\(\int \frac {x^3 (a+b x^2)^p}{(c+d x)^{3/2}} \, dx\) [94]

Optimal result

Integrand size = 22, antiderivative size = 487 \[ \int \frac {x^3 \left (a+b x^2\right )^p}{(c+d x)^{3/2}} \, dx=\frac {2 c^3 \left (a+b x^2\right )^{1+p}}{d^2 \left (b c^2+a d^2\right ) \sqrt {c+d x}}+\frac {2 \sqrt {c+d x} \left (a+b x^2\right )^{1+p}}{b d^2 (5+4 p)}-\frac {2 \left (a^2 d^4-a b c^2 d^2 (13+12 p)-8 b^2 c^4 \left (3+5 p+2 p^2\right )\right ) \sqrt {c+d x} \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{b d^4 \left (b c^2+a d^2\right ) (5+4 p)}-\frac {2 c \left (a d^2 (9+8 p)+8 b c^2 \left (3+5 p+2 p^2\right )\right ) (c+d x)^{3/2} \left (a+b x^2\right )^p \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{3 d^4 \left (b c^2+a d^2\right ) (5+4 p)} \] Output:

2*c^3*(b*x^2+a)^(p+1)/d^2/(a*d^2+b*c^2)/(d*x+c)^(1/2)+2*(d*x+c)^(1/2)*(b*x 
^2+a)^(p+1)/b/d^2/(5+4*p)-2*(a^2*d^4-a*b*c^2*d^2*(13+12*p)-8*b^2*c^4*(2*p^ 
2+5*p+3))*(d*x+c)^(1/2)*(b*x^2+a)^p*AppellF1(1/2,-p,-p,3/2,(d*x+c)/(c-(-a) 
^(1/2)*d/b^(1/2)),(d*x+c)/(c+(-a)^(1/2)*d/b^(1/2)))/b/d^4/(a*d^2+b*c^2)/(5 
+4*p)/((1-(d*x+c)/(c-(-a)^(1/2)*d/b^(1/2)))^p)/((1-(d*x+c)/(c+(-a)^(1/2)*d 
/b^(1/2)))^p)-2/3*c*(a*d^2*(9+8*p)+8*b*c^2*(2*p^2+5*p+3))*(d*x+c)^(3/2)*(b 
*x^2+a)^p*AppellF1(3/2,-p,-p,5/2,(d*x+c)/(c-(-a)^(1/2)*d/b^(1/2)),(d*x+c)/ 
(c+(-a)^(1/2)*d/b^(1/2)))/d^4/(a*d^2+b*c^2)/(5+4*p)/((1-(d*x+c)/(c-(-a)^(1 
/2)*d/b^(1/2)))^p)/((1-(d*x+c)/(c+(-a)^(1/2)*d/b^(1/2)))^p)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 1.29 (sec) , antiderivative size = 487, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {603, 27, 2185, 27, 719, 514, 150}

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.

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

\(\Big \downarrow \) 603

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\frac {2 \int -\frac {\left (a d \left (a d^2-4 b c^2 (p+1)\right )+b c \left (8 b \left (2 p^2+5 p+3\right ) c^2+a d^2 (8 p+9)\right ) x\right ) \left (b x^2+a\right )^p}{2 \sqrt {c+d x}}dx}{b d^2 (4 p+5)}+\frac {2 \sqrt {c+d x} \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{b d^2 (4 p+5)}}{a d^2+b c^2}+\frac {2 c^3 \left (a+b x^2\right )^{p+1}}{d^2 \sqrt {c+d x} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 \sqrt {c+d x} \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{b d^2 (4 p+5)}-\frac {\int \frac {\left (a d \left (a d^2-4 b c^2 (p+1)\right )+b c \left (8 b \left (2 p^2+5 p+3\right ) c^2+a d^2 (8 p+9)\right ) x\right ) \left (b x^2+a\right )^p}{\sqrt {c+d x}}dx}{b d^2 (4 p+5)}}{a d^2+b c^2}+\frac {2 c^3 \left (a+b x^2\right )^{p+1}}{d^2 \sqrt {c+d x} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {\frac {2 \sqrt {c+d x} \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{b d^2 (4 p+5)}-\frac {\frac {\left (a^2 d^4-a b c^2 d^2 (12 p+13)-8 b^2 c^4 \left (2 p^2+5 p+3\right )\right ) \int \frac {\left (b x^2+a\right )^p}{\sqrt {c+d x}}dx}{d}+\frac {b c \left (a d^2 (8 p+9)+8 b c^2 \left (2 p^2+5 p+3\right )\right ) \int \sqrt {c+d x} \left (b x^2+a\right )^pdx}{d}}{b d^2 (4 p+5)}}{a d^2+b c^2}+\frac {2 c^3 \left (a+b x^2\right )^{p+1}}{d^2 \sqrt {c+d x} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 514

