Integrand size = 18, antiderivative size = 108 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^4} \, dx=-\frac {a (c+d x)^{1+n}}{3 c x^3}-\frac {b (c+d x)^{1+n}}{d (1-n) x^2}+\frac {d \left (6 b c^2+a d^2 \left (2-3 n+n^2\right )\right ) (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (3,1+n,2+n,1+\frac {d x}{c}\right )}{3 c^4 \left (1-n^2\right )} \] Output:
-1/3*a*(d*x+c)^(1+n)/c/x^3-b*(d*x+c)^(1+n)/d/(1-n)/x^2+1/3*d*(6*b*c^2+a*d^ 2*(n^2-3*n+2))*(d*x+c)^(1+n)*hypergeom([3, 1+n],[2+n],1+d*x/c)/c^4/(-n^2+1 )
Time = 0.01 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.57 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^4} \, dx=\frac {d (c+d x)^{1+n} \left (b c^2 \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {d x}{c}\right )+a d^2 \operatorname {Hypergeometric2F1}\left (4,1+n,2+n,1+\frac {d x}{c}\right )\right )}{c^4 (1+n)} \] Input:
Integrate[((c + d*x)^n*(a + b*x^2))/x^4,x]
Output:
(d*(c + d*x)^(1 + n)*(b*c^2*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (d*x)/c ] + a*d^2*Hypergeometric2F1[4, 1 + n, 2 + n, 1 + (d*x)/c]))/(c^4*(1 + n))
Time = 0.23 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {520, 87, 75}
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.
| \(\displaystyle \int \frac {\left (a+b x^2\right ) (c+d x)^n}{x^4} \, dx\) |
| \(\Big \downarrow \) 520 |
| \(\displaystyle -\frac {\int \frac {(a d (2-n)-3 b c x) (c+d x)^n}{x^3}dx}{3 c}-\frac {a (c+d x)^{n+1}}{3 c x^3}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle -\frac {-\frac {\left (a d^2 (1-n) (2-n)+6 b c^2\right ) \int \frac {(c+d x)^n}{x^2}dx}{2 c}-\frac {a d (2-n) (c+d x)^{n+1}}{2 c x^2}}{3 c}-\frac {a (c+d x)^{n+1}}{3 c x^3}\) |
| \(\Big \downarrow \) 75 |
| \(\displaystyle -\frac {-\frac {d (c+d x)^{n+1} \left (a d^2 (1-n) (2-n)+6 b c^2\right ) \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {d x}{c}+1\right )}{2 c^3 (n+1)}-\frac {a d (2-n) (c+d x)^{n+1}}{2 c x^2}}{3 c}-\frac {a (c+d x)^{n+1}}{3 c x^3}\) |
Input:
Int[((c + d*x)^n*(a + b*x^2))/x^4,x]
Output:
-1/3*(a*(c + d*x)^(1 + n))/(c*x^3) - (-1/2*(a*d*(2 - n)*(c + d*x)^(1 + n)) /(c*x^2) - (d*(6*b*c^2 + a*d^2*(1 - n)*(2 - n))*(c + d*x)^(1 + n)*Hypergeo metric2F1[2, 1 + n, 2 + n, 1 + (d*x)/c])/(2*c^3*(1 + n)))/(3*c)
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, e*x, x], R = Pol ynomialRemainder[(a + b*x^2)^p, e*x, x]}, Simp[R*(e*x)^(m + 1)*((c + d*x)^( n + 1)/((m + 1)*(e*c))), x] + Simp[1/((m + 1)*(e*c)) Int[(e*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(e*c)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr eeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[m, -1] && !IntegerQ[n]
\[\int \frac {\left (d x +c \right )^{n} \left (b \,x^{2}+a \right )}{x^{4}}d x\]
Input:
int((d*x+c)^n*(b*x^2+a)/x^4,x)
Output:
int((d*x+c)^n*(b*x^2+a)/x^4,x)
\[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )} {\left (d x + c\right )}^{n}}{x^{4}} \,d x } \] Input:
integrate((d*x+c)^n*(b*x^2+a)/x^4,x, algorithm="fricas")
Output:
integral((b*x^2 + a)*(d*x + c)^n/x^4, x)
Leaf count of result is larger than twice the leaf count of optimal. 2926 vs. \(2 (87) = 174\).
