\(\int x^3 (a+b \sqrt {c+d x})^p \, dx\) [139]

Optimal result

Integrand size = 19, antiderivative size = 350 \[ \int x^3 \left (a+b \sqrt {c+d x}\right )^p \, dx=-\frac {2 a \left (a^2-b^2 c\right )^3 \left (a+b \sqrt {c+d x}\right )^{1+p}}{b^8 d^4 (1+p)}+\frac {2 \left (a^2-b^2 c\right )^2 \left (7 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{2+p}}{b^8 d^4 (2+p)}-\frac {6 a \left (7 a^2-3 b^2 c\right ) \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{3+p}}{b^8 d^4 (3+p)}+\frac {2 \left (35 a^4-30 a^2 b^2 c+3 b^4 c^2\right ) \left (a+b \sqrt {c+d x}\right )^{4+p}}{b^8 d^4 (4+p)}-\frac {10 a \left (7 a^2-3 b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{5+p}}{b^8 d^4 (5+p)}+\frac {6 \left (7 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{6+p}}{b^8 d^4 (6+p)}-\frac {14 a \left (a+b \sqrt {c+d x}\right )^{7+p}}{b^8 d^4 (7+p)}+\frac {2 \left (a+b \sqrt {c+d x}\right )^{8+p}}{b^8 d^4 (8+p)} \] Output:

-2*a*(-b^2*c+a^2)^3*(a+b*(d*x+c)^(1/2))^(p+1)/b^8/d^4/(p+1)+2*(-b^2*c+a^2) 
^2*(-b^2*c+7*a^2)*(a+b*(d*x+c)^(1/2))^(2+p)/b^8/d^4/(2+p)-6*a*(-3*b^2*c+7* 
a^2)*(-b^2*c+a^2)*(a+b*(d*x+c)^(1/2))^(3+p)/b^8/d^4/(3+p)+2*(3*b^4*c^2-30* 
a^2*b^2*c+35*a^4)*(a+b*(d*x+c)^(1/2))^(4+p)/b^8/d^4/(4+p)-10*a*(-3*b^2*c+7 
*a^2)*(a+b*(d*x+c)^(1/2))^(5+p)/b^8/d^4/(5+p)+6*(-b^2*c+7*a^2)*(a+b*(d*x+c 
)^(1/2))^(6+p)/b^8/d^4/(6+p)-14*a*(a+b*(d*x+c)^(1/2))^(7+p)/b^8/d^4/(7+p)+ 
2*(a+b*(d*x+c)^(1/2))^(8+p)/b^8/d^4/(8+p)
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 555, normalized size of antiderivative = 1.59 \[ \int x^3 \left (a+b \sqrt {c+d x}\right )^p \, dx=-\frac {2 \left (a+b \sqrt {c+d x}\right )^{1+p} \left (5040 a^7-5040 a^6 b (1+p) \sqrt {c+d x}+360 a^5 b^2 \left (6 c \left (-7+p+p^2\right )+7 d \left (2+3 p+p^2\right ) x\right )-120 a^4 b^3 (1+p) \sqrt {c+d x} \left (2 c \left (-63-5 p+2 p^2\right )+7 d \left (6+5 p+p^2\right ) x\right )+6 a^3 b^4 \left (8 c^2 \left (315-124 p-139 p^2-14 p^3+p^4\right )+40 c d \left (-42-61 p-16 p^2+4 p^3+p^4\right ) x+35 d^2 \left (24+50 p+35 p^2+10 p^3+p^4\right ) x^2\right )-6 a^2 b^5 (1+p) \sqrt {c+d x} \left (-24 c^2 \left (-105-24 p+5 p^2+p^3\right )+4 c d \left (-420-386 p-94 p^2-p^3+p^4\right ) x+7 d^2 \left (120+154 p+71 p^2+14 p^3+p^4\right ) x^2\right )-b^7 \left (105+176 p+86 p^2+16 p^3+p^4\right ) \sqrt {c+d x} \left (-48 c^3+24 c^2 d (2+p) x-6 c d^2 \left (8+6 p+p^2\right ) x^2+d^3 \left (48+44 p+12 p^2+p^3\right ) x^3\right )+a b^6 \left (48 c^3 \left (-105+103 p+138 p^2+38 p^3+3 p^4\right )-24 c^2 d \left (-210-283 p-21 p^2+74 p^3+24 p^4+2 p^5\right ) x+6 c d^2 \left (-840-1726 p-1151 p^2-265 p^3+10 p^4+11 p^5+p^6\right ) x^2+7 d^3 \left (720+1764 p+1624 p^2+735 p^3+175 p^4+21 p^5+p^6\right ) x^3\right )\right )}{b^8 d^4 (1+p) (2+p) (3+p) (4+p) (5+p) (6+p) (7+p) (8+p)} \] Input:

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

Output:

(-2*(a + b*Sqrt[c + d*x])^(1 + p)*(5040*a^7 - 5040*a^6*b*(1 + p)*Sqrt[c + 
d*x] + 360*a^5*b^2*(6*c*(-7 + p + p^2) + 7*d*(2 + 3*p + p^2)*x) - 120*a^4* 
b^3*(1 + p)*Sqrt[c + d*x]*(2*c*(-63 - 5*p + 2*p^2) + 7*d*(6 + 5*p + p^2)*x 
) + 6*a^3*b^4*(8*c^2*(315 - 124*p - 139*p^2 - 14*p^3 + p^4) + 40*c*d*(-42 
- 61*p - 16*p^2 + 4*p^3 + p^4)*x + 35*d^2*(24 + 50*p + 35*p^2 + 10*p^3 + p 
^4)*x^2) - 6*a^2*b^5*(1 + p)*Sqrt[c + d*x]*(-24*c^2*(-105 - 24*p + 5*p^2 + 
 p^3) + 4*c*d*(-420 - 386*p - 94*p^2 - p^3 + p^4)*x + 7*d^2*(120 + 154*p + 
 71*p^2 + 14*p^3 + p^4)*x^2) - b^7*(105 + 176*p + 86*p^2 + 16*p^3 + p^4)*S 
qrt[c + d*x]*(-48*c^3 + 24*c^2*d*(2 + p)*x - 6*c*d^2*(8 + 6*p + p^2)*x^2 + 
 d^3*(48 + 44*p + 12*p^2 + p^3)*x^3) + a*b^6*(48*c^3*(-105 + 103*p + 138*p 
^2 + 38*p^3 + 3*p^4) - 24*c^2*d*(-210 - 283*p - 21*p^2 + 74*p^3 + 24*p^4 + 
 2*p^5)*x + 6*c*d^2*(-840 - 1726*p - 1151*p^2 - 265*p^3 + 10*p^4 + 11*p^5 
+ p^6)*x^2 + 7*d^3*(720 + 1764*p + 1624*p^2 + 735*p^3 + 175*p^4 + 21*p^5 + 
 p^6)*x^3)))/(b^8*d^4*(1 + p)*(2 + p)*(3 + p)*(4 + p)*(5 + p)*(6 + p)*(7 + 
 p)*(8 + p))
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {896, 25, 1732, 522, 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.

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

\(\Big \downarrow \) 896

\(\displaystyle \frac {\int d^3 x^3 \left (a+b \sqrt {c+d x}\right )^pd(c+d x)}{d^4}\)

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1732

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

\(\Big \downarrow \) 522

\(\displaystyle -\frac {2 \int \left (\frac {a \left (a^2-b^2 c\right )^3 \left (a+b \sqrt {c+d x}\right )^p}{b^7}+\frac {\left (b^2 c-7 a^2\right ) \left (b^2 c-a^2\right )^2 \left (a+b \sqrt {c+d x}\right )^{p+1}}{b^7}+\frac {3 \left (7 a^5-10 b^2 c a^3+3 b^4 c^2 a\right ) \left (a+b \sqrt {c+d x}\right )^{p+2}}{b^7}+\frac {\left (-35 a^4+30 b^2 c a^2-3 b^4 c^2\right ) \left (a+b \sqrt {c+d x}\right )^{p+3}}{b^7}-\frac {5 \left (3 a b^2 c-7 a^3\right ) \left (a+b \sqrt {c+d x}\right )^{p+4}}{b^7}+\frac {3 \left (b^2 c-7 a^2\right ) \left (a+b \sqrt {c+d x}\right )^{p+5}}{b^7}+\frac {7 a \left (a+b \sqrt {c+d x}\right )^{p+6}}{b^7}-\frac {\left (a+b \sqrt {c+d x}\right )^{p+7}}{b^7}\right )d\sqrt {c+d x}}{d^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \left (\frac {a \left (a^2-b^2 c\right )^3 \left (a+b \sqrt {c+d x}\right )^{p+1}}{b^8 (p+1)}-\frac {\left (a^2-b^2 c\right )^2 \left (7 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{p+2}}{b^8 (p+2)}+\frac {3 a \left (7 a^2-3 b^2 c\right ) \left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{p+3}}{b^8 (p+3)}+\frac {5 a \left (7 a^2-3 b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{p+5}}{b^8 (p+5)}-\frac {3 \left (7 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )^{p+6}}{b^8 (p+6)}-\frac {\left (35 a^4-30 a^2 b^2 c+3 b^4 c^2\right ) \left (a+b \sqrt {c+d x}\right )^{p+4}}{b^8 (p+4)}+\frac {7 a \left (a+b \sqrt {c+d x}\right )^{p+7}}{b^8 (p+7)}-\frac {\left (a+b \sqrt {c+d x}\right )^{p+8}}{b^8 (p+8)}\right )}{d^4}\)

