2-2 p} \left (a+b x^2\right )^{p+1}}{4 e}}{5 b}+\frac {d^3 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^2}\) |
Input:
Int[(e*x)^(1 - 2*p)*(c + d*x)^3*(a + b*x^2)^p,x]
Output:
(d^3*(e*x)^(3 - 2*p)*(a + b*x^2)^(1 + p))/(5*b*e^2) + ((15*c*d^2*(e*x)^(2 - 2*p)*(a + b*x^2)^(1 + p))/(4*e) + ((2*d*(15*b*c^2 - a*d^2*(3 - 2*p))*(e* x)^(3 - 2*p)*(a + b*x^2)^p*Hypergeometric2F1[(3 - 2*p)/2, -p, (5 - 2*p)/2, -((b*x^2)/a)])/(e^2*(3 - 2*p)*(1 + (b*x^2)/a)^p) + (5*c*(2*b*c^2 - 3*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)])/(2*e*(1 - p)*(1 + (b*x^2)/a)^p))/2)/(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])
\[\int \left (e x \right )^{1-2 p} \left (d x +c \right )^{3} \left (b \,x^{2}+a \right )^{p}d x\]
Input:
int((e*x)^(1-2*p)*(d*x+c)^3*(b*x^2+a)^p,x)
Output:
int((e*x)^(1-2*p)*(d*x+c)^3*(b*x^2+a)^p,x)
\[ \int (e x)^{1-2 p} (c+d x)^3 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{3} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p + 1} \,d x } \] Input:
integrate((e*x)^(1-2*p)*(d*x+c)^3*(b*x^2+a)^p,x, algorithm="fricas")
Output:
integral((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*(b*x^2 + a)^p*(e*x)^(-2 *p + 1), x)
Result contains complex when optimal does not.
Time = 98.26 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.89 \[ \int (e x)^{1-2 p} (c+d x)^3 \left (a+b x^2\right )^p \, dx=\frac {a^{p} c^{3} e^{1 - 2 p} x^{2 - 2 p} \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (2 - p\right )} + \frac {3 a^{p} c^{2} d e^{1 - 2 p} x^{3 - 2 p} \Gamma \left (\frac {3}{2} - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {3}{2} - p \\ \frac {5}{2} - p \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {5}{2} - p\right )} + \frac {3 a^{p} c d^{2} e^{1 - 2 p} x^{4 - 2 p} \Gamma \left (2 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 2 - p \\ 3 - p \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (3 - p\right )} + \frac {a^{p} d^{3} e^{1 - 2 p} x^{5 - 2 p} \Gamma \left (\frac {5}{2} - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {5}{2} - p \\ \frac {7}{2} - p \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {7}{2} - p\right )} \] Input:
integrate((e*x)**(1-2*p)*(d*x+c)**3*(b*x**2+a)**p,x)
Output:
a**p*c**3*e**(1 - 2*p)*x**(2 - 2*p)*gamma(1 - p)*hyper((-p, 1 - p), (2 - p ,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(2 - p)) + 3*a**p*c**2*d*e**(1 - 2*p )*x**(3 - 2*p)*gamma(3/2 - p)*hyper((-p, 3/2 - p), (5/2 - p,), b*x**2*exp_ polar(I*pi)/a)/(2*gamma(5/2 - p)) + 3*a**p*c*d**2*e**(1 - 2*p)*x**(4 - 2*p )*gamma(2 - p)*hyper((-p, 2 - p), (3 - p,), b*x**2*exp_polar(I*pi)/a)/(2*g amma(3 - p)) + a**p*d**3*e**(1 - 2*p)*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*gamma(7/2 - p))
\[ \int (e x)^{1-2 p} (c+d x)^3 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{3} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p + 1} \,d x } \] Input:
integrate((e*x)^(1-2*p)*(d*x+c)^3*(b*x^2+a)^p,x, algorithm="maxima")
Output:
integrate((d*x + c)^3*(b*x^2 + a)^p*(e*x)^(-2*p + 1), x)
