Integrand size = 26, antiderivative size = 387 \[ \int (d+e x)^{-4-2 p} (e+f x) \left (a+c x^2\right )^p \, dx=-\frac {\left (e^2-d f\right ) (d+e x)^{-3-2 p} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right ) (3+2 p)}-\frac {\left (a e^2 f (3+2 p)-c d \left (d f-2 e^2 (2+p)\right )\right ) (d+e x)^{-2 (1+p)} \left (a+c x^2\right )^{1+p}}{2 \left (c d^2+a e^2\right )^2 (1+p) (3+2 p)}+\frac {c e \left (a e^2-2 a d f (2+p)-c d^2 (3+2 p)\right ) \left (\sqrt {-a}-\sqrt {c} x\right ) \left (-\frac {\left (\sqrt {c} d+\sqrt {-a} e\right ) \left (\sqrt {-a}+\sqrt {c} x\right )}{\left (\sqrt {c} d-\sqrt {-a} e\right ) \left (\sqrt {-a}-\sqrt {c} x\right )}\right )^{-p} (d+e x)^{-1-2 p} \left (a+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,\frac {2 \sqrt {-a} \sqrt {c} (d+e x)}{\left (\sqrt {c} d-\sqrt {-a} e\right ) \left (\sqrt {-a}-\sqrt {c} x\right )}\right )}{\left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right )^2 (1+2 p) (3+2 p)} \] Output:
-(-d*f+e^2)*(e*x+d)^(-3-2*p)*(c*x^2+a)^(p+1)/(a*e^2+c*d^2)/(3+2*p)-1/2*(a* e^2*f*(3+2*p)-c*d*(d*f-2*e^2*(2+p)))*(c*x^2+a)^(p+1)/(a*e^2+c*d^2)^2/(p+1) /(3+2*p)/((e*x+d)^(2*p+2))+c*e*(a*e^2-2*a*d*f*(2+p)-c*d^2*(3+2*p))*((-a)^( 1/2)-c^(1/2)*x)*(e*x+d)^(-1-2*p)*(c*x^2+a)^p*hypergeom([-p, -1-2*p],[-2*p] ,2*(-a)^(1/2)*c^(1/2)*(e*x+d)/(c^(1/2)*d-(-a)^(1/2)*e)/((-a)^(1/2)-c^(1/2) *x))/(c^(1/2)*d+(-a)^(1/2)*e)/(a*e^2+c*d^2)^2/(1+2*p)/(3+2*p)/((-(c^(1/2)* d+(-a)^(1/2)*e)*((-a)^(1/2)+c^(1/2)*x)/(c^(1/2)*d-(-a)^(1/2)*e)/((-a)^(1/2 )-c^(1/2)*x))^p)
\[ \int (d+e x)^{-4-2 p} (e+f x) \left (a+c x^2\right )^p \, dx=\int (d+e x)^{-4-2 p} (e+f x) \left (a+c x^2\right )^p \, dx \] Input:
Integrate[(d + e*x)^(-4 - 2*p)*(e + f*x)*(a + c*x^2)^p,x]
Output:
Integrate[(d + e*x)^(-4 - 2*p)*(e + f*x)*(a + c*x^2)^p, x]
Time = 0.54 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {689, 25, 679, 489}
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.
