\(\int \frac {(a+b x^4)^p}{c+d x} \, dx\) [221]

Optimal result

Integrand size = 17, antiderivative size = 257 \[ \int \frac {\left (a+b x^4\right )^p}{c+d x} \, dx=\frac {x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},-p,1,\frac {5}{4},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{c}-\frac {d x^2 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{2 c^2}+\frac {d^2 x^3 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},-p,1,\frac {7}{4},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{3 c^3}-\frac {d^3 \left (a+b x^4\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^4 \left (a+b x^4\right )}{b c^4+a d^4}\right )}{4 \left (b c^4+a d^4\right ) (1+p)} \] Output:

x*(b*x^4+a)^p*AppellF1(1/4,1,-p,5/4,d^4*x^4/c^4,-b*x^4/a)/c/((1+b*x^4/a)^p 
)-1/2*d*x^2*(b*x^4+a)^p*AppellF1(1/2,1,-p,3/2,d^4*x^4/c^4,-b*x^4/a)/c^2/(( 
1+b*x^4/a)^p)+1/3*d^2*x^3*(b*x^4+a)^p*AppellF1(3/4,1,-p,7/4,d^4*x^4/c^4,-b 
*x^4/a)/c^3/((1+b*x^4/a)^p)-1/4*d^3*(b*x^4+a)^(p+1)*hypergeom([1, p+1],[2+ 
p],d^4*(b*x^4+a)/(a*d^4+b*c^4))/(a*d^4+b*c^4)/(p+1)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.588, Rules used = {2584, 1791, 1569, 1577, 504, 334, 333, 353, 78, 2009}

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.

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

\(\Big \downarrow \) 2584

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

\(\Big \downarrow \) 1791

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

\(\Big \downarrow \) 1569

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

\(\Big \downarrow \) 1577

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

\(\Big \downarrow \) 504

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 78

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

\(\Big \downarrow \) 2009

\(\displaystyle c \left (\frac {d^2 x^3 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},-p,1,\frac {7}{4},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{3 c^4}+\frac {x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},-p,1,\frac {5}{4},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{c^2}\right )-\frac {1}{2} d \left (\frac {x^2 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{c^2}+\frac {d^2 \left (a+b x^4\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^4 \left (b x^4+a\right )}{b c^4+a d^4}\right )}{2 (p+1) \left (a d^4+b c^4\right )}\right )\)

Input:

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

Output:

c*((x*(a + b*x^4)^p*AppellF1[1/4, -p, 1, 5/4, -((b*x^4)/a), (d^4*x^4)/c^4] 
)/(c^2*(1 + (b*x^4)/a)^p) + (d^2*x^3*(a + b*x^4)^p*AppellF1[3/4, -p, 1, 7/ 
4, -((b*x^4)/a), (d^4*x^4)/c^4])/(3*c^4*(1 + (b*x^4)/a)^p)) - (d*((x^2*(a 
+ b*x^4)^p*AppellF1[1/2, -p, 1, 3/2, -((b*x^4)/a), (d^4*x^4)/c^4])/(c^2*(1 
 + (b*x^4)/a)^p) + (d^2*(a + b*x^4)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 
+ p, (d^4*(a + b*x^4))/(b*c^4 + a*d^4)])/(2*(b*c^4 + a*d^4)*(1 + p))))/2
 

Defintions of rubi rules used

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 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 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(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]
 

rule 504

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
 

rule 1569

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int 
[ExpandIntegrand[(a + c*x^4)^p, (d/(d^2 - e^2*x^4) - e*(x^2/(d^2 - e^2*x^4) 
))^(-q), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  ! 
IntegerQ[p] && ILtQ[q, 0]
 

rule 1577

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; Free 
Q[{a, c, d, e, p, q}, x]
 

rule 1791

Int[((A_) + (B_.)*(x_)^(m_.))*((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)* 
(x_)^(n2_))^(p_.), x_Symbol] :> Simp[A   Int[(d + e*x^n)^q*(a + c*x^(2*n))^ 
p, x], x] + Simp[B   Int[x^m*(d + e*x^n)^q*(a + c*x^(2*n))^p, x], x] /; Fre 
eQ[{a, c, d, e, A, B, m, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[m - n + 1, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2584

Int[((c_) + (d_.)*(x_)^(n_.))^(q_)*((a_) + (b_.)*(x_)^(nn_.))^(p_), x_Symbo 
l] :> Int[ExpandToSum[(c - d*x^n)^(-q), x]*((a + b*x^nn)^p/(c^2 - d^2*x^(2* 
n))^(-q)), x] /; FreeQ[{a, b, c, d, n, nn, p}, x] &&  !IntegerQ[p] && ILtQ[ 
q, 0] && IGtQ[Log[2, nn/n], 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^4\right )^p}{c+d x} \, dx=\frac {\left (b \,x^{4}+a \right )^{p}+4 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{b d \,x^{5}+b c \,x^{4}+a d x +a c}d x \right ) a d p -4 \left (\int \frac {\left (b \,x^{4}+a \right )^{p} x^{3}}{b d \,x^{5}+b c \,x^{4}+a d x +a c}d x \right ) b c p}{4 d p} \] Input:

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

Output:

((a + b*x**4)**p + 4*int((a + b*x**4)**p/(a*c + a*d*x + b*c*x**4 + b*d*x** 
5),x)*a*d*p - 4*int(((a + b*x**4)**p*x**3)/(a*c + a*d*x + b*c*x**4 + b*d*x 
**5),x)*b*c*p)/(4*d*p)