\[ \int (e x)^{1-2 p} (c+d x)^3 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{3} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p + 1} \,d x } \] Input:
integrate((e*x)^(1-2*p)*(d*x+c)^3*(b*x^2+a)^p,x, algorithm="giac")
Output:
integrate((d*x + c)^3*(b*x^2 + a)^p*(e*x)^(-2*p + 1), x)
Timed out. \[ \int (e x)^{1-2 p} (c+d x)^3 \left (a+b x^2\right )^p \, dx=\int {\left (e\,x\right )}^{1-2\,p}\,{\left (b\,x^2+a\right )}^p\,{\left (c+d\,x\right )}^3 \,d x \] Input:
int((e*x)^(1 - 2*p)*(a + b*x^2)^p*(c + d*x)^3,x)
Output:
int((e*x)^(1 - 2*p)*(a + b*x^2)^p*(c + d*x)^3, x)
\[ \int (e x)^{1-2 p} (c+d x)^3 \left (a+b x^2\right )^p \, dx=\frac {e \left (16 \left (b \,x^{2}+a \right )^{p} a^{2} d^{3} p^{2} x -24 \left (b \,x^{2}+a \right )^{p} a^{2} d^{3} p x +120 \left (b \,x^{2}+a \right )^{p} a b \,c^{2} d p x +45 \left (b \,x^{2}+a \right )^{p} a b c \,d^{2} p \,x^{2}+8 \left (b \,x^{2}+a \right )^{p} a b \,d^{3} p \,x^{3}+30 \left (b \,x^{2}+a \right )^{p} b^{2} c^{3} x^{2}+60 \left (b \,x^{2}+a \right )^{p} b^{2} c^{2} d \,x^{3}+45 \left (b \,x^{2}+a \right )^{p} b^{2} c \,d^{2} x^{4}+12 \left (b \,x^{2}+a \right )^{p} b^{2} d^{3} x^{5}+32 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{3} d^{3} p^{3}-64 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{3} d^{3} p^{2}+24 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{3} d^{3} p +240 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{2} b \,c^{2} d \,p^{2}-120 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{2} b \,c^{2} d p +90 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p} x}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{2} b c \,d^{2} p^{2}-90 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p} x}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{2} b c \,d^{2} p +60 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p} x}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a \,b^{2} c^{3} p \right )}{60 x^{2 p} e^{2 p} b^{2}} \] Input:
int((e*x)^(1-2*p)*(d*x+c)^3*(b*x^2+a)^p,x)
Output:
(e*(16*(a + b*x**2)**p*a**2*d**3*p**2*x - 24*(a + b*x**2)**p*a**2*d**3*p*x + 120*(a + b*x**2)**p*a*b*c**2*d*p*x + 45*(a + b*x**2)**p*a*b*c*d**2*p*x* *2 + 8*(a + b*x**2)**p*a*b*d**3*p*x**3 + 30*(a + b*x**2)**p*b**2*c**3*x**2 + 60*(a + b*x**2)**p*b**2*c**2*d*x**3 + 45*(a + b*x**2)**p*b**2*c*d**2*x* *4 + 12*(a + b*x**2)**p*b**2*d**3*x**5 + 32*x**(2*p)*int((a + b*x**2)**p/( x**(2*p)*a + x**(2*p)*b*x**2),x)*a**3*d**3*p**3 - 64*x**(2*p)*int((a + b*x **2)**p/(x**(2*p)*a + x**(2*p)*b*x**2),x)*a**3*d**3*p**2 + 24*x**(2*p)*int ((a + b*x**2)**p/(x**(2*p)*a + x**(2*p)*b*x**2),x)*a**3*d**3*p + 240*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*p* *2 - 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*p + 90*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**2*p**2 - 90*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**2*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*p))/(60*x**(2*p)*e* *(2*p)*b**2)