Integrand size = 33, antiderivative size = 27 \[ \int (a d+b d x)^m \left (a^2 c+2 a b c x+b^2 c x^2\right )^2 \, dx=\frac {c^2 (a d+b d x)^{5+m}}{b d^5 (5+m)} \] Output:
c^2*(b*d*x+a*d)^(5+m)/b/d^5/(5+m)
Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int (a d+b d x)^m \left (a^2 c+2 a b c x+b^2 c x^2\right )^2 \, dx=\frac {c^2 (a+b x)^5 (d (a+b x))^m}{b (5+m)} \] Input:
Integrate[(a*d + b*d*x)^m*(a^2*c + 2*a*b*c*x + b^2*c*x^2)^2,x]
Output:
(c^2*(a + b*x)^5*(d*(a + b*x))^m)/(b*(5 + m))
Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1098, 27, 35, 17}
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.
\(\displaystyle \int \left (a^2 c+2 a b c x+b^2 c x^2\right )^2 (a d+b d x)^m \, dx\) |
\(\Big \downarrow \) 1098 |
\(\displaystyle \frac {\int b^4 c^4 (a+b x)^4 (a d+b x d)^mdx}{b^4 c^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^2 \int (a+b x)^4 (a d+b x d)^mdx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle \frac {c^2 \int (a d+b x d)^{m+4}dx}{d^4}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {c^2 (a d+b d x)^{m+5}}{b d^5 (m+5)}\) |
Input:
Int[(a*d + b*d*x)^m*(a^2*c + 2*a*b*c*x + b^2*c*x^2)^2,x]
Output:
(c^2*(a*d + b*d*x)^(5 + m))/(b*d^5*(5 + m))
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} , x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && !(IntegerQ[n] && SimplerQ[a + b*x, c + d*x])
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[ {a, b, c, d, e, m}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 1.34 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70
method | result | size |
gosper | \(\frac {\left (b x +a \right ) \left (b d x +a d \right )^{m} c^{2} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}{b \left (5+m \right )}\) | \(46\) |
orering | \(\frac {\left (b x +a \right ) \left (b d x +a d \right )^{m} \left (b^{2} c \,x^{2}+2 a b c x +a^{2} c \right )^{2}}{b \left (5+m \right )}\) | \(47\) |
risch | \(\frac {c^{2} \left (b^{5} x^{5}+5 a \,b^{4} x^{4}+10 a^{2} b^{3} x^{3}+10 a^{3} b^{2} x^{2}+5 a^{4} b x +a^{5}\right ) \left (d \left (b x +a \right )\right )^{m}}{b \left (5+m \right )}\) | \(71\) |
parallelrisch | \(\frac {x^{5} \left (d \left (b x +a \right )\right )^{m} a \,b^{5} c^{2}+5 x^{4} \left (d \left (b x +a \right )\right )^{m} a^{2} b^{4} c^{2}+10 x^{3} \left (d \left (b x +a \right )\right )^{m} a^{3} b^{3} c^{2}+10 x^{2} \left (d \left (b x +a \right )\right )^{m} a^{4} b^{2} c^{2}+5 x \left (d \left (b x +a \right )\right )^{m} a^{5} b \,c^{2}+\left (d \left (b x +a \right )\right )^{m} a^{6} c^{2}}{a \left (5+m \right ) b}\) | \(138\) |
norman | \(\frac {a^{5} c^{2} {\mathrm e}^{m \ln \left (b d x +a d \right )}}{b \left (5+m \right )}+\frac {b^{4} c^{2} x^{5} {\mathrm e}^{m \ln \left (b d x +a d \right )}}{5+m}+\frac {5 a^{4} c^{2} x \,{\mathrm e}^{m \ln \left (b d x +a d \right )}}{5+m}+\frac {5 a \,b^{3} c^{2} x^{4} {\mathrm e}^{m \ln \left (b d x +a d \right )}}{5+m}+\frac {10 a^{2} b^{2} c^{2} x^{3} {\mathrm e}^{m \ln \left (b d x +a d \right )}}{5+m}+\frac {10 a^{3} b \,c^{2} x^{2} {\mathrm e}^{m \ln \left (b d x +a d \right )}}{5+m}\) | \(171\) |
Input:
int((b*d*x+a*d)^m*(b^2*c*x^2+2*a*b*c*x+a^2*c)^2,x,method=_RETURNVERBOSE)
Output:
(b*x+a)*(b*d*x+a*d)^m*c^2*(b^2*x^2+2*a*b*x+a^2)^2/b/(5+m)
Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (27) = 54\).
Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.26 \[ \int (a d+b d x)^m \left (a^2 c+2 a b c x+b^2 c x^2\right )^2 \, dx=\frac {{\left (b^{5} c^{2} x^{5} + 5 \, a b^{4} c^{2} x^{4} + 10 \, a^{2} b^{3} c^{2} x^{3} + 10 \, a^{3} b^{2} c^{2} x^{2} + 5 \, a^{4} b c^{2} x + a^{5} c^{2}\right )} {\left (b d x + a d\right )}^{m}}{b m + 5 \, b} \] Input:
integrate((b*d*x+a*d)^m*(b^2*c*x^2+2*a*b*c*x+a^2*c)^2,x, algorithm="fricas ")
Output:
(b^5*c^2*x^5 + 5*a*b^4*c^2*x^4 + 10*a^2*b^3*c^2*x^3 + 10*a^3*b^2*c^2*x^2 + 5*a^4*b*c^2*x + a^5*c^2)*(b*d*x + a*d)^m/(b*m + 5*b)
Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (22) = 44\).
Time = 0.55 (sec) , antiderivative size = 214, normalized size of antiderivative = 7.93 \[ \int (a d+b d x)^m \left (a^2 c+2 a b c x+b^2 c x^2\right )^2 \, dx=\begin {cases} \frac {c^{2} x}{a d^{5}} & \text {for}\: b = 0 \wedge m = -5 \\a^{4} c^{2} x \left (a d\right )^{m} & \text {for}\: b = 0 \\\frac {c^{2} \log {\left (\frac {a}{b} + x \right )}}{b d^{5}} & \text {for}\: m = -5 \\\frac {a^{5} c^{2} \left (a d + b d x\right )^{m}}{b m + 5 b} + \frac {5 a^{4} b c^{2} x \left (a d + b d x\right )^{m}}{b m + 5 b} + \frac {10 a^{3} b^{2} c^{2} x^{2} \left (a d + b d x\right )^{m}}{b m + 5 b} + \frac {10 a^{2} b^{3} c^{2} x^{3} \left (a d + b d x\right )^{m}}{b m + 5 b} + \frac {5 a b^{4} c^{2} x^{4} \left (a d + b d x\right )^{m}}{b m + 5 b} + \frac {b^{5} c^{2} x^{5} \left (a d + b d x\right )^{m}}{b m + 5 b} & \text {otherwise} \end {cases} \] Input:
integrate((b*d*x+a*d)**m*(b**2*c*x**2+2*a*b*c*x+a**2*c)**2,x)
Output:
Piecewise((c**2*x/(a*d**5), Eq(b, 0) & Eq(m, -5)), (a**4*c**2*x*(a*d)**m, Eq(b, 0)), (c**2*log(a/b + x)/(b*d**5), Eq(m, -5)), (a**5*c**2*(a*d + b*d* x)**m/(b*m + 5*b) + 5*a**4*b*c**2*x*(a*d + b*d*x)**m/(b*m + 5*b) + 10*a**3 *b**2*c**2*x**2*(a*d + b*d*x)**m/(b*m + 5*b) + 10*a**2*b**3*c**2*x**3*(a*d + b*d*x)**m/(b*m + 5*b) + 5*a*b**4*c**2*x**4*(a*d + b*d*x)**m/(b*m + 5*b) + b**5*c**2*x**5*(a*d + b*d*x)**m/(b*m + 5*b), True))
Leaf count of result is larger than twice the leaf count of optimal. 457 vs. \(2 (27) = 54\).
