\(\int \frac {(a+b x^2)^p}{x (c+d x)} \, dx\) [32]

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 20, antiderivative size = 176 \[ \int \frac {\left (a+b x^2\right )^p}{x (c+d x)} \, dx=-\frac {d x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2}+\frac {d^2 \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 c \left (b c^2+a d^2\right ) (1+p)}-\frac {\left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a c (1+p)} \] Output:

-d*x*(b*x^2+a)^p*AppellF1(1/2,1,-p,3/2,d^2*x^2/c^2,-b*x^2/a)/c^2/((1+b*x^2 
/a)^p)+1/2*d^2*(b*x^2+a)^(p+1)*hypergeom([1, p+1],[2+p],d^2*(b*x^2+a)/(a*d 
^2+b*c^2))/c/(a*d^2+b*c^2)/(p+1)-1/2*(b*x^2+a)^(p+1)*hypergeom([1, p+1],[2 
+p],1+b*x^2/a)/a/c/(p+1)
 

Mathematica [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^2\right )^p}{x (c+d x)} \, dx=\frac {\left (a+b x^2\right )^p \left (-\left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )+\left (1+\frac {a}{b x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {a}{b x^2}\right )\right )}{2 c p} \] Input:

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

Output:

((a + b*x^2)^p*(-(AppellF1[-2*p, -p, -p, 1 - 2*p, (c - Sqrt[-(a/b)]*d)/(c 
+ d*x), (c + Sqrt[-(a/b)]*d)/(c + d*x)]/(((d*(-Sqrt[-(a/b)] + x))/(c + d*x 
))^p*((d*(Sqrt[-(a/b)] + x))/(c + d*x))^p)) + Hypergeometric2F1[-p, -p, 1 
- p, -(a/(b*x^2))]/(1 + a/(b*x^2))^p))/(2*c*p)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {621, 334, 333, 354, 97, 75, 78}

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.

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

\(\Big \downarrow \) 621

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

\(\displaystyle c \int \frac {\left (b x^2+a\right )^p}{x \left (c^2-d^2 x^2\right )}dx-\frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2}\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} c \int \frac {\left (b x^2+a\right )^p}{x^2 \left (c^2-d^2 x^2\right )}dx^2-\frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2}\)

\(\Big \downarrow \) 97

\(\displaystyle \frac {1}{2} c \left (\frac {d^2 \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx^2}{c^2}+\frac {\int \frac {\left (b x^2+a\right )^p}{x^2}dx^2}{c^2}\right )-\frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {1}{2} c \left (\frac {d^2 \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx^2}{c^2}-\frac {\left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a c^2 (p+1)}\right )-\frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {1}{2} c \left (\frac {d^2 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{c^2 (p+1) \left (a d^2+b c^2\right )}-\frac {\left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a c^2 (p+1)}\right )-\frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2}\)

Input:

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

Output:

-((d*x*(a + b*x^2)^p*AppellF1[1/2, -p, 1, 3/2, -((b*x^2)/a), (d^2*x^2)/c^2 
])/(c^2*(1 + (b*x^2)/a)^p)) + (c*((d^2*(a + b*x^2)^(1 + p)*Hypergeometric2 
F1[1, 1 + p, 2 + p, (d^2*(a + b*x^2))/(b*c^2 + a*d^2)])/(c^2*(b*c^2 + a*d^ 
2)*(1 + p)) - ((a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + 
(b*x^2)/a])/(a*c^2*(1 + p))))/2
 

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 78
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 97
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[b/(b*c - a*d)   Int[(e + f*x)^p/(a + b*x), x], x] - Simp[d/(b*c 
 - a*d)   Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p}, 
 x] &&  !IntegerQ[p]
 

rule 333
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 354
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 621
Int[((x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] 
:> Simp[c   Int[x^m*((a + b*x^2)^p/(c^2 - d^2*x^2)), x], x] - Simp[d   Int[ 
x^(m + 1)*((a + b*x^2)^p/(c^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, m, 
p}, x]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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