\(\int \frac {(c+d x)^3 (a+b x^2)^p}{x^2} \, dx\) [26]

Optimal result

Integrand size = 20, antiderivative size = 159 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )^p}{x^2} \, dx=\frac {d^3 \left (a+b x^2\right )^{1+p}}{2 b (1+p)}-\frac {c^3 \left (a+b x^2\right )^{1+p}}{a x}+\frac {c \left (3 a d^2+b c^2 (1+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{a}-\frac {3 c^2 d \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a (1+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.97 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )^p}{x^2} \, dx=\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (-2 a b c^3 (1+p) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )+d x \left (6 a b c d (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+\left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p \left (a d^2-3 b c^2 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )\right )\right )\right )}{2 a b (1+p) x} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, 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, 90, 75, 359, 238, 237}

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.

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

\(\Big \downarrow \) 543

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 75

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

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 238

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

\(\Big \downarrow \) 237

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

Input:

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

Output:

-((c^3*(a + b*x^2)^(1 + p))/(a*x)) + (c*(3*a*d^2 + b*c^2*(1 + 2*p))*x*(a + 
 b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(a*(1 + (b*x^2)/a 
)^p) + (d*((d^2*(a + b*x^2)^(1 + p))/(b*(1 + p)) - (3*c^2*(a + b*x^2)^(1 + 
 p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(a*(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 90

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

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [A] (verification not implemented)

Time = 5.49 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^3 \left (a+b x^2\right )^p}{x^2} \, dx=- \frac {a^{p} c^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} + 3 a^{p} c d^{2} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} - \frac {3 b^{p} c^{2} d x^{2 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + d^{3} \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

-a**p*c**3*hyper((-1/2, -p), (1/2,), b*x**2*exp_polar(I*pi)/a)/x + 3*a**p* 
c*d**2*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) - 3*b**p*c**2* 
d*x**(2*p)*gamma(-p)*hyper((-p, -p), (1 - p,), a*exp_polar(I*pi)/(b*x**2)) 
/(2*gamma(1 - p)) + d**3*Piecewise((a**p*x**2/2, Eq(b, 0)), (Piecewise(((a 
 + b*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(a + b*x**2), True))/(2*b), T 
rue))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(12*(a + b*x**2)**p*a*c*d**2*p**3 + 12*(a + b*x**2)**p*a*c*d**2*p**2 + 4*( 
a + b*x**2)**p*a*d**3*p**3*x - (a + b*x**2)**p*a*d**3*p*x + 4*(a + b*x**2) 
**p*b*c**3*p**3 + 6*(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 + 12*(a + b*x**2)**p*b*c**2*d*p**2 
*x - 3*(a + b*x**2)**p*b*c**2*d*p*x - 3*(a + b*x**2)**p*b*c**2*d*x + 12*(a 
 + b*x**2)**p*b*c*d**2*p**3*x**2 + 6*(a + b*x**2)**p*b*c*d**2*p**2*x**2 - 
6*(a + b*x**2)**p*b*c*d**2*p*x**2 + 4*(a + b*x**2)**p*b*d**3*p**3*x**3 - ( 
a + b*x**2)**p*b*d**3*p*x**3 + 48*int((a + b*x**2)**p/(4*a*p**2*x**2 - a*x 
**2 + 4*b*p**2*x**4 - b*x**4),x)*a**2*c*d**2*p**5*x + 48*int((a + b*x**2)* 
*p/(4*a*p**2*x**2 - a*x**2 + 4*b*p**2*x**4 - b*x**4),x)*a**2*c*d**2*p**4*x 
 - 12*int((a + b*x**2)**p/(4*a*p**2*x**2 - a*x**2 + 4*b*p**2*x**4 - b*x**4 
),x)*a**2*c*d**2*p**3*x - 12*int((a + b*x**2)**p/(4*a*p**2*x**2 - a*x**2 + 
 4*b*p**2*x**4 - b*x**4),x)*a**2*c*d**2*p**2*x + 32*int((a + b*x**2)**p/(4 
*a*p**2*x**2 - a*x**2 + 4*b*p**2*x**4 - b*x**4),x)*a*b*c**3*p**6*x + 48*in 
t((a + b*x**2)**p/(4*a*p**2*x**2 - a*x**2 + 4*b*p**2*x**4 - b*x**4),x)*a*b 
*c**3*p**5*x + 8*int((a + b*x**2)**p/(4*a*p**2*x**2 - a*x**2 + 4*b*p**2*x* 
*4 - b*x**4),x)*a*b*c**3*p**4*x - 12*int((a + b*x**2)**p/(4*a*p**2*x**2 - 
a*x**2 + 4*b*p**2*x**4 - b*x**4),x)*a*b*c**3*p**3*x - 4*int((a + b*x**2)** 
p/(4*a*p**2*x**2 - a*x**2 + 4*b*p**2*x**4 - b*x**4),x)*a*b*c**3*p**2*x + 9 
6*int((a + b*x**2)**p/(4*a*p**2*x - a*x + 4*b*p**2*x**3 - b*x**3),x)*a*...