\(\int \frac {(c+d x)^3 (a+b x^2)^p}{x^4} \, dx\) [28]

Optimal result

Integrand size = 20, antiderivative size = 181 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )^p}{x^4} \, dx=-\frac {c^3 \left (a+b x^2\right )^{1+p}}{3 a x^3}+\frac {d^3 \left (a+b x^2\right )^{1+p}}{2 b p x^2}-\frac {c \left (9 a d^2-b c^2 (1-2 p)\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{3 a x}+\frac {d \left (a d^2+3 b c^2 p\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a^2 p (1+p)} \] Output:

-1/3*c^3*(b*x^2+a)^(p+1)/a/x^3+1/2*d^3*(b*x^2+a)^(p+1)/b/p/x^2-1/3*c*(9*a* 
d^2-b*c^2*(1-2*p))*(b*x^2+a)^p*hypergeom([-1/2, -p],[1/2],-b*x^2/a)/a/x/(( 
1+b*x^2/a)^p)+1/2*d*(3*b*c^2*p+a*d^2)*(b*x^2+a)^(p+1)*hypergeom([2, p+1],[ 
2+p],1+b*x^2/a)/a^2/p/(p+1)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.96 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )^p}{x^4} \, dx=-\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (2 a^2 c^3 (1+p) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-p,-\frac {1}{2},-\frac {b x^2}{a}\right )+3 d x^2 \left (6 a^2 c d (1+p) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )+x \left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p \left (a d^2 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )-3 b c^2 \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,1+\frac {b x^2}{a}\right )\right )\right )\right )}{6 a^2 (1+p) x^3} \] Input:

Integrate[((c + d*x)^3*(a + b*x^2)^p)/x^4,x]
 

Output:

-1/6*((a + b*x^2)^p*(2*a^2*c^3*(1 + p)*Hypergeometric2F1[-3/2, -p, -1/2, - 
((b*x^2)/a)] + 3*d*x^2*(6*a^2*c*d*(1 + p)*Hypergeometric2F1[-1/2, -p, 1/2, 
 -((b*x^2)/a)] + x*(a + b*x^2)*(1 + (b*x^2)/a)^p*(a*d^2*Hypergeometric2F1[ 
1, 1 + p, 2 + p, 1 + (b*x^2)/a] - 3*b*c^2*Hypergeometric2F1[2, 1 + p, 2 + 
p, 1 + (b*x^2)/a]))))/(a^2*(1 + p)*x^3*(1 + (b*x^2)/a)^p)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {543, 354, 27, 87, 75, 359, 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.

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

\(\Big \downarrow \) 543

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 75

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

\(\Big \downarrow \) 359

\(\displaystyle \frac {c \left (9 a d^2-b c^2 (1-2 p)\right ) \int \frac {\left (b x^2+a\right )^p}{x^2}dx}{3 a}+\frac {1}{2} d \left (-\frac {\left (a+b x^2\right )^{p+1} \left (a d^2+3 b c^2 p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a^2 (p+1)}-\frac {3 c^2 \left (a+b x^2\right )^{p+1}}{a x^2}\right )-\frac {c^3 \left (a+b x^2\right )^{p+1}}{3 a x^3}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {c \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (9 a d^2-b c^2 (1-2 p)\right ) \int \frac {\left (\frac {b x^2}{a}+1\right )^p}{x^2}dx}{3 a}+\frac {1}{2} d \left (-\frac {\left (a+b x^2\right )^{p+1} \left (a d^2+3 b c^2 p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a^2 (p+1)}-\frac {3 c^2 \left (a+b x^2\right )^{p+1}}{a x^2}\right )-\frac {c^3 \left (a+b x^2\right )^{p+1}}{3 a x^3}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {1}{2} d \left (-\frac {\left (a+b x^2\right )^{p+1} \left (a d^2+3 b c^2 p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a^2 (p+1)}-\frac {3 c^2 \left (a+b x^2\right )^{p+1}}{a x^2}\right )-\frac {c^3 \left (a+b x^2\right )^{p+1}}{3 a x^3}-\frac {c \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (9 a d^2-b c^2 (1-2 p)\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{3 a x}\)

Input:

Int[((c + d*x)^3*(a + b*x^2)^p)/x^4,x]
 

Output:

-1/3*(c^3*(a + b*x^2)^(1 + p))/(a*x^3) - (c*(9*a*d^2 - b*c^2*(1 - 2*p))*(a 
 + b*x^2)^p*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)])/(3*a*x*(1 + (b 
*x^2)/a)^p) + (d*((-3*c^2*(a + b*x^2)^(1 + p))/(a*x^2) - ((a*d^2 + 3*b*c^2 
*p)*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a]) 
/(a^2*(1 + p))))/2
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 87

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]))))
 

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 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 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

rule 543

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), 
 {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ 
(n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; 
 FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] &&  !IntegerQ[2*p] 
&&  !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
 

Maple [F]

\[\int \frac {\left (d x +c \right )^{3} \left (b \,x^{2}+a \right )^{p}}{x^{4}}d x\]

Input:

int((d*x+c)^3*(b*x^2+a)^p/x^4,x)
 

Output:

int((d*x+c)^3*(b*x^2+a)^p/x^4,x)
 

Fricas [F]

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

integrate((d*x+c)^3*(b*x^2+a)^p/x^4,x, algorithm="fricas")
 