Input:

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

Output:

(-2*((a*(a^2 - b^2*c)^3*(a + b*Sqrt[c + d*x])^(1 + p))/(b^8*(1 + p)) - ((a 
^2 - b^2*c)^2*(7*a^2 - b^2*c)*(a + b*Sqrt[c + d*x])^(2 + p))/(b^8*(2 + p)) 
 + (3*a*(7*a^2 - 3*b^2*c)*(a^2 - b^2*c)*(a + b*Sqrt[c + d*x])^(3 + p))/(b^ 
8*(3 + p)) - ((35*a^4 - 30*a^2*b^2*c + 3*b^4*c^2)*(a + b*Sqrt[c + d*x])^(4 
 + p))/(b^8*(4 + p)) + (5*a*(7*a^2 - 3*b^2*c)*(a + b*Sqrt[c + d*x])^(5 + p 
))/(b^8*(5 + p)) - (3*(7*a^2 - b^2*c)*(a + b*Sqrt[c + d*x])^(6 + p))/(b^8* 
(6 + p)) + (7*a*(a + b*Sqrt[c + d*x])^(7 + p))/(b^8*(7 + p)) - (a + b*Sqrt 
[c + d*x])^(8 + p)/(b^8*(8 + p))))/d^4
 

Defintions of rubi rules used

rule 25

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

rule 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

rule 896

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff 
icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1)   Subst[Int[Si 
mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; 
 FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
 

rule 1732

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symb 
ol] :> With[{g = Denominator[n]}, Simp[g   Subst[Int[x^(g - 1)*(d + e*x^(g* 
n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p, q} 
, x] && EqQ[n2, 2*n] && FractionQ[n]
 

rule 2009

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

Maple [F]

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

Input:

int(x^3*(a+b*(d*x+c)^(1/2))^p,x)
 

Output:

int(x^3*(a+b*(d*x+c)^(1/2))^p,x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1416 vs. \(2 (335) = 670\).

Time = 0.21 (sec) , antiderivative size = 1416, normalized size of antiderivative = 4.05 \[ \int x^3 \left (a+b \sqrt {c+d x}\right )^p \, dx=\text {Too large to display} \] Input:

integrate(x^3*(a+b*(d*x+c)^(1/2))^p,x, algorithm="fricas")
 

Output:

-2*(5040*b^8*c^4 - 20160*a^2*b^6*c^3 + 30240*a^4*b^4*c^2 - 20160*a^6*b^2*c 
 + 5040*a^8 + 48*(b^8*c^4 + 6*a^2*b^6*c^3 + a^4*b^4*c^2)*p^4 - (b^8*d^4*p^ 
7 + 28*b^8*d^4*p^6 + 322*b^8*d^4*p^5 + 1960*b^8*d^4*p^4 + 6769*b^8*d^4*p^3 
 + 13132*b^8*d^4*p^2 + 13068*b^8*d^4*p + 5040*b^8*d^4)*x^4 + 384*(2*b^8*c^ 
4 + 7*a^2*b^6*c^3 - 3*a^4*b^4*c^2)*p^3 - (b^8*c*d^3*p^7 + (22*b^8*c - 7*a^ 
2*b^6)*d^3*p^6 + 5*(38*b^8*c - 21*a^2*b^6)*d^3*p^5 + 5*(164*b^8*c - 119*a^ 
2*b^6)*d^3*p^4 + (1849*b^8*c - 1575*a^2*b^6)*d^3*p^3 + 2*(1019*b^8*c - 959 
*a^2*b^6)*d^3*p^2 + 840*(b^8*c - a^2*b^6)*d^3*p)*x^3 + 48*(86*b^8*c^4 + 81 
*a^2*b^6*c^3 - 124*a^4*b^4*c^2 + 45*a^6*b^2*c)*p^2 + 6*(18*b^8*c^2*d^2*p^5 
 + (b^8*c^2 + a^2*b^6*c)*d^2*p^6 + (118*b^8*c^2 - 95*a^2*b^6*c + 35*a^4*b^ 
4)*d^2*p^4 + 6*(58*b^8*c^2 - 80*a^2*b^6*c + 35*a^4*b^4)*d^2*p^3 + (457*b^8 
*c^2 - 806*a^2*b^6*c + 385*a^4*b^4)*d^2*p^2 + 210*(b^8*c^2 - 2*a^2*b^6*c + 
 a^4*b^4)*d^2*p)*x^2 + 192*(44*b^8*c^4 - 71*a^2*b^6*c^3 + 54*a^4*b^4*c^2 - 
 15*a^6*b^2*c)*p - 24*((b^8*c^3 + 3*a^2*b^6*c^2)*d*p^5 + 2*(8*b^8*c^3 + 9* 
a^2*b^6*c^2 - 5*a^4*b^4*c)*d*p^4 + (86*b^8*c^3 - 57*a^2*b^6*c^2 + 15*a^4*b 
^4*c)*d*p^3 + (176*b^8*c^3 - 387*a^2*b^6*c^2 + 340*a^4*b^4*c - 105*a^6*b^2 
)*d*p^2 + 105*(b^8*c^3 - 3*a^2*b^6*c^2 + 3*a^4*b^4*c - a^6*b^2)*d*p)*x + ( 
192*(a*b^7*c^3 + a^3*b^5*c^2)*p^4 + 96*(27*a*b^7*c^3 + 2*a^3*b^5*c^2 - 5*a 
^5*b^3*c)*p^3 - (a*b^7*d^3*p^7 + 21*a*b^7*d^3*p^6 + 175*a*b^7*d^3*p^5 + 73 
5*a*b^7*d^3*p^4 + 1624*a*b^7*d^3*p^3 + 1764*a*b^7*d^3*p^2 + 720*a*b^7*d...
 

Sympy [F]

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

integrate(x**3*(a+b*(d*x+c)**(1/2))**p,x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 728 vs. \(2 (335) = 670\).

Time = 0.05 (sec) , antiderivative size = 728, normalized size of antiderivative = 2.08 \[ \int x^3 \left (a+b \sqrt {c+d x}\right )^p \, dx =\text {Too large to display} \] Input:

integrate(x^3*(a+b*(d*x+c)^(1/2))^p,x, algorithm="maxima")
 

Output:

-2*(((d*x + c)*b^2*(p + 1) + sqrt(d*x + c)*a*b*p - a^2)*(sqrt(d*x + c)*b + 
 a)^p*c^3/((p^2 + 3*p + 2)*b^2) - 3*((p^3 + 6*p^2 + 11*p + 6)*(d*x + c)^2* 
b^4 + (p^3 + 3*p^2 + 2*p)*(d*x + c)^(3/2)*a*b^3 - 3*(p^2 + p)*(d*x + c)*a^ 
2*b^2 + 6*sqrt(d*x + c)*a^3*b*p - 6*a^4)*(sqrt(d*x + c)*b + a)^p*c^2/((p^4 
 + 10*p^3 + 35*p^2 + 50*p + 24)*b^4) + 3*((p^5 + 15*p^4 + 85*p^3 + 225*p^2 
 + 274*p + 120)*(d*x + c)^3*b^6 + (p^5 + 10*p^4 + 35*p^3 + 50*p^2 + 24*p)* 
(d*x + c)^(5/2)*a*b^5 - 5*(p^4 + 6*p^3 + 11*p^2 + 6*p)*(d*x + c)^2*a^2*b^4 
 + 20*(p^3 + 3*p^2 + 2*p)*(d*x + c)^(3/2)*a^3*b^3 - 60*(p^2 + p)*(d*x + c) 
*a^4*b^2 + 120*sqrt(d*x + c)*a^5*b*p - 120*a^6)*(sqrt(d*x + c)*b + a)^p*c/ 
((p^6 + 21*p^5 + 175*p^4 + 735*p^3 + 1624*p^2 + 1764*p + 720)*b^6) - ((p^7 
 + 28*p^6 + 322*p^5 + 1960*p^4 + 6769*p^3 + 13132*p^2 + 13068*p + 5040)*(d 
*x + c)^4*b^8 + (p^7 + 21*p^6 + 175*p^5 + 735*p^4 + 1624*p^3 + 1764*p^2 + 
720*p)*(d*x + c)^(7/2)*a*b^7 - 7*(p^6 + 15*p^5 + 85*p^4 + 225*p^3 + 274*p^ 
2 + 120*p)*(d*x + c)^3*a^2*b^6 + 42*(p^5 + 10*p^4 + 35*p^3 + 50*p^2 + 24*p 
)*(d*x + c)^(5/2)*a^3*b^5 - 210*(p^4 + 6*p^3 + 11*p^2 + 6*p)*(d*x + c)^2*a 
^4*b^4 + 840*(p^3 + 3*p^2 + 2*p)*(d*x + c)^(3/2)*a^5*b^3 - 2520*(p^2 + p)* 
(d*x + c)*a^6*b^2 + 5040*sqrt(d*x + c)*a^7*b*p - 5040*a^8)*(sqrt(d*x + c)* 
b + a)^p/((p^8 + 36*p^7 + 546*p^6 + 4536*p^5 + 22449*p^4 + 67284*p^3 + 118 
124*p^2 + 109584*p + 40320)*b^8))/d^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5699 vs. \(2 (335) = 670\).

