3.2847 \(\int (c+d x)^5 (a+b (c+d x)^2)^p \, dx\)

Optimal. Leaf size=93 \[ \frac {a^2 \left (a+b (c+d x)^2\right )^{p+1}}{2 b^3 d (p+1)}-\frac {a \left (a+b (c+d x)^2\right )^{p+2}}{b^3 d (p+2)}+\frac {\left (a+b (c+d x)^2\right )^{p+3}}{2 b^3 d (p+3)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {372, 266, 43} \[ \frac {a^2 \left (a+b (c+d x)^2\right )^{p+1}}{2 b^3 d (p+1)}-\frac {a \left (a+b (c+d x)^2\right )^{p+2}}{b^3 d (p+2)}+\frac {\left (a+b (c+d x)^2\right )^{p+3}}{2 b^3 d (p+3)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^5*(a + b*(c + d*x)^2)^p,x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 372

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rubi steps

\begin {align*} \int (c+d x)^5 \left (a+b (c+d x)^2\right )^p \, dx &=\frac {\operatorname {Subst}\left (\int x^5 \left (a+b x^2\right )^p \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int x^2 (a+b x)^p \, dx,x,(c+d x)^2\right )}{2 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^2 (a+b x)^p}{b^2}-\frac {2 a (a+b x)^{1+p}}{b^2}+\frac {(a+b x)^{2+p}}{b^2}\right ) \, dx,x,(c+d x)^2\right )}{2 d}\\ &=\frac {a^2 \left (a+b (c+d x)^2\right )^{1+p}}{2 b^3 d (1+p)}-\frac {a \left (a+b (c+d x)^2\right )^{2+p}}{b^3 d (2+p)}+\frac {\left (a+b (c+d x)^2\right )^{3+p}}{2 b^3 d (3+p)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 73, normalized size = 0.78 \[ \frac {\left (a+b (c+d x)^2\right )^{p+1} \left (\frac {a^2}{p+1}-\frac {2 a \left (a+b (c+d x)^2\right )}{p+2}+\frac {\left (a+b (c+d x)^2\right )^2}{p+3}\right )}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^5*(a + b*(c + d*x)^2)^p,x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.71, size = 438, normalized size = 4.71 \[ \frac {{\left (2 \, b^{3} c^{6} + {\left (b^{3} d^{6} p^{2} + 3 \, b^{3} d^{6} p + 2 \, b^{3} d^{6}\right )} x^{6} + 6 \, {\left (b^{3} c d^{5} p^{2} + 3 \, b^{3} c d^{5} p + 2 \, b^{3} c d^{5}\right )} x^{5} + {\left (30 \, b^{3} c^{2} d^{4} + {\left (15 \, b^{3} c^{2} + a b^{2}\right )} d^{4} p^{2} + {\left (45 \, b^{3} c^{2} + a b^{2}\right )} d^{4} p\right )} x^{4} + 4 \, {\left (10 \, b^{3} c^{3} d^{3} + {\left (5 \, b^{3} c^{3} + a b^{2} c\right )} d^{3} p^{2} + {\left (15 \, b^{3} c^{3} + a b^{2} c\right )} d^{3} p\right )} x^{3} + 2 \, a^{3} + {\left (b^{3} c^{6} + a b^{2} c^{4}\right )} p^{2} + {\left (30 \, b^{3} c^{4} d^{2} + 3 \, {\left (5 \, b^{3} c^{4} + 2 \, a b^{2} c^{2}\right )} d^{2} p^{2} + {\left (45 \, b^{3} c^{4} + 6 \, a b^{2} c^{2} - 2 \, a^{2} b\right )} d^{2} p\right )} x^{2} + {\left (3 \, b^{3} c^{6} + a b^{2} c^{4} - 2 \, a^{2} b c^{2}\right )} p + 2 \, {\left (6 \, b^{3} c^{5} d + {\left (3 \, b^{3} c^{5} + 2 \, a b^{2} c^{3}\right )} d p^{2} + {\left (9 \, b^{3} c^{5} + 2 \, a b^{2} c^{3} - 2 \, a^{2} b c\right )} d p\right )} x\right )} {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p}}{2 \, {\left (b^{3} d p^{3} + 6 \, b^{3} d p^{2} + 11 \, b^{3} d p + 6 \, b^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^5*(a+b*(d*x+c)^2)^p,x, algorithm="fricas")

