\(\int (a x^m+b x^{1+6 m})^5 \, dx\) [336]

Optimal result

Integrand size = 17, antiderivative size = 27 \[ \int \left (a x^m+b x^{1+6 m}\right )^5 \, dx=\frac {\left (a+b x^{1+5 m}\right )^6}{6 b (1+5 m)} \]

[Out]

1/6*(a+b*x^(1+5*m))^6/b/(1+5*m)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1607, 267} \[ \int \left (a x^m+b x^{1+6 m}\right )^5 \, dx=\frac {\left (a+b x^{5 m+1}\right )^6}{6 b (5 m+1)} \]

[In]

Int[(a*x^m + b*x^(1 + 6*m))^5,x]

[Out]

(a + b*x^(1 + 5*m))^6/(6*b*(1 + 5*m))

Rule 267

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

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int x^{5 m} \left (a+b x^{1+5 m}\right )^5 \, dx \\ & = \frac {\left (a+b x^{1+5 m}\right )^6}{6 b (1+5 m)} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(88\) vs. \(2(27)=54\).

Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.26 \[ \int \left (a x^m+b x^{1+6 m}\right )^5 \, dx=\frac {x^{1+5 m} \left (6 a^5+15 a^4 b x^{1+5 m}+20 a^3 b^2 x^{2+10 m}+15 a^2 b^3 x^{3+15 m}+6 a b^4 x^{4+20 m}+b^5 x^{5+25 m}\right )}{6+30 m} \]

[In]

Integrate[(a*x^m + b*x^(1 + 6*m))^5,x]

[Out]

(x^(1 + 5*m)*(6*a^5 + 15*a^4*b*x^(1 + 5*m) + 20*a^3*b^2*x^(2 + 10*m) + 15*a^2*b^3*x^(3 + 15*m) + 6*a*b^4*x^(4
+ 20*m) + b^5*x^(5 + 25*m)))/(6 + 30*m)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(119\) vs. \(2(25)=50\).

Time = 2.24 (sec) , antiderivative size = 120, normalized size of antiderivative = 4.44

method	result	size
parallelrisch	\(\frac {6 x \,x^{5 m} a^{5}+15 x \,x^{4 m} x^{1+6 m} a^{4} b +20 x \,x^{3 m} x^{2+12 m} a^{3} b^{2}+15 x \,x^{2 m} x^{3+18 m} a^{2} b^{3}+6 x \,x^{m} x^{4+24 m} a \,b^{4}+x \,x^{5+30 m} b^{5}}{6+30 m}\)	\(120\)
risch	\(\frac {b^{5} x^{6} x^{30 m}}{6+30 m}+\frac {a \,b^{4} x^{5} x^{25 m}}{1+5 m}+\frac {5 a^{2} b^{3} x^{4} x^{20 m}}{2 \left (1+5 m \right )}+\frac {10 a^{3} b^{2} x^{3} x^{15 m}}{3 \left (1+5 m \right )}+\frac {5 a^{4} b \,x^{2} x^{10 m}}{2 \left (1+5 m \right )}+\frac {a^{5} x \,x^{5 m}}{1+5 m}\)	\(126\)

[In]

int((x^m*a+b*x^(1+6*m))^5,x,method=_RETURNVERBOSE)

[Out]

1/6*(6*x*(x^m)^5*a^5+15*x*(x^m)^4*x^(1+6*m)*a^4*b+20*x*(x^m)^3*(x^(1+6*m))^2*a^3*b^2+15*x*(x^m)^2*(x^(1+6*m))^
3*a^2*b^3+6*x*x^m*(x^(1+6*m))^4*a*b^4+x*(x^(1+6*m))^5*b^5)/(1+5*m)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (25) = 50\).

Time = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.44 \[ \int \left (a x^m+b x^{1+6 m}\right )^5 \, dx=\frac {b^{5} x^{6} x^{30 \, m} + 6 \, a b^{4} x^{5} x^{25 \, m} + 15 \, a^{2} b^{3} x^{4} x^{20 \, m} + 20 \, a^{3} b^{2} x^{3} x^{15 \, m} + 15 \, a^{4} b x^{2} x^{10 \, m} + 6 \, a^{5} x x^{5 \, m}}{6 \, {\left (5 \, m + 1\right )}} \]

[In]

integrate((a*x^m+b*x^(1+6*m))^5,x, algorithm="fricas")

[Out]

1/6*(b^5*x^6*x^(30*m) + 6*a*b^4*x^5*x^(25*m) + 15*a^2*b^3*x^4*x^(20*m) + 20*a^3*b^2*x^3*x^(15*m) + 15*a^4*b*x^
2*x^(10*m) + 6*a^5*x*x^(5*m))/(5*m + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (19) = 38\).