Time = 0.07 (sec) , antiderivative size = 457, normalized size of antiderivative = 16.93 \[ \int (a d+b d x)^m \left (a^2 c+2 a b c x+b^2 c x^2\right )^2 \, dx=\frac {4 \, {\left (b^{2} d^{m} {\left (m + 1\right )} x^{2} + a b d^{m} m x - a^{2} d^{m}\right )} {\left (b x + a\right )}^{m} a^{3} c^{2}}{{\left (m^{2} + 3 \, m + 2\right )} b} + \frac {6 \, {\left ({\left (m^{2} + 3 \, m + 2\right )} b^{3} d^{m} x^{3} + {\left (m^{2} + m\right )} a b^{2} d^{m} x^{2} - 2 \, a^{2} b d^{m} m x + 2 \, a^{3} d^{m}\right )} {\left (b x + a\right )}^{m} a^{2} c^{2}}{{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} b} + \frac {{\left (b d x + a d\right )}^{m + 1} a^{4} c^{2}}{b d {\left (m + 1\right )}} + \frac {4 \, {\left ({\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} b^{4} d^{m} x^{4} + {\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} a b^{3} d^{m} x^{3} - 3 \, {\left (m^{2} + m\right )} a^{2} b^{2} d^{m} x^{2} + 6 \, a^{3} b d^{m} m x - 6 \, a^{4} d^{m}\right )} {\left (b x + a\right )}^{m} a c^{2}}{{\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} b} + \frac {{\left ({\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} b^{5} d^{m} x^{5} + {\left (m^{4} + 6 \, m^{3} + 11 \, m^{2} + 6 \, m\right )} a b^{4} d^{m} x^{4} - 4 \, {\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} a^{2} b^{3} d^{m} x^{3} + 12 \, {\left (m^{2} + m\right )} a^{3} b^{2} d^{m} x^{2} - 24 \, a^{4} b d^{m} m x + 24 \, a^{5} d^{m}\right )} {\left (b x + a\right )}^{m} c^{2}}{{\left (m^{5} + 15 \, m^{4} + 85 \, m^{3} + 225 \, m^{2} + 274 \, m + 120\right )} b} \] Input:
integrate((b*d*x+a*d)^m*(b^2*c*x^2+2*a*b*c*x+a^2*c)^2,x, algorithm="maxima ")
Output:
4*(b^2*d^m*(m + 1)*x^2 + a*b*d^m*m*x - a^2*d^m)*(b*x + a)^m*a^3*c^2/((m^2 + 3*m + 2)*b) + 6*((m^2 + 3*m + 2)*b^3*d^m*x^3 + (m^2 + m)*a*b^2*d^m*x^2 - 2*a^2*b*d^m*m*x + 2*a^3*d^m)*(b*x + a)^m*a^2*c^2/((m^3 + 6*m^2 + 11*m + 6 )*b) + (b*d*x + a*d)^(m + 1)*a^4*c^2/(b*d*(m + 1)) + 4*((m^3 + 6*m^2 + 11* m + 6)*b^4*d^m*x^4 + (m^3 + 3*m^2 + 2*m)*a*b^3*d^m*x^3 - 3*(m^2 + m)*a^2*b ^2*d^m*x^2 + 6*a^3*b*d^m*m*x - 6*a^4*d^m)*(b*x + a)^m*a*c^2/((m^4 + 10*m^3 + 35*m^2 + 50*m + 24)*b) + ((m^4 + 10*m^3 + 35*m^2 + 50*m + 24)*b^5*d^m*x ^5 + (m^4 + 6*m^3 + 11*m^2 + 6*m)*a*b^4*d^m*x^4 - 4*(m^3 + 3*m^2 + 2*m)*a^ 2*b^3*d^m*x^3 + 12*(m^2 + m)*a^3*b^2*d^m*x^2 - 24*a^4*b*d^m*m*x + 24*a^5*d ^m)*(b*x + a)^m*c^2/((m^5 + 15*m^4 + 85*m^3 + 225*m^2 + 274*m + 120)*b)
Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (27) = 54\).