[Out]

1/2*(2*b^3*c^6 + (b^3*d^6*p^2 + 3*b^3*d^6*p + 2*b^3*d^6)*x^6 + 6*(b^3*c*d^5*p^2 + 3*b^3*c*d^5*p + 2*b^3*c*d^5)
*x^5 + (30*b^3*c^2*d^4 + (15*b^3*c^2 + a*b^2)*d^4*p^2 + (45*b^3*c^2 + a*b^2)*d^4*p)*x^4 + 4*(10*b^3*c^3*d^3 +
(5*b^3*c^3 + a*b^2*c)*d^3*p^2 + (15*b^3*c^3 + a*b^2*c)*d^3*p)*x^3 + 2*a^3 + (b^3*c^6 + a*b^2*c^4)*p^2 + (30*b^
3*c^4*d^2 + 3*(5*b^3*c^4 + 2*a*b^2*c^2)*d^2*p^2 + (45*b^3*c^4 + 6*a*b^2*c^2 - 2*a^2*b)*d^2*p)*x^2 + (3*b^3*c^6
+ a*b^2*c^4 - 2*a^2*b*c^2)*p + 2*(6*b^3*c^5*d + (3*b^3*c^5 + 2*a*b^2*c^3)*d*p^2 + (9*b^3*c^5 + 2*a*b^2*c^3 -
2*a^2*b*c)*d*p)*x)*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p/(b^3*d*p^3 + 6*b^3*d*p^2 + 11*b^3*d*p + 6*b^3*d)

________________________________________________________________________________________

giac [B]  time = 0.25, size = 1290, normalized size = 13.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^5*(a+b*(d*x+c)^2)^p,x, algorithm="giac")

[Out]

1/2*((b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*d^6*p^2*x^6 + 6*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c*d^5
*p^2*x^5 + 3*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*d^6*p*x^6 + 15*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^
3*c^2*d^4*p^2*x^4 + 18*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c*d^5*p*x^5 + 2*(b*d^2*x^2 + 2*b*c*d*x + b*c^
2 + a)^p*b^3*d^6*x^6 + 20*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^3*d^3*p^2*x^3 + 45*(b*d^2*x^2 + 2*b*c*d*
x + b*c^2 + a)^p*b^3*c^2*d^4*p*x^4 + 12*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c*d^5*x^5 + 15*(b*d^2*x^2 +
2*b*c*d*x + b*c^2 + a)^p*b^3*c^4*d^2*p^2*x^2 + 60*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^3*d^3*p*x^3 + 30
*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^2*d^4*x^4 + (b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*d^4*p^2*x
^4 + 6*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^5*d*p^2*x + 45*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^
4*d^2*p*x^2 + 40*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^3*d^3*x^3 + 4*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)
^p*a*b^2*c*d^3*p^2*x^3 + (b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*d^4*p*x^4 + (b*d^2*x^2 + 2*b*c*d*x + b*c^
2 + a)^p*b^3*c^6*p^2 + 18*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^5*d*p*x + 30*(b*d^2*x^2 + 2*b*c*d*x + b*
c^2 + a)^p*b^3*c^4*d^2*x^2 + 6*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^2*d^2*p^2*x^2 + 4*(b*d^2*x^2 + 2*
b*c*d*x + b*c^2 + a)^p*a*b^2*c*d^3*p*x^3 + 3*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^6*p + 12*(b*d^2*x^2 +
2*b*c*d*x + b*c^2 + a)^p*b^3*c^5*d*x + 4*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^3*d*p^2*x + 6*(b*d^2*x
^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^2*d^2*p*x^2 + 2*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*b^3*c^6 + (b*d^2*x
^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^4*p^2 + 4*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^3*d*p*x + (b*d^2
*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a*b^2*c^4*p - 2*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a^2*b*d^2*p*x^2 - 4*(b*d
^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a^2*b*c*d*p*x - 2*(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a^2*b*c^2*p + 2*(b*d
^2*x^2 + 2*b*c*d*x + b*c^2 + a)^p*a^3)/(b^3*d*p^3 + 6*b^3*d*p^2 + 11*b^3*d*p + 6*b^3*d)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 289, normalized size = 3.11 \[ \frac {\left (b^{2} d^{4} p^{2} x^{4}+4 b^{2} c \,d^{3} p^{2} x^{3}+3 b^{2} d^{4} p \,x^{4}+6 b^{2} c^{2} d^{2} p^{2} x^{2}+12 b^{2} c \,d^{3} p \,x^{3}+2 d^{4} x^{4} b^{2}+4 b^{2} c^{3} d \,p^{2} x +18 b^{2} c^{2} d^{2} p \,x^{2}+8 c \,d^{3} x^{3} b^{2}+b^{2} c^{4} p^{2}+12 b^{2} c^{3} d p x +12 b^{2} c^{2} d^{2} x^{2}-2 a b \,d^{2} p \,x^{2}+3 b^{2} c^{4} p +8 b^{2} c^{3} d x -4 a b c d p x -2 a b \,d^{2} x^{2}+2 b^{2} c^{4}-2 a b \,c^{2} p -4 a b c d x -2 a b \,c^{2}+2 a^{2}\right ) \left (b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a \right )^{p +1}}{2 \left (p^{3}+6 p^{2}+11 p +6\right ) b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^5*(a+b*(d*x+c)^2)^p,x)

