\(\int \frac {(c+d x) (a+b x^2)^p}{x^3} \, dx\) [9]

Optimal result

Integrand size = 18, antiderivative size = 92 \[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{x^3} \, dx=-\frac {d \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}+\frac {b c \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a^2 (1+p)} \] Output:

-d*(b*x^2+a)^p*hypergeom([-1/2, -p],[1/2],-b*x^2/a)/x/((1+b*x^2/a)^p)+1/2* 
b*c*(b*x^2+a)^(p+1)*hypergeom([2, p+1],[2+p],1+b*x^2/a)/a^2/(p+1)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.97 \[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{x^3} \, dx=\frac {1}{2} \left (a+b x^2\right )^p \left (-\frac {2 d \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}+\frac {b c \left (a+b x^2\right ) \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,1+\frac {b x^2}{a}\right )}{a^2 (1+p)}\right ) \] Input:

Integrate[((c + d*x)*(a + b*x^2)^p)/x^3,x]
 

Output:

((a + b*x^2)^p*((-2*d*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)])/(x*( 
1 + (b*x^2)/a)^p) + (b*c*(a + b*x^2)*Hypergeometric2F1[2, 1 + p, 2 + p, 1 
+ (b*x^2)/a])/(a^2*(1 + p))))/2
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {542, 243, 75, 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.

\(\displaystyle \int \frac {(c+d x) \left (a+b x^2\right )^p}{x^3} \, dx\)

\(\Big \downarrow \) 542

\(\displaystyle c \int \frac {\left (b x^2+a\right )^p}{x^3}dx+d \int \frac {\left (b x^2+a\right )^p}{x^2}dx\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{2} c \int \frac {\left (b x^2+a\right )^p}{x^4}dx^2+d \int \frac {\left (b x^2+a\right )^p}{x^2}dx\)

\(\Big \downarrow \) 75

\(\displaystyle d \int \frac {\left (b x^2+a\right )^p}{x^2}dx+\frac {b c \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (2,p+1,p+2,\frac {b x^2}{a}+1\right )}{2 a^2 (p+1)}\)

\(\Big \downarrow \) 279

\(\displaystyle d \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \frac {\left (\frac {b x^2}{a}+1\right )^p}{x^2}dx+\frac {b c \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (2,p+1,p+2,\frac {b x^2}{a}+1\right )}{2 a^2 (p+1)}\)

\(\Big \downarrow \) 278

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

Input:

Int[((c + d*x)*(a + b*x^2)^p)/x^3,x]
 

Output:

-((d*(a + b*x^2)^p*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)])/(x*(1 + 
 (b*x^2)/a)^p)) + (b*c*(a + b*x^2)^(1 + p)*Hypergeometric2F1[2, 1 + p, 2 + 
 p, 1 + (b*x^2)/a])/(2*a^2*(1 + p))
 

Defintions of rubi rules used

rule 75

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])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 278

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])
 

rule 279

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])
 

rule 542

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[c   Int[x^m*(a + b*x^2)^p, x], x] + Simp[d   Int[x^(m + 1)*(a + b*x^2 
)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && IntegerQ[m] &&  !IntegerQ[2*p]
 

Maple [F]

\[\int \frac {\left (d x +c \right ) \left (b \,x^{2}+a \right )^{p}}{x^{3}}d x\]

Input:

int((d*x+c)*(b*x^2+a)^p/x^3,x)
 

Output:

int((d*x+c)*(b*x^2+a)^p/x^3,x)
 

Fricas [F]

\[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (d x + c\right )} {\left (b x^{2} + a\right )}^{p}}{x^{3}} \,d x } \] Input:

integrate((d*x+c)*(b*x^2+a)^p/x^3,x, algorithm="fricas")
 

Output:

integral((d*x + c)*(b*x^2 + a)^p/x^3, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.76 \[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{x^3} \, dx=- \frac {a^{p} d {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} - \frac {b^{p} c x^{2 p - 2} \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (2 - p\right )} \] Input:

integrate((d*x+c)*(b*x**2+a)**p/x**3,x)
 

Output:

-a**p*d*hyper((-1/2, -p), (1/2,), b*x**2*exp_polar(I*pi)/a)/x - b**p*c*x** 
(2*p - 2)*gamma(1 - p)*hyper((-p, 1 - p), (2 - p,), a*exp_polar(I*pi)/(b*x 
**2))/(2*gamma(2 - p))
 

Maxima [F]

\[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (d x + c\right )} {\left (b x^{2} + a\right )}^{p}}{x^{3}} \,d x } \] Input:

integrate((d*x+c)*(b*x^2+a)^p/x^3,x, algorithm="maxima")
 

Output:

integrate((d*x + c)*(b*x^2 + a)^p/x^3, x)
 

Giac [F]

\[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{x^3} \, dx=\int { \frac {{\left (d x + c\right )} {\left (b x^{2} + a\right )}^{p}}{x^{3}} \,d x } \] Input:

integrate((d*x+c)*(b*x^2+a)^p/x^3,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((d*x + c)*(b*x^2 + a)^p/x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{x^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,\left (c+d\,x\right )}{x^3} \,d x \] Input:

int(((a + b*x^2)^p*(c + d*x))/x^3,x)
 

Output:

int(((a + b*x^2)^p*(c + d*x))/x^3, x)
 

Reduce [F]

\[ \int \frac {(c+d x) \left (a+b x^2\right )^p}{x^3} \, dx=\frac {-2 \left (b \,x^{2}+a \right )^{p} c p +\left (b \,x^{2}+a \right )^{p} c +2 \left (b \,x^{2}+a \right )^{p} d x +8 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{2 b \,p^{2} x^{4}-3 b p \,x^{4}+2 a \,p^{2} x^{2}+b \,x^{4}-3 a p \,x^{2}+a \,x^{2}}d x \right ) a d \,p^{3} x^{2}-12 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{2 b \,p^{2} x^{4}-3 b p \,x^{4}+2 a \,p^{2} x^{2}+b \,x^{4}-3 a p \,x^{2}+a \,x^{2}}d x \right ) a d \,p^{2} x^{2}+4 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{2 b \,p^{2} x^{4}-3 b p \,x^{4}+2 a \,p^{2} x^{2}+b \,x^{4}-3 a p \,x^{2}+a \,x^{2}}d x \right ) a d p \,x^{2}+8 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{2 b p \,x^{3}-b \,x^{3}+2 a p x -a x}d x \right ) b c \,p^{3} x^{2}-8 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{2 b p \,x^{3}-b \,x^{3}+2 a p x -a x}d x \right ) b c \,p^{2} x^{2}+2 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{2 b p \,x^{3}-b \,x^{3}+2 a p x -a x}d x \right ) b c p \,x^{2}}{2 x^{2} \left (2 p -1\right )} \] Input:

int((d*x+c)*(b*x^2+a)^p/x^3,x)
 

Output:

( - 2*(a + b*x**2)**p*c*p + (a + b*x**2)**p*c + 2*(a + b*x**2)**p*d*x + 8* 
int((a + b*x**2)**p/(2*a*p**2*x**2 - 3*a*p*x**2 + a*x**2 + 2*b*p**2*x**4 - 
 3*b*p*x**4 + b*x**4),x)*a*d*p**3*x**2 - 12*int((a + b*x**2)**p/(2*a*p**2* 
x**2 - 3*a*p*x**2 + a*x**2 + 2*b*p**2*x**4 - 3*b*p*x**4 + b*x**4),x)*a*d*p 
**2*x**2 + 4*int((a + b*x**2)**p/(2*a*p**2*x**2 - 3*a*p*x**2 + a*x**2 + 2* 
b*p**2*x**4 - 3*b*p*x**4 + b*x**4),x)*a*d*p*x**2 + 8*int((a + b*x**2)**p/( 
2*a*p*x - a*x + 2*b*p*x**3 - b*x**3),x)*b*c*p**3*x**2 - 8*int((a + b*x**2) 
**p/(2*a*p*x - a*x + 2*b*p*x**3 - b*x**3),x)*b*c*p**2*x**2 + 2*int((a + b* 
x**2)**p/(2*a*p*x - a*x + 2*b*p*x**3 - b*x**3),x)*b*c*p*x**2)/(2*x**2*(2*p 
 - 1))