Time = 0.63 (sec) , antiderivative size = 197, normalized size of antiderivative = 7.30 \[ \int \left (a x^m+b x^{1+6 m}\right )^5 \, dx=\begin {cases} \frac {6 a^{5} x x^{5 m}}{30 m + 6} + \frac {15 a^{4} b x x^{4 m} x^{6 m + 1}}{30 m + 6} + \frac {20 a^{3} b^{2} x x^{3 m} x^{12 m + 2}}{30 m + 6} + \frac {15 a^{2} b^{3} x x^{2 m} x^{18 m + 3}}{30 m + 6} + \frac {6 a b^{4} x x^{m} x^{24 m + 4}}{30 m + 6} + \frac {b^{5} x x^{30 m + 5}}{30 m + 6} & \text {for}\: m \neq - \frac {1}{5} \\a^{5} \log {\left (x \right )} + 5 a^{4} b \log {\left (x \right )} + 10 a^{3} b^{2} \log {\left (x \right )} + 10 a^{2} b^{3} \log {\left (x \right )} + 5 a b^{4} \log {\left (x \right )} + b^{5} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate((a*x**m+b*x**(1+6*m))**5,x)

[Out]

Piecewise((6*a**5*x*x**(5*m)/(30*m + 6) + 15*a**4*b*x*x**(4*m)*x**(6*m + 1)/(30*m + 6) + 20*a**3*b**2*x*x**(3*
m)*x**(12*m + 2)/(30*m + 6) + 15*a**2*b**3*x*x**(2*m)*x**(18*m + 3)/(30*m + 6) + 6*a*b**4*x*x**m*x**(24*m + 4)
/(30*m + 6) + b**5*x*x**(30*m + 5)/(30*m + 6), Ne(m, -1/5)), (a**5*log(x) + 5*a**4*b*log(x) + 10*a**3*b**2*log
(x) + 10*a**2*b**3*log(x) + 5*a*b**4*log(x) + b**5*log(x), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (25) = 50\).

Time = 0.20 (sec) , antiderivative size = 121, normalized size of antiderivative = 4.48 \[ \int \left (a x^m+b x^{1+6 m}\right )^5 \, dx=\frac {b^{5} x^{30 \, m + 6}}{6 \, {\left (5 \, m + 1\right )}} + \frac {a b^{4} x^{25 \, m + 5}}{5 \, m + 1} + \frac {5 \, a^{2} b^{3} x^{20 \, m + 4}}{2 \, {\left (5 \, m + 1\right )}} + \frac {10 \, a^{3} b^{2} x^{15 \, m + 3}}{3 \, {\left (5 \, m + 1\right )}} + \frac {5 \, a^{4} b x^{10 \, m + 2}}{2 \, {\left (5 \, m + 1\right )}} + \frac {a^{5} x^{5 \, m + 1}}{5 \, m + 1} \]

[In]

integrate((a*x^m+b*x^(1+6*m))^5,x, algorithm="maxima")

[Out]

1/6*b^5*x^(30*m + 6)/(5*m + 1) + a*b^4*x^(25*m + 5)/(5*m + 1) + 5/2*a^2*b^3*x^(20*m + 4)/(5*m + 1) + 10/3*a^3*
b^2*x^(15*m + 3)/(5*m + 1) + 5/2*a^4*b*x^(10*m + 2)/(5*m + 1) + a^5*x^(5*m + 1)/(5*m + 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (25) = 50\).

Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.44 \[ \int \left (a x^m+b x^{1+6 m}\right )^5 \, dx=\frac {b^{5} x^{6} x^{30 \, m} + 6 \, a b^{4} x^{5} x^{25 \, m} + 15 \, a^{2} b^{3} x^{4} x^{20 \, m} + 20 \, a^{3} b^{2} x^{3} x^{15 \, m} + 15 \, a^{4} b x^{2} x^{10 \, m} + 6 \, a^{5} x x^{5 \, m}}{6 \, {\left (5 \, m + 1\right )}} \]

[In]

integrate((a*x^m+b*x^(1+6*m))^5,x, algorithm="giac")

[Out]

1/6*(b^5*x^6*x^(30*m) + 6*a*b^4*x^5*x^(25*m) + 15*a^2*b^3*x^4*x^(20*m) + 20*a^3*b^2*x^3*x^(15*m) + 15*a^4*b*x^
2*x^(10*m) + 6*a^5*x*x^(5*m))/(5*m + 1)

Mupad [B] (verification not implemented)

Time = 9.34 (sec) , antiderivative size = 124, normalized size of antiderivative = 4.59 \[ \int \left (a x^m+b x^{1+6 m}\right )^5 \, dx=\frac {b^5\,x^{30\,m}\,x^6}{30\,m+6}+\frac {a^5\,x\,x^{5\,m}}{5\,m+1}+\frac {5\,a^4\,b\,x^{10\,m}\,x^2}{10\,m+2}+\frac {a\,b^4\,x^{25\,m}\,x^5}{5\,m+1}+\frac {5\,a^2\,b^3\,x^{20\,m}\,x^4}{10\,m+2}+\frac {10\,a^3\,b^2\,x^{15\,m}\,x^3}{15\,m+3} \]

[In]

int((a*x^m + b*x^(6*m + 1))^5,x)

[Out]

(b^5*x^(30*m)*x^6)/(30*m + 6) + (a^5*x*x^(5*m))/(5*m + 1) + (5*a^4*b*x^(10*m)*x^2)/(10*m + 2) + (a*b^4*x^(25*m
)*x^5)/(5*m + 1) + (5*a^2*b^3*x^(20*m)*x^4)/(10*m + 2) + (10*a^3*b^2*x^(15*m)*x^3)/(15*m + 3)