[Out]

1/2*(b*d^2*x^2+2*b*c*d*x+b*c^2+a)^(p+1)*(b^2*d^4*p^2*x^4+4*b^2*c*d^3*p^2*x^3+3*b^2*d^4*p*x^4+6*b^2*c^2*d^2*p^2
*x^2+12*b^2*c*d^3*p*x^3+2*b^2*d^4*x^4+4*b^2*c^3*d*p^2*x+18*b^2*c^2*d^2*p*x^2+8*b^2*c*d^3*x^3+b^2*c^4*p^2+12*b^
2*c^3*d*p*x+12*b^2*c^2*d^2*x^2-2*a*b*d^2*p*x^2+3*b^2*c^4*p+8*b^2*c^3*d*x-4*a*b*c*d*p*x-2*a*b*d^2*x^2+2*b^2*c^4
-2*a*b*c^2*p-4*a*b*c*d*x-2*a*b*c^2+2*a^2)/b^3/d/(p^3+6*p^2+11*p+6)

________________________________________________________________________________________

maxima [B]  time = 0.99, size = 300, normalized size = 3.23 \[ \frac {{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} d^{6} x^{6} + 6 \, {\left (p^{2} + 3 \, p + 2\right )} b^{3} c d^{5} x^{5} + {\left (p^{2} + 3 \, p + 2\right )} b^{3} c^{6} + {\left (p^{2} + p\right )} a b^{2} c^{4} - 2 \, a^{2} b c^{2} p + {\left (15 \, {\left (p^{2} + 3 \, p + 2\right )} b^{3} c^{2} d^{4} + {\left (p^{2} + p\right )} a b^{2} d^{4}\right )} x^{4} + 4 \, {\left (5 \, {\left (p^{2} + 3 \, p + 2\right )} b^{3} c^{3} d^{3} + {\left (p^{2} + p\right )} a b^{2} c d^{3}\right )} x^{3} + 2 \, a^{3} + {\left (15 \, {\left (p^{2} + 3 \, p + 2\right )} b^{3} c^{4} d^{2} + 6 \, {\left (p^{2} + p\right )} a b^{2} c^{2} d^{2} - 2 \, a^{2} b d^{2} p\right )} x^{2} + 2 \, {\left (3 \, {\left (p^{2} + 3 \, p + 2\right )} b^{3} c^{5} d + 2 \, {\left (p^{2} + p\right )} a b^{2} c^{3} d - 2 \, a^{2} b c d p\right )} x\right )} {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{p}}{2 \, {\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^5*(a+b*(d*x+c)^2)^p,x, algorithm="maxima")

[Out]