Time = 11.67 (sec) , antiderivative size = 2926, normalized size of antiderivative = 27.09 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^4} \, dx=\text {Too large to display} \] Input:
integrate((d*x+c)**n*(b*x**2+a)/x**4,x)
Output:
-2*a*c**3*d**(n + 4)*n**4*(c/d + x)**(n + 1)*lerchphi(1 + d*x/c, 1, n + 1) *gamma(n + 1)/(12*c**7*gamma(n + 2) + 18*c**6*d*x*gamma(n + 2) - 18*c**5*d **2*(c/d + x)**2*gamma(n + 2) + 6*c**4*d**3*(c/d + x)**3*gamma(n + 2)) + 4 *a*c**3*d**(n + 4)*n**3*(c/d + x)**(n + 1)*lerchphi(1 + d*x/c, 1, n + 1)*g amma(n + 1)/(12*c**7*gamma(n + 2) + 18*c**6*d*x*gamma(n + 2) - 18*c**5*d** 2*(c/d + x)**2*gamma(n + 2) + 6*c**4*d**3*(c/d + x)**3*gamma(n + 2)) + a*c **3*d**(n + 4)*n**3*(c/d + x)**(n + 1)*gamma(n + 1)/(12*c**7*gamma(n + 2) + 18*c**6*d*x*gamma(n + 2) - 18*c**5*d**2*(c/d + x)**2*gamma(n + 2) + 6*c* *4*d**3*(c/d + x)**3*gamma(n + 2)) + 2*a*c**3*d**(n + 4)*n**2*(c/d + x)**( n + 1)*lerchphi(1 + d*x/c, 1, n + 1)*gamma(n + 1)/(12*c**7*gamma(n + 2) + 18*c**6*d*x*gamma(n + 2) - 18*c**5*d**2*(c/d + x)**2*gamma(n + 2) + 6*c**4 *d**3*(c/d + x)**3*gamma(n + 2)) - 2*a*c**3*d**(n + 4)*n**2*(c/d + x)**(n + 1)*gamma(n + 1)/(12*c**7*gamma(n + 2) + 18*c**6*d*x*gamma(n + 2) - 18*c* *5*d**2*(c/d + x)**2*gamma(n + 2) + 6*c**4*d**3*(c/d + x)**3*gamma(n + 2)) - 4*a*c**3*d**(n + 4)*n*(c/d + x)**(n + 1)*lerchphi(1 + d*x/c, 1, n + 1)* gamma(n + 1)/(12*c**7*gamma(n + 2) + 18*c**6*d*x*gamma(n + 2) - 18*c**5*d* *2*(c/d + x)**2*gamma(n + 2) + 6*c**4*d**3*(c/d + x)**3*gamma(n + 2)) - 3* a*c**3*d**(n + 4)*n*(c/d + x)**(n + 1)*gamma(n + 1)/(12*c**7*gamma(n + 2) + 18*c**6*d*x*gamma(n + 2) - 18*c**5*d**2*(c/d + x)**2*gamma(n + 2) + 6*c* *4*d**3*(c/d + x)**3*gamma(n + 2)) - 3*a*c**2*d*d**(n + 4)*n**4*x*(c/d ...
\[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )} {\left (d x + c\right )}^{n}}{x^{4}} \,d x } \] Input:
integrate((d*x+c)^n*(b*x^2+a)/x^4,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)*(d*x + c)^n/x^4, x)
\[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )} {\left (d x + c\right )}^{n}}{x^{4}} \,d x } \] Input:
integrate((d*x+c)^n*(b*x^2+a)/x^4,x, algorithm="giac")
Output:
integrate((b*x^2 + a)*(d*x + c)^n/x^4, x)
Timed out. \[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^4} \, dx=\int \frac {\left (b\,x^2+a\right )\,{\left (c+d\,x\right )}^n}{x^4} \,d x \] Input:
int(((a + b*x^2)*(c + d*x)^n)/x^4,x)
Output:
int(((a + b*x^2)*(c + d*x)^n)/x^4, x)
\[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^4} \, dx=\frac {-2 \left (d x +c \right )^{n} a \,c^{2}-\left (d x +c \right )^{n} a c d n x -\left (d x +c \right )^{n} a \,d^{2} n^{2} x^{2}+2 \left (d x +c \right )^{n} a \,d^{2} n \,x^{2}-6 \left (d x +c \right )^{n} b \,c^{2} x^{2}+\left (\int \frac {\left (d x +c \right )^{n}}{d \,x^{2}+c x}d x \right ) a \,d^{3} n^{3} x^{3}-3 \left (\int \frac {\left (d x +c \right )^{n}}{d \,x^{2}+c x}d x \right ) a \,d^{3} n^{2} x^{3}+2 \left (\int \frac {\left (d x +c \right )^{n}}{d \,x^{2}+c x}d x \right ) a \,d^{3} n \,x^{3}+6 \left (\int \frac {\left (d x +c \right )^{n}}{d \,x^{2}+c x}d x \right ) b \,c^{2} d n \,x^{3}}{6 c^{2} x^{3}} \] Input:
int((d*x+c)^n*(b*x^2+a)/x^4,x)
Output:
( - 2*(c + d*x)**n*a*c**2 - (c + d*x)**n*a*c*d*n*x - (c + d*x)**n*a*d**2*n **2*x**2 + 2*(c + d*x)**n*a*d**2*n*x**2 - 6*(c + d*x)**n*b*c**2*x**2 + int ((c + d*x)**n/(c*x + d*x**2),x)*a*d**3*n**3*x**3 - 3*int((c + d*x)**n/(c*x + d*x**2),x)*a*d**3*n**2*x**3 + 2*int((c + d*x)**n/(c*x + d*x**2),x)*a*d* *3*n*x**3 + 6*int((c + d*x)**n/(c*x + d*x**2),x)*b*c**2*d*n*x**3)/(6*c**2* x**3)