Integrand size = 15, antiderivative size = 87 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {a^4 x^{1+m}}{1+m}+\frac {8 a^3 b x^{\frac {3}{2}+m}}{3+2 m}+\frac {6 a^2 b^2 x^{2+m}}{2+m}+\frac {8 a b^3 x^{\frac {5}{2}+m}}{5+2 m}+\frac {b^4 x^{3+m}}{3+m} \] Output:
a^4*x^(1+m)/(1+m)+8*a^3*b*x^(3/2+m)/(3+2*m)+6*a^2*b^2*x^(2+m)/(2+m)+8*a*b^ 3*x^(5/2+m)/(5+2*m)+b^4*x^(3+m)/(3+m)
Time = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.90 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=x^{1+m} \left (\frac {a^4}{1+m}+\frac {8 a^3 b \sqrt {x}}{3+2 m}+\frac {6 a^2 b^2 x}{2+m}+\frac {8 a b^3 x^{3/2}}{5+2 m}+\frac {b^4 x^2}{3+m}\right ) \] Input:
Integrate[(a + b*Sqrt[x])^4*x^m,x]
Output:
x^(1 + m)*(a^4/(1 + m) + (8*a^3*b*Sqrt[x])/(3 + 2*m) + (6*a^2*b^2*x)/(2 + m) + (8*a*b^3*x^(3/2))/(5 + 2*m) + (b^4*x^2)/(3 + m))
Time = 0.39 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {802, 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.
\(\displaystyle \int x^m \left (a+b \sqrt {x}\right )^4 \, dx\) |
\(\Big \downarrow \) 802 |
\(\displaystyle \int \left (a^4 x^m+4 a^3 b x^{m+\frac {1}{2}}+6 a^2 b^2 x^{m+1}+4 a b^3 x^{m+\frac {3}{2}}+b^4 x^{m+2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^4 x^{m+1}}{m+1}+\frac {8 a^3 b x^{m+\frac {3}{2}}}{2 m+3}+\frac {6 a^2 b^2 x^{m+2}}{m+2}+\frac {8 a b^3 x^{m+\frac {5}{2}}}{2 m+5}+\frac {b^4 x^{m+3}}{m+3}\) |
Input:
Int[(a + b*Sqrt[x])^4*x^m,x]
Output:
(a^4*x^(1 + m))/(1 + m) + (8*a^3*b*x^(3/2 + m))/(3 + 2*m) + (6*a^2*b^2*x^( 2 + m))/(2 + m) + (8*a*b^3*x^(5/2 + m))/(5 + 2*m) + (b^4*x^(3 + m))/(3 + m )
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[Exp andIntegrand[(c*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(505\) vs. \(2(83)=166\).
Time = 0.73 (sec) , antiderivative size = 506, normalized size of antiderivative = 5.82
method | result | size |
orering | \(\frac {\left (-8 b^{6} m^{4} x^{3}+24 a^{2} b^{4} m^{4} x^{2}-54 b^{6} m^{3} x^{3}-24 a^{4} b^{2} m^{4} x +174 a^{2} b^{4} m^{3} x^{2}-133 b^{6} m^{2} x^{3}+8 a^{6} m^{4}-186 a^{4} b^{2} m^{3} x +435 a^{2} b^{4} m^{2} x^{2}-141 x^{3} b^{6} m +66 a^{6} m^{3}-495 a^{4} b^{2} m^{2} x +453 x^{2} b^{4} a^{2} m -54 b^{6} x^{3}+193 a^{6} m^{2}-519 x \,b^{2} a^{4} m +168 a^{2} b^{4} x^{2}+231 a^{6} m -180 a^{4} b^{2} x +90 a^{6}\right ) x \left (a +b \sqrt {x}\right )^{4} x^{m}}{\left (2 m^{2}+5 m +3\right ) \left (2 m^{2}+9 m +10\right ) \left (3+m \right ) \left (-b^{2} x +a^{2}\right )^{3}}-\frac {2 \left (-2 b^{6} m^{3} x^{3}+6 a^{2} b^{4} m^{3} x^{2}-9 b^{6} m^{2} x^{3}-6 a^{4} b^{2} m^{3} x +33 a^{2} b^{4} m^{2} x^{2}-13 x^{3} b^{6} m +2 a^{6} m^{3}-39 a^{4} b^{2} m^{2} x +51 x^{2} b^{4} a^{2} m -6 b^{6} x^{3}+15 a^{6} m^{2}-75 x \,b^{2} a^{4} m +24 a^{2} b^{4} x^{2}+37 a^{6} m -36 a^{4} b^{2} x +30 a^{6}\right ) x^{2} \left (\frac {2 \left (a +b \sqrt {x}\right )^{3} x^{m} b}{\sqrt {x}}+\frac {\left (a +b \sqrt {x}\right )^{4} x^{m} m}{x}\right )}{\left (2 m^{2}+5 m +3\right ) \left (2 m^{2}+9 m +10\right ) \left (3+m \right ) \left (-b^{2} x +a^{2}\right )^{3}}\) | \(506\) |
Input:
int((a+b*x^(1/2))^4*x^m,x,method=_RETURNVERBOSE)
Output:
(-8*b^6*m^4*x^3+24*a^2*b^4*m^4*x^2-54*b^6*m^3*x^3-24*a^4*b^2*m^4*x+174*a^2 *b^4*m^3*x^2-133*b^6*m^2*x^3+8*a^6*m^4-186*a^4*b^2*m^3*x+435*a^2*b^4*m^2*x ^2-141*b^6*m*x^3+66*a^6*m^3-495*a^4*b^2*m^2*x+453*a^2*b^4*m*x^2-54*b^6*x^3 +193*a^6*m^2-519*a^4*b^2*m*x+168*a^2*b^4*x^2+231*a^6*m-180*a^4*b^2*x+90*a^ 6)*x/(2*m^2+5*m+3)/(2*m^2+9*m+10)/(3+m)/(-b^2*x+a^2)^3*(a+b*x^(1/2))^4*x^m -2*(-2*b^6*m^3*x^3+6*a^2*b^4*m^3*x^2-9*b^6*m^2*x^3-6*a^4*b^2*m^3*x+33*a^2* b^4*m^2*x^2-13*b^6*m*x^3+2*a^6*m^3-39*a^4*b^2*m^2*x+51*a^2*b^4*m*x^2-6*b^6 *x^3+15*a^6*m^2-75*a^4*b^2*m*x+24*a^2*b^4*x^2+37*a^6*m-36*a^4*b^2*x+30*a^6 )/(2*m^2+5*m+3)/(2*m^2+9*m+10)/(3+m)/(-b^2*x+a^2)^3*x^2*(2*(a+b*x^(1/2))^3 *x^m*b/x^(1/2)+(a+b*x^(1/2))^4*x^m*m/x)
Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (83) = 166\).
Time = 0.09 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.99 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {{\left ({\left (4 \, b^{4} m^{4} + 28 \, b^{4} m^{3} + 71 \, b^{4} m^{2} + 77 \, b^{4} m + 30 \, b^{4}\right )} x^{3} + 6 \, {\left (4 \, a^{2} b^{2} m^{4} + 32 \, a^{2} b^{2} m^{3} + 91 \, a^{2} b^{2} m^{2} + 108 \, a^{2} b^{2} m + 45 \, a^{2} b^{2}\right )} x^{2} + {\left (4 \, a^{4} m^{4} + 36 \, a^{4} m^{3} + 119 \, a^{4} m^{2} + 171 \, a^{4} m + 90 \, a^{4}\right )} x + 8 \, {\left ({\left (2 \, a b^{3} m^{4} + 15 \, a b^{3} m^{3} + 40 \, a b^{3} m^{2} + 45 \, a b^{3} m + 18 \, a b^{3}\right )} x^{2} + {\left (2 \, a^{3} b m^{4} + 17 \, a^{3} b m^{3} + 52 \, a^{3} b m^{2} + 67 \, a^{3} b m + 30 \, a^{3} b\right )} x\right )} \sqrt {x}\right )} x^{m}}{4 \, m^{5} + 40 \, m^{4} + 155 \, m^{3} + 290 \, m^{2} + 261 \, m + 90} \] Input:
integrate((a+b*x^(1/2))^4*x^m,x, algorithm="fricas")
Output:
((4*b^4*m^4 + 28*b^4*m^3 + 71*b^4*m^2 + 77*b^4*m + 30*b^4)*x^3 + 6*(4*a^2* b^2*m^4 + 32*a^2*b^2*m^3 + 91*a^2*b^2*m^2 + 108*a^2*b^2*m + 45*a^2*b^2)*x^ 2 + (4*a^4*m^4 + 36*a^4*m^3 + 119*a^4*m^2 + 171*a^4*m + 90*a^4)*x + 8*((2* a*b^3*m^4 + 15*a*b^3*m^3 + 40*a*b^3*m^2 + 45*a*b^3*m + 18*a*b^3)*x^2 + (2* a^3*b*m^4 + 17*a^3*b*m^3 + 52*a^3*b*m^2 + 67*a^3*b*m + 30*a^3*b)*x)*sqrt(x ))*x^m/(4*m^5 + 40*m^4 + 155*m^3 + 290*m^2 + 261*m + 90)
Time = 1.49 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.72 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=a^{4} \left (\begin {cases} \frac {x^{m + 1}}{m + 1} & \text {for}\: m \neq -1 \\\log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + 8 a^{3} b \left (\begin {cases} \frac {x^{\frac {3}{2}} x^{m}}{2 m + 3} & \text {for}\: m \neq - \frac {3}{2} \\x^{\frac {3}{2}} x^{m} \log {\left (\sqrt {x} \right )} & \text {otherwise} \end {cases}\right ) + 6 a^{2} b^{2} \left (\begin {cases} \frac {x^{2} x^{m}}{m + 2} & \text {for}\: m \neq -2 \\x^{2} x^{m} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + 8 a b^{3} \left (\begin {cases} \frac {x^{\frac {5}{2}} x^{m}}{2 m + 5} & \text {for}\: m \neq - \frac {5}{2} \\x^{\frac {5}{2}} x^{m} \log {\left (\sqrt {x} \right )} & \text {otherwise} \end {cases}\right ) + b^{4} \left (\begin {cases} \frac {x^