Output:

integral((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*(b*x^2 + a)^p/x^4, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 10.55 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )^p}{x^4} \, dx=- \frac {a^{p} c^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - p \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 x^{3}} - \frac {3 a^{p} c d^{2} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} - \frac {3 b^{p} c^{2} d x^{2 p - 2} \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (2 - p\right )} - \frac {b^{p} d^{3} x^{2 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} \] Input:

integrate((d*x+c)**3*(b*x**2+a)**p/x**4,x)
 

Output:

-a**p*c**3*hyper((-3/2, -p), (-1/2,), b*x**2*exp_polar(I*pi)/a)/(3*x**3) - 
 3*a**p*c*d**2*hyper((-1/2, -p), (1/2,), b*x**2*exp_polar(I*pi)/a)/x - 3*b 
**p*c**2*d*x**(2*p - 2)*gamma(1 - p)*hyper((-p, 1 - p), (2 - p,), a*exp_po 
lar(I*pi)/(b*x**2))/(2*gamma(2 - p)) - b**p*d**3*x**(2*p)*gamma(-p)*hyper( 
(-p, -p), (1 - p,), a*exp_polar(I*pi)/(b*x**2))/(2*gamma(1 - p))
 

Maxima [F]

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

integrate((d*x+c)^3*(b*x^2+a)^p/x^4,x, algorithm="maxima")
 

Output:

integrate((d*x + c)^3*(b*x^2 + a)^p/x^4, x)
 

Giac [F]

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

integrate((d*x+c)^3*(b*x^2+a)^p/x^4,x, algorithm="giac")
 

Output:

integrate((d*x + c)^3*(b*x^2 + a)^p/x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )^p}{x^4} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,{\left (c+d\,x\right )}^3}{x^4} \,d x \] Input:

int(((a + b*x^2)^p*(c + d*x)^3)/x^4,x)
 

Output:

int(((a + b*x^2)^p*(c + d*x)^3)/x^4, x)
 

Reduce [F]

\[ \int \frac {(c+d x)^3 \left (a+b x^2\right )^p}{x^4} \, dx=\text {too large to display} \] Input:

int((d*x+c)^3*(b*x^2+a)^p/x^4,x)
 

Output:

(12*(a + b*x**2)**p*a*c*d**2*p**2 + 4*(a + b*x**2)**p*b*c**3*p**2 - 2*(a + 
 b*x**2)**p*b*c**3*p - 12*(a + b*x**2)**p*b*c**2*d*p**3*x + 24*(a + b*x**2 
)**p*b*c**2*d*p**2*x - 9*(a + b*x**2)**p*b*c**2*d*p*x + 12*(a + b*x**2)**p 
*b*c*d**2*p**2*x**2 - 18*(a + b*x**2)**p*b*c*d**2*p*x**2 + 4*(a + b*x**2)* 
*p*b*d**3*p**2*x**3 - 8*(a + b*x**2)**p*b*d**3*p*x**3 + 3*(a + b*x**2)**p* 
b*d**3*x**3 + 144*int((a + b*x**2)**p/(4*a*p**3*x**4 - 12*a*p**2*x**4 + 11 
*a*p*x**4 - 3*a*x**4 + 4*b*p**3*x**6 - 12*b*p**2*x**6 + 11*b*p*x**6 - 3*b* 
x**6),x)*a**2*c*d**2*p**5*x**3 - 432*int((a + b*x**2)**p/(4*a*p**3*x**4 - 
12*a*p**2*x**4 + 11*a*p*x**4 - 3*a*x**4 + 4*b*p**3*x**6 - 12*b*p**2*x**6 + 
 11*b*p*x**6 - 3*b*x**6),x)*a**2*c*d**2*p**4*x**3 + 396*int((a + b*x**2)** 
p/(4*a*p**3*x**4 - 12*a*p**2*x**4 + 11*a*p*x**4 - 3*a*x**4 + 4*b*p**3*x**6 
 - 12*b*p**2*x**6 + 11*b*p*x**6 - 3*b*x**6),x)*a**2*c*d**2*p**3*x**3 - 108 
*int((a + b*x**2)**p/(4*a*p**3*x**4 - 12*a*p**2*x**4 + 11*a*p*x**4 - 3*a*x 
**4 + 4*b*p**3*x**6 - 12*b*p**2*x**6 + 11*b*p*x**6 - 3*b*x**6),x)*a**2*c*d 
**2*p**2*x**3 + 32*int((a + b*x**2)**p/(4*a*p**3*x**4 - 12*a*p**2*x**4 + 1 
1*a*p*x**4 - 3*a*x**4 + 4*b*p**3*x**6 - 12*b*p**2*x**6 + 11*b*p*x**6 - 3*b 
*x**6),x)*a*b*c**3*p**6*x**3 - 112*int((a + b*x**2)**p/(4*a*p**3*x**4 - 12 
*a*p**2*x**4 + 11*a*p*x**4 - 3*a*x**4 + 4*b*p**3*x**6 - 12*b*p**2*x**6 + 1 
1*b*p*x**6 - 3*b*x**6),x)*a*b*c**3*p**5*x**3 + 136*int((a + b*x**2)**p/(4* 
a*p**3*x**4 - 12*a*p**2*x**4 + 11*a*p*x**4 - 3*a*x**4 + 4*b*p**3*x**6 -...