Time = 0.25 (sec) , antiderivative size = 5699, normalized size of antiderivative = 16.28 \[ \int x^3 \left (a+b \sqrt {c+d x}\right )^p \, dx=\text {Too large to display} \] Input:

integrate(x^3*(a+b*(d*x+c)^(1/2))^p,x, algorithm="giac")
 

Output:

-2*((sqrt(d*x + c)*b + a)^2*(sqrt(d*x + c)*b + a)^p*b^6*c^3*p^7 - (sqrt(d* 
x + c)*b + a)*(sqrt(d*x + c)*b + a)^p*a*b^6*c^3*p^7 + 34*(sqrt(d*x + c)*b 
+ a)^2*(sqrt(d*x + c)*b + a)^p*b^6*c^3*p^6 - 35*(sqrt(d*x + c)*b + a)*(sqr 
t(d*x + c)*b + a)^p*a*b^6*c^3*p^6 - 3*(sqrt(d*x + c)*b + a)^4*(sqrt(d*x + 
c)*b + a)^p*b^4*c^2*p^7 + 9*(sqrt(d*x + c)*b + a)^3*(sqrt(d*x + c)*b + a)^ 
p*a*b^4*c^2*p^7 - 9*(sqrt(d*x + c)*b + a)^2*(sqrt(d*x + c)*b + a)^p*a^2*b^ 
4*c^2*p^7 + 3*(sqrt(d*x + c)*b + a)*(sqrt(d*x + c)*b + a)^p*a^3*b^4*c^2*p^ 
7 + 478*(sqrt(d*x + c)*b + a)^2*(sqrt(d*x + c)*b + a)^p*b^6*c^3*p^5 - 511* 
(sqrt(d*x + c)*b + a)*(sqrt(d*x + c)*b + a)^p*a*b^6*c^3*p^5 - 96*(sqrt(d*x 
 + c)*b + a)^4*(sqrt(d*x + c)*b + a)^p*b^4*c^2*p^6 + 297*(sqrt(d*x + c)*b 
+ a)^3*(sqrt(d*x + c)*b + a)^p*a*b^4*c^2*p^6 - 306*(sqrt(d*x + c)*b + a)^2 
*(sqrt(d*x + c)*b + a)^p*a^2*b^4*c^2*p^6 + 105*(sqrt(d*x + c)*b + a)*(sqrt 
(d*x + c)*b + a)^p*a^3*b^4*c^2*p^6 + 3*(sqrt(d*x + c)*b + a)^6*(sqrt(d*x + 
 c)*b + a)^p*b^2*c*p^7 - 15*(sqrt(d*x + c)*b + a)^5*(sqrt(d*x + c)*b + a)^ 
p*a*b^2*c*p^7 + 30*(sqrt(d*x + c)*b + a)^4*(sqrt(d*x + c)*b + a)^p*a^2*b^2 
*c*p^7 - 30*(sqrt(d*x + c)*b + a)^3*(sqrt(d*x + c)*b + a)^p*a^3*b^2*c*p^7 
+ 15*(sqrt(d*x + c)*b + a)^2*(sqrt(d*x + c)*b + a)^p*a^4*b^2*c*p^7 - 3*(sq 
rt(d*x + c)*b + a)*(sqrt(d*x + c)*b + a)^p*a^5*b^2*c*p^7 + 3580*(sqrt(d*x 
+ c)*b + a)^2*(sqrt(d*x + c)*b + a)^p*b^6*c^3*p^4 - 4025*(sqrt(d*x + c)*b 
+ a)*(sqrt(d*x + c)*b + a)^p*a*b^6*c^3*p^4 - 1254*(sqrt(d*x + c)*b + a)...
 