\(\displaystyle \frac {\frac {2 \sqrt {c+d x} \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{b d^2 (4 p+5)}-\frac {\frac {\left (a+b x^2\right )^p \left (a^2 d^4-a b c^2 d^2 (12 p+13)-8 b^2 c^4 \left (2 p^2+5 p+3\right )\right ) \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \int \frac {\left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p}{\sqrt {c+d x}}d(c+d x)}{d^2}+\frac {b c \left (a+b x^2\right )^p \left (a d^2 (8 p+9)+8 b c^2 \left (2 p^2+5 p+3\right )\right ) \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \int \sqrt {c+d x} \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^p \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^pd(c+d x)}{d^2}}{b d^2 (4 p+5)}}{a d^2+b c^2}+\frac {2 c^3 \left (a+b x^2\right )^{p+1}}{d^2 \sqrt {c+d x} \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\frac {2 \sqrt {c+d x} \left (a d^2+b c^2\right ) \left (a+b x^2\right )^{p+1}}{b d^2 (4 p+5)}-\frac {\frac {2 \sqrt {c+d x} \left (a+b x^2\right )^p \left (a^2 d^4-a b c^2 d^2 (12 p+13)-8 b^2 c^4 \left (2 p^2+5 p+3\right )\right ) \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{d^2}+\frac {2 b c (c+d x)^{3/2} \left (a+b x^2\right )^p \left (a d^2 (8 p+9)+8 b c^2 \left (2 p^2+5 p+3\right )\right ) \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-p} \left (1-\frac {c+d x}{\frac {\sqrt {-a} d}{\sqrt {b}}+c}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{3 d^2}}{b d^2 (4 p+5)}}{a d^2+b c^2}+\frac {2 c^3 \left (a+b x^2\right )^{p+1}}{d^2 \sqrt {c+d x} \left (a d^2+b c^2\right )}\)

Input:

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

Output:

(2*c^3*(a + b*x^2)^(1 + p))/(d^2*(b*c^2 + a*d^2)*Sqrt[c + d*x]) + ((2*(b*c 
^2 + a*d^2)*Sqrt[c + d*x]*(a + b*x^2)^(1 + p))/(b*d^2*(5 + 4*p)) - ((2*(a^ 
2*d^4 - a*b*c^2*d^2*(13 + 12*p) - 8*b^2*c^4*(3 + 5*p + 2*p^2))*Sqrt[c + d* 
x]*(a + b*x^2)^p*AppellF1[1/2, -p, -p, 3/2, (c + d*x)/(c - (Sqrt[-a]*d)/Sq 
rt[b]), (c + d*x)/(c + (Sqrt[-a]*d)/Sqrt[b])])/(d^2*(1 - (c + d*x)/(c - (S 
qrt[-a]*d)/Sqrt[b]))^p*(1 - (c + d*x)/(c + (Sqrt[-a]*d)/Sqrt[b]))^p) + (2* 
b*c*(a*d^2*(9 + 8*p) + 8*b*c^2*(3 + 5*p + 2*p^2))*(c + d*x)^(3/2)*(a + b*x 
^2)^p*AppellF1[3/2, -p, -p, 5/2, (c + d*x)/(c - (Sqrt[-a]*d)/Sqrt[b]), (c 
+ d*x)/(c + (Sqrt[-a]*d)/Sqrt[b])])/(3*d^2*(1 - (c + d*x)/(c - (Sqrt[-a]*d 
)/Sqrt[b]))^p*(1 - (c + d*x)/(c + (Sqrt[-a]*d)/Sqrt[b]))^p))/(b*d^2*(5 + 4 
*p)))/(b*c^2 + a*d^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 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 514