Time = 0.13 (sec) , antiderivative size = 138, normalized size of antiderivative = 5.11 \[ \int (a d+b d x)^m \left (a^2 c+2 a b c x+b^2 c x^2\right )^2 \, dx=\frac {{\left (b d x + a d\right )}^{m} b^{5} c^{2} x^{5} + 5 \, {\left (b d x + a d\right )}^{m} a b^{4} c^{2} x^{4} + 10 \, {\left (b d x + a d\right )}^{m} a^{2} b^{3} c^{2} x^{3} + 10 \, {\left (b d x + a d\right )}^{m} a^{3} b^{2} c^{2} x^{2} + 5 \, {\left (b d x + a d\right )}^{m} a^{4} b c^{2} x + {\left (b d x + a d\right )}^{m} a^{5} c^{2}}{b m + 5 \, b} \] Input:
integrate((b*d*x+a*d)^m*(b^2*c*x^2+2*a*b*c*x+a^2*c)^2,x, algorithm="giac")
Output:
((b*d*x + a*d)^m*b^5*c^2*x^5 + 5*(b*d*x + a*d)^m*a*b^4*c^2*x^4 + 10*(b*d*x + a*d)^m*a^2*b^3*c^2*x^3 + 10*(b*d*x + a*d)^m*a^3*b^2*c^2*x^2 + 5*(b*d*x + a*d)^m*a^4*b*c^2*x + (b*d*x + a*d)^m*a^5*c^2)/(b*m + 5*b)
Time = 5.60 (sec) , antiderivative size = 109, normalized size of antiderivative = 4.04 \[ \int (a d+b d x)^m \left (a^2 c+2 a b c x+b^2 c x^2\right )^2 \, dx={\left (a\,d+b\,d\,x\right )}^m\,\left (\frac {5\,a^4\,c^2\,x}{m+5}+\frac {a^5\,c^2}{b\,\left (m+5\right )}+\frac {b^4\,c^2\,x^5}{m+5}+\frac {10\,a^3\,b\,c^2\,x^2}{m+5}+\frac {5\,a\,b^3\,c^2\,x^4}{m+5}+\frac {10\,a^2\,b^2\,c^2\,x^3}{m+5}\right ) \] Input:
int((a*d + b*d*x)^m*(a^2*c + b^2*c*x^2 + 2*a*b*c*x)^2,x)
Output:
(a*d + b*d*x)^m*((5*a^4*c^2*x)/(m + 5) + (a^5*c^2)/(b*(m + 5)) + (b^4*c^2* x^5)/(m + 5) + (10*a^3*b*c^2*x^2)/(m + 5) + (5*a*b^3*c^2*x^4)/(m + 5) + (1 0*a^2*b^2*c^2*x^3)/(m + 5))
Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.63 \[ \int (a d+b d x)^m \left (a^2 c+2 a b c x+b^2 c x^2\right )^2 \, dx=\frac {\left (b d x +a d \right )^{m} c^{2} \left (b^{5} x^{5}+5 a \,b^{4} x^{4}+10 a^{2} b^{3} x^{3}+10 a^{3} b^{2} x^{2}+5 a^{4} b x +a^{5}\right )}{b \left (m +5\right )} \] Input:
int((b*d*x+a*d)^m*(b^2*c*x^2+2*a*b*c*x+a^2*c)^2,x)
Output:
((a*d + b*d*x)**m*c**2*(a**5 + 5*a**4*b*x + 10*a**3*b**2*x**2 + 10*a**2*b* *3*x**3 + 5*a*b**4*x**4 + b**5*x**5))/(b*(m + 5))