Mupad [F(-1)]

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

int(x^3*(a + b*(c + d*x)^(1/2))^p,x)
 

Output:

int(x^3*(a + b*(c + d*x)^(1/2))^p, x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 1909, normalized size of antiderivative = 5.45 \[ \int x^3 \left (a+b \sqrt {c+d x}\right )^p \, dx =\text {Too large to display} \] Input:

int(x^3*(a+b*(d*x+c)^(1/2))^p,x)
 

Output:

(2*(sqrt(c + d*x)*b + a)**p*(5040*sqrt(c + d*x)*a**7*b*p + 480*sqrt(c + d* 
x)*a**5*b**3*c*p**3 - 2880*sqrt(c + d*x)*a**5*b**3*c*p**2 - 18480*sqrt(c + 
 d*x)*a**5*b**3*c*p + 840*sqrt(c + d*x)*a**5*b**3*d*p**3*x + 2520*sqrt(c + 
 d*x)*a**5*b**3*d*p**2*x + 1680*sqrt(c + d*x)*a**5*b**3*d*p*x - 192*sqrt(c 
 + d*x)*a**3*b**5*c**2*p**4 - 192*sqrt(c + d*x)*a**3*b**5*c**2*p**3 + 9408 
*sqrt(c + d*x)*a**3*b**5*c**2*p**2 + 24528*sqrt(c + d*x)*a**3*b**5*c**2*p 
+ 24*sqrt(c + d*x)*a**3*b**5*c*d*p**5*x - 240*sqrt(c + d*x)*a**3*b**5*c*d* 
p**4*x - 3240*sqrt(c + d*x)*a**3*b**5*c*d*p**3*x - 7680*sqrt(c + d*x)*a**3 
*b**5*c*d*p**2*x - 4704*sqrt(c + d*x)*a**3*b**5*c*d*p*x + 42*sqrt(c + d*x) 
*a**3*b**5*d**2*p**5*x**2 + 420*sqrt(c + d*x)*a**3*b**5*d**2*p**4*x**2 + 1 
470*sqrt(c + d*x)*a**3*b**5*d**2*p**3*x**2 + 2100*sqrt(c + d*x)*a**3*b**5* 
d**2*p**2*x**2 + 1008*sqrt(c + d*x)*a**3*b**5*d**2*p*x**2 - 192*sqrt(c + d 
*x)*a*b**7*c**3*p**4 - 2592*sqrt(c + d*x)*a*b**7*c**3*p**3 - 10752*sqrt(c 
+ d*x)*a*b**7*c**3*p**2 - 13392*sqrt(c + d*x)*a*b**7*c**3*p + 72*sqrt(c + 
d*x)*a*b**7*c**2*d*p**5*x + 1008*sqrt(c + d*x)*a*b**7*c**2*d*p**4*x + 4608 
*sqrt(c + d*x)*a*b**7*c**2*d*p**3*x + 7848*sqrt(c + d*x)*a*b**7*c**2*d*p** 
2*x + 4176*sqrt(c + d*x)*a*b**7*c**2*d*p*x - 12*sqrt(c + d*x)*a*b**7*c*d** 
2*p**6*x**2 - 198*sqrt(c + d*x)*a*b**7*c*d**2*p**5*x**2 - 1200*sqrt(c + d* 
x)*a*b**7*c*d**2*p**4*x**2 - 3330*sqrt(c + d*x)*a*b**7*c*d**2*p**3*x**2 - 
4188*sqrt(c + d*x)*a*b**7*c*d**2*p**2*x**2 - 1872*sqrt(c + d*x)*a*b**7*...