\(\int \frac {(a+b x^2)^p}{x^4 (c+d x)} \, dx\) [38]

Optimal result

Integrand size = 20, antiderivative size = 217 \[ \int \frac {\left (a+b x^2\right )^p}{x^4 (c+d x)} \, dx=\frac {d \left (a+b x^2\right )^{1+p}}{2 a c^2 x^2}-\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (-\frac {3}{2},-p,1,-\frac {1}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{3 c x^3}-\frac {d^5 \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 c^4 \left (b c^2+a d^2\right ) (1+p)}+\frac {d \left (a d^2+b c^2 p\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a^2 c^4 (1+p)} \] Output:

1/2*d*(b*x^2+a)^(p+1)/a/c^2/x^2-1/3*(b*x^2+a)^p*AppellF1(-3/2,1,-p,-1/2,d^ 
2*x^2/c^2,-b*x^2/a)/c/x^3/((1+b*x^2/a)^p)-1/2*d^5*(b*x^2+a)^(p+1)*hypergeo 
m([1, p+1],[2+p],d^2*(b*x^2+a)/(a*d^2+b*c^2))/c^4/(a*d^2+b*c^2)/(p+1)+1/2* 
d*(b*c^2*p+a*d^2)*(b*x^2+a)^(p+1)*hypergeom([1, p+1],[2+p],1+b*x^2/a)/a^2/ 
c^4/(p+1)
 

Mathematica [A] (warning: unable to verify)

Time = 0.73 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a+b x^2\right )^p}{x^4 (c+d x)} \, dx=\frac {\left (a+b x^2\right )^p \left (\frac {3 d^3 \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{p}-\frac {2 c^3 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-p,-\frac {1}{2},-\frac {b x^2}{a}\right )}{x^3}-\frac {6 c d^2 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}-\frac {3 c^2 d \left (1+\frac {a}{b x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {a}{b x^2}\right )}{(-1+p) x^2}-\frac {3 d^3 \left (1+\frac {a}{b x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {a}{b x^2}\right )}{p}\right )}{6 c^4} \] Input:

Integrate[(a + b*x^2)^p/(x^4*(c + d*x)),x]
 

Output:

((a + b*x^2)^p*((3*d^3*AppellF1[-2*p, -p, -p, 1 - 2*p, (c - Sqrt[-(a/b)]*d 
)/(c + d*x), (c + Sqrt[-(a/b)]*d)/(c + d*x)])/(p*((d*(-Sqrt[-(a/b)] + x))/ 
(c + d*x))^p*((d*(Sqrt[-(a/b)] + x))/(c + d*x))^p) - (2*c^3*Hypergeometric 
2F1[-3/2, -p, -1/2, -((b*x^2)/a)])/(x^3*(1 + (b*x^2)/a)^p) - (6*c*d^2*Hype 
rgeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)])/(x*(1 + (b*x^2)/a)^p) - (3*c^2 
*d*Hypergeometric2F1[1 - p, -p, 2 - p, -(a/(b*x^2))])/((-1 + p)*(1 + a/(b* 
x^2))^p*x^2) - (3*d^3*Hypergeometric2F1[-p, -p, 1 - p, -(a/(b*x^2))])/(p*( 
1 + a/(b*x^2))^p)))/(6*c^4)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {621, 354, 114, 25, 174, 75, 78, 395, 394}

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.

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

\(\Big \downarrow \) 621

\(\displaystyle c \int \frac {\left (b x^2+a\right )^p}{x^4 \left (c^2-d^2 x^2\right )}dx-d \int \frac {\left (b x^2+a\right )^p}{x^3 \left (c^2-d^2 x^2\right )}dx\)

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 75

\(\displaystyle c \int \frac {\left (b x^2+a\right )^p}{x^4 \left (c^2-d^2 x^2\right )}dx-\frac {1}{2} d \left (\frac {\frac {a d^4 \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx^2}{c^2}-\frac {\left (a+b x^2\right )^{p+1} \left (\frac {a d^2}{c^2}+b p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a (p+1)}}{a c^2}-\frac {\left (a+b x^2\right )^{p+1}}{a c^2 x^2}\right )\)

\(\Big \downarrow \) 78

\(\displaystyle c \int \frac {\left (b x^2+a\right )^p}{x^4 \left (c^2-d^2 x^2\right )}dx-\frac {1}{2} d \left (\frac {\frac {a d^4 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{c^2 (p+1) \left (a d^2+b c^2\right )}-\frac {\left (a+b x^2\right )^{p+1} \left (\frac {a d^2}{c^2}+b p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a (p+1)}}{a c^2}-\frac {\left (a+b x^2\right )^{p+1}}{a c^2 x^2}\right )\)

\(\Big \downarrow \) 395

\(\displaystyle c \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \frac {\left (\frac {b x^2}{a}+1\right )^p}{x^4 \left (c^2-d^2 x^2\right )}dx-\frac {1}{2} d \left (\frac {\frac {a d^4 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{c^2 (p+1) \left (a d^2+b c^2\right )}-\frac {\left (a+b x^2\right )^{p+1} \left (\frac {a d^2}{c^2}+b p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a (p+1)}}{a c^2}-\frac {\left (a+b x^2\right )^{p+1}}{a c^2 x^2}\right )\)

\(\Big \downarrow \) 394

\(\displaystyle -\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (-\frac {3}{2},-p,1,-\frac {1}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{3 c x^3}-\frac {1}{2} d \left (\frac {\frac {a d^4 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{c^2 (p+1) \left (a d^2+b c^2\right )}-\frac {\left (a+b x^2\right )^{p+1} \left (\frac {a d^2}{c^2}+b p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a (p+1)}}{a c^2}-\frac {\left (a+b x^2\right )^{p+1}}{a c^2 x^2}\right )\)

Input:

Int[(a + b*x^2)^p/(x^4*(c + d*x)),x]
 

Output:

-1/3*((a + b*x^2)^p*AppellF1[-3/2, -p, 1, -1/2, -((b*x^2)/a), (d^2*x^2)/c^ 
2])/(c*x^3*(1 + (b*x^2)/a)^p) - (d*(-((a + b*x^2)^(1 + p)/(a*c^2*x^2)) + ( 
(a*d^4*(a + b*x^2)^(1 + p)*Hypergeometric2F1[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*d^2)/c^2 + b* 
p)*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/ 
(a*(1 + p)))/(a*c^2)))/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 114

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

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 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

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

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]

\[\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{4} \left (d x +c \right )}d x\]

Input:

int((b*x^2+a)^p/x^4/(d*x+c),x)
 

Output:

int((b*x^2+a)^p/x^4/(d*x+c),x)
 

Fricas [F]

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

integrate((b*x^2+a)^p/x^4/(d*x+c),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^p/(d*x^5 + c*x^4), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{x^4 (c+d x)} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+a)**p/x**4/(d*x+c),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((b*x^2+a)^p/x^4/(d*x+c),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate((b*x^2 + a)^p/((d*x + c)*x^4), x)
 

Giac [F]

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

integrate((b*x^2+a)^p/x^4/(d*x+c),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^p/((d*x + c)*x^4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{x^4 (c+d x)} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p}{x^4\,\left (c+d\,x\right )} \,d x \] Input:

int((a + b*x^2)^p/(x^4*(c + d*x)),x)
 

Output:

int((a + b*x^2)^p/(x^4*(c + d*x)), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^p}{x^4 (c+d x)} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{p}}{d \,x^{5}+c \,x^{4}}d x \] Input:

int((b*x^2+a)^p/x^4/(d*x+c),x)
 

Output:

int((a + b*x**2)**p/(c*x**4 + d*x**5),x)