\(\displaystyle \int (e+f x) \left (a+c x^2\right )^p (d+e x)^{-2 p-4} \, dx\) |
\(\Big \downarrow \) 689 |
\(\displaystyle -\frac {\int -(d+e x)^{-2 p-3} \left (e (c d+a f) (2 p+3)-c \left (e^2-d f\right ) x\right ) \left (c x^2+a\right )^pdx}{(2 p+3) \left (a e^2+c d^2\right )}-\frac {\left (e^2-d f\right ) \left (a+c x^2\right )^{p+1} (d+e x)^{-2 p-3}}{(2 p+3) \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int (d+e x)^{-2 p-3} \left (e (c d+a f) (2 p+3)-c \left (e^2-d f\right ) x\right ) \left (c x^2+a\right )^pdx}{(2 p+3) \left (a e^2+c d^2\right )}-\frac {\left (e^2-d f\right ) \left (a+c x^2\right )^{p+1} (d+e x)^{-2 p-3}}{(2 p+3) \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 679 |
\(\displaystyle \frac {-\frac {c e \left (-2 a d f (p+2)+a e^2-c d^2 (2 p+3)\right ) \int (d+e x)^{-2 (p+1)} \left (c x^2+a\right )^pdx}{a e^2+c d^2}-\frac {\left (a+c x^2\right )^{p+1} (d+e x)^{-2 (p+1)} \left (a e^2 f (2 p+3)-c d \left (d f-2 e^2 (p+2)\right )\right )}{2 (p+1) \left (a e^2+c d^2\right )}}{(2 p+3) \left (a e^2+c d^2\right )}-\frac {\left (e^2-d f\right ) \left (a+c x^2\right )^{p+1} (d+e x)^{-2 p-3}}{(2 p+3) \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 489 |
\(\displaystyle \frac {\frac {c e \left (\sqrt {-a}-\sqrt {c} x\right ) \left (a+c x^2\right )^p (d+e x)^{-2 p-1} \left (-2 a d f (p+2)+a e^2-c d^2 (2 p+3)\right ) \left (-\frac {\left (\sqrt {-a}+\sqrt {c} x\right ) \left (\sqrt {-a} e+\sqrt {c} d\right )}{\left (\sqrt {-a}-\sqrt {c} x\right ) \left (\sqrt {c} d-\sqrt {-a} e\right )}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-2 p-1,-p,-2 p,\frac {2 \sqrt {-a} \sqrt {c} (d+e x)}{\left (\sqrt {c} d-\sqrt {-a} e\right ) \left (\sqrt {-a}-\sqrt {c} x\right )}\right )}{(2 p+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}-\frac {\left (a+c x^2\right )^{p+1} (d+e x)^{-2 (p+1)} \left (a e^2 f (2 p+3)-c d \left (d f-2 e^2 (p+2)\right )\right )}{2 (p+1) \left (a e^2+c d^2\right )}}{(2 p+3) \left (a e^2+c d^2\right )}-\frac {\left (e^2-d f\right ) \left (a+c x^2\right )^{p+1} (d+e x)^{-2 p-3}}{(2 p+3) \left (a e^2+c d^2\right )}\) |
Input:
Int[(d + e*x)^(-4 - 2*p)*(e + f*x)*(a + c*x^2)^p,x]
Output:
-(((e^2 - d*f)*(d + e*x)^(-3 - 2*p)*(a + c*x^2)^(1 + p))/((c*d^2 + a*e^2)* (3 + 2*p))) + (-1/2*((a*e^2*f*(3 + 2*p) - c*d*(d*f - 2*e^2*(2 + p)))*(a + c*x^2)^(1 + p))/((c*d^2 + a*e^2)*(1 + p)*(d + e*x)^(2*(1 + p))) + (c*e*(a* e^2 - 2*a*d*f*(2 + p) - c*d^2*(3 + 2*p))*(Sqrt[-a] - Sqrt[c]*x)*(d + e*x)^ (-1 - 2*p)*(a + c*x^2)^p*Hypergeometric2F1[-1 - 2*p, -p, -2*p, (2*Sqrt[-a] *Sqrt[c]*(d + e*x))/((Sqrt[c]*d - Sqrt[-a]*e)*(Sqrt[-a] - Sqrt[c]*x))])/(( Sqrt[c]*d + Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 + 2*p)*(-(((Sqrt[c]*d + Sqrt[-a ]*e)*(Sqrt[-a] + Sqrt[c]*x))/((Sqrt[c]*d - Sqrt[-a]*e)*(Sqrt[-a] - Sqrt[c] *x))))^p))/((c*d^2 + a*e^2)*(3 + 2*p))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {q = Rt[(-a)*b, 2]}, Simp[(q - b*x)*(c + d*x)^(n + 1)*((a + b*x^2)^p/((n + 1)*(b*c + d*q)*((b*c + d*q)*((q + b*x)/((b*c - d*q)*(-q + b*x))))^p))*Hyper geometric2F1[n + 1, -p, n + 2, 2*b*q*((c + d*x)/((b*c - d*q)*(q - b*x)))], x]] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[n + 2*p + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 )/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/( (m + 1)*(c*d^2 + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 + a*e^2)) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && ILtQ[Sim plify[m + 2*p + 3], 0] && NeQ[m, -1]