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Rt[-a/b, 2]}, Simp[(a + b*x^2)^p/(d*(1 - (c + d*x)/(c - d*q))^p*(1 - ( 
c + d*x)/(c + d*q))^p)   Subst[Int[x^n*Simp[1 - x/(c + d*q), x]^p*Simp[1 - 
x/(c - d*q), x]^p, x], x, c + d*x], x]] /; FreeQ[{a, b, c, d, n, p}, x] && 
NeQ[b*c^2 + a*d^2, 0]
 

rule 603

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[x^m, c + d*x, x], R = PolynomialRemainde 
r[x^m, c + d*x, x]}, Simp[d*R*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 
1)*(b*c^2 + a*d^2))), x] + Simp[1/((n + 1)*(b*c^2 + a*d^2))   Int[(c + d*x) 
^(n + 1)*(a + b*x^2)^p*ExpandToSum[(n + 1)*(b*c^2 + a*d^2)*Qx + b*c*R*(n + 
1) - b*d*R*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IGt 
Q[m, 1] && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 2185

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x 
)^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p 
)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d 
, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && 
True) &&  !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 
 1/2, 0]))
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 8*sqrt(c + d*x)*(a + b*x**2)**p*a**2*d**3*p - 6*sqrt(c + d*x)*(a + b*x 
**2)**p*a**2*d**3 + 24*sqrt(c + d*x)*(a + b*x**2)**p*a*b*c**2*d*p + 36*sqr 
t(c + d*x)*(a + b*x**2)**p*a*b*c**2*d + 32*sqrt(c + d*x)*(a + b*x**2)**p*a 
*b*c*d**2*p**2*x + 24*sqrt(c + d*x)*(a + b*x**2)**p*a*b*c*d**2*p*x + 32*sq 
rt(c + d*x)*(a + b*x**2)**p*b**2*c**3*p**2*x + 80*sqrt(c + d*x)*(a + b*x** 
2)**p*b**2*c**3*p*x + 48*sqrt(c + d*x)*(a + b*x**2)**p*b**2*c**3*x - 32*sq 
rt(c + d*x)*(a + b*x**2)**p*b**2*c**2*d*p**2*x**2 - 56*sqrt(c + d*x)*(a + 
b*x**2)**p*b**2*c**2*d*p*x**2 - 12*sqrt(c + d*x)*(a + b*x**2)**p*b**2*c**2 
*d*x**2 + 32*sqrt(c + d*x)*(a + b*x**2)**p*b**2*c*d**2*p**2*x**3 + 32*sqrt 
(c + d*x)*(a + b*x**2)**p*b**2*c*d**2*p*x**3 + 6*sqrt(c + d*x)*(a + b*x**2 
)**p*b**2*c*d**2*x**3 + 1024*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(64* 
a*c**2*p**3 + 144*a*c**2*p**2 + 92*a*c**2*p + 15*a*c**2 + 128*a*c*d*p**3*x 
 + 288*a*c*d*p**2*x + 184*a*c*d*p*x + 30*a*c*d*x + 64*a*d**2*p**3*x**2 + 1 
44*a*d**2*p**2*x**2 + 92*a*d**2*p*x**2 + 15*a*d**2*x**2 + 64*b*c**2*p**3*x 
**2 + 144*b*c**2*p**2*x**2 + 92*b*c**2*p*x**2 + 15*b*c**2*x**2 + 128*b*c*d 
*p**3*x**3 + 288*b*c*d*p**2*x**3 + 184*b*c*d*p*x**3 + 30*b*c*d*x**3 + 64*b 
*d**2*p**3*x**4 + 144*b*d**2*p**2*x**4 + 92*b*d**2*p*x**4 + 15*b*d**2*x**4 
),x)*a**2*b*c*d**4*p**5 + 2816*int((sqrt(c + d*x)*(a + b*x**2)**p*x**2)/(6 
4*a*c**2*p**3 + 144*a*c**2*p**2 + 92*a*c**2*p + 15*a*c**2 + 128*a*c*d*p**3 
*x + 288*a*c*d*p**2*x + 184*a*c*d*p*x + 30*a*c*d*x + 64*a*d**2*p**3*x**...