1/2*((p^2 + 3*p + 2)*b^3*d^6*x^6 + 6*(p^2 + 3*p + 2)*b^3*c*d^5*x^5 + (p^2 + 3*p + 2)*b^3*c^6 + (p^2 + p)*a*b^2
*c^4 - 2*a^2*b*c^2*p + (15*(p^2 + 3*p + 2)*b^3*c^2*d^4 + (p^2 + p)*a*b^2*d^4)*x^4 + 4*(5*(p^2 + 3*p + 2)*b^3*c
^3*d^3 + (p^2 + p)*a*b^2*c*d^3)*x^3 + 2*a^3 + (15*(p^2 + 3*p + 2)*b^3*c^4*d^2 + 6*(p^2 + p)*a*b^2*c^2*d^2 - 2*
a^2*b*d^2*p)*x^2 + 2*(3*(p^2 + 3*p + 2)*b^3*c^5*d + 2*(p^2 + p)*a*b^2*c^3*d - 2*a^2*b*c*d*p)*x)*(b*d^2*x^2 + 2
*b*c*d*x + b*c^2 + a)^p/((p^3 + 6*p^2 + 11*p + 6)*b^3*d)

________________________________________________________________________________________

mupad [B]  time = 1.57, size = 401, normalized size = 4.31 \[ {\left (a+b\,{\left (c+d\,x\right )}^2\right )}^p\,\left (\frac {d^5\,x^6\,\left (p^2+3\,p+2\right )}{2\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {\left (b\,c^2+a\right )\,\left (2\,a^2-2\,a\,b\,c^2\,p-2\,a\,b\,c^2+b^2\,c^4\,p^2+3\,b^2\,c^4\,p+2\,b^2\,c^4\right )}{2\,b^3\,d\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {c\,x\,\left (-2\,a^2\,p+2\,a\,b\,c^2\,p^2+2\,a\,b\,c^2\,p+3\,b^2\,c^4\,p^2+9\,b^2\,c^4\,p+6\,b^2\,c^4\right )}{b^2\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {d\,x^2\,\left (-2\,a^2\,p+6\,a\,b\,c^2\,p^2+6\,a\,b\,c^2\,p+15\,b^2\,c^4\,p^2+45\,b^2\,c^4\,p+30\,b^2\,c^4\right )}{2\,b^2\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {3\,c\,d^4\,x^5\,\left (p^2+3\,p+2\right )}{p^3+6\,p^2+11\,p+6}+\frac {d^3\,x^4\,\left (p+1\right )\,\left (a\,p+30\,b\,c^2+15\,b\,c^2\,p\right )}{2\,b\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {2\,c\,d^2\,x^3\,\left (p+1\right )\,\left (a\,p+10\,b\,c^2+5\,b\,c^2\,p\right )}{b\,\left (p^3+6\,p^2+11\,p+6\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c + d*x)^2)^p*(c + d*x)^5,x)

[Out]

(a + b*(c + d*x)^2)^p*((d^5*x^6*(3*p + p^2 + 2))/(2*(11*p + 6*p^2 + p^3 + 6)) + ((a + b*c^2)*(2*a^2 + 2*b^2*c^
4 + 3*b^2*c^4*p + b^2*c^4*p^2 - 2*a*b*c^2 - 2*a*b*c^2*p))/(2*b^3*d*(11*p + 6*p^2 + p^3 + 6)) + (c*x*(6*b^2*c^4
- 2*a^2*p + 9*b^2*c^4*p + 3*b^2*c^4*p^2 + 2*a*b*c^2*p + 2*a*b*c^2*p^2))/(b^2*(11*p + 6*p^2 + p^3 + 6)) + (d*x
^2*(30*b^2*c^4 - 2*a^2*p + 45*b^2*c^4*p + 15*b^2*c^4*p^2 + 6*a*b*c^2*p + 6*a*b*c^2*p^2))/(2*b^2*(11*p + 6*p^2
+ p^3 + 6)) + (3*c*d^4*x^5*(3*p + p^2 + 2))/(11*p + 6*p^2 + p^3 + 6) + (d^3*x^4*(p + 1)*(a*p + 30*b*c^2 + 15*b
*c^2*p))/(2*b*(11*p + 6*p^2 + p^3 + 6)) + (2*c*d^2*x^3*(p + 1)*(a*p + 10*b*c^2 + 5*b*c^2*p))/(b*(11*p + 6*p^2
+ p^3 + 6)))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**5*(a+b*(d*x+c)**2)**p,x)

[Out]

Timed out

________________________________________________________________________________________