\[\int \left (e x +d \right )^{-4-2 p} \left (f x +e \right ) \left (c \,x^{2}+a \right )^{p}d x\]
Input:
int((e*x+d)^(-4-2*p)*(f*x+e)*(c*x^2+a)^p,x)
Output:
int((e*x+d)^(-4-2*p)*(f*x+e)*(c*x^2+a)^p,x)
\[ \int (d+e x)^{-4-2 p} (e+f x) \left (a+c x^2\right )^p \, dx=\int { {\left (f x + e\right )} {\left (c x^{2} + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 4} \,d x } \] Input:
integrate((e*x+d)^(-4-2*p)*(f*x+e)*(c*x^2+a)^p,x, algorithm="fricas")
Output:
integral((f*x + e)*(c*x^2 + a)^p*(e*x + d)^(-2*p - 4), x)
Timed out. \[ \int (d+e x)^{-4-2 p} (e+f x) \left (a+c x^2\right )^p \, dx=\text {Timed out} \] Input:
integrate((e*x+d)**(-4-2*p)*(f*x+e)*(c*x**2+a)**p,x)
Output:
Timed out
\[ \int (d+e x)^{-4-2 p} (e+f x) \left (a+c x^2\right )^p \, dx=\int { {\left (f x + e\right )} {\left (c x^{2} + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 4} \,d x } \] Input:
integrate((e*x+d)^(-4-2*p)*(f*x+e)*(c*x^2+a)^p,x, algorithm="maxima")
Output:
integrate((f*x + e)*(c*x^2 + a)^p*(e*x + d)^(-2*p - 4), x)
\[ \int (d+e x)^{-4-2 p} (e+f x) \left (a+c x^2\right )^p \, dx=\int { {\left (f x + e\right )} {\left (c x^{2} + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 4} \,d x } \] Input:
integrate((e*x+d)^(-4-2*p)*(f*x+e)*(c*x^2+a)^p,x, algorithm="giac")
Output:
integrate((f*x + e)*(c*x^2 + a)^p*(e*x + d)^(-2*p - 4), x)
Timed out. \[ \int (d+e x)^{-4-2 p} (e+f x) \left (a+c x^2\right )^p \, dx=\int \frac {\left (e+f\,x\right )\,{\left (c\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p+4}} \,d x \] Input:
int(((e + f*x)*(a + c*x^2)^p)/(d + e*x)^(2*p + 4),x)
Output:
int(((e + f*x)*(a + c*x^2)^p)/(d + e*x)^(2*p + 4), x)
\[ \int (d+e x)^{-4-2 p} (e+f x) \left (a+c x^2\right )^p \, dx=\left (\int \frac {\left (c \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{2 p} d^{4}+4 \left (e x +d \right )^{2 p} d^{3} e x +6 \left (e x +d \right )^{2 p} d^{2} e^{2} x^{2}+4 \left (e x +d \right )^{2 p} d \,e^{3} x^{3}+\left (e x +d \right )^{2 p} e^{4} x^{4}}d x \right ) e +\left (\int \frac {\left (c \,x^{2}+a \right )^{p} x}{\left (e x +d \right )^{2 p} d^{4}+4 \left (e x +d \right )^{2 p} d^{3} e x +6 \left (e x +d \right )^{2 p} d^{2} e^{2} x^{2}+4 \left (e x +d \right )^{2 p} d \,e^{3} x^{3}+\left (e x +d \right )^{2 p} e^{4} x^{4}}d x \right ) f \] Input:
int((e*x+d)^(-4-2*p)*(f*x+e)*(c*x^2+a)^p,x)
Output:
int((a + c*x**2)**p/((d + e*x)**(2*p)*d**4 + 4*(d + e*x)**(2*p)*d**3*e*x + 6*(d + e*x)**(2*p)*d**2*e**2*x**2 + 4*(d + e*x)**(2*p)*d*e**3*x**3 + (d + e*x)**(2*p)*e**4*x**4),x)*e + int(((a + c*x**2)**p*x)/((d + e*x)**(2*p)*d **4 + 4*(d + e*x)**(2*p)*d**3*e*x + 6*(d + e*x)**(2*p)*d**2*e**2*x**2 + 4* (d + e*x)**(2*p)*d*e**3*x**3 + (d + e*x)**(2*p)*e**4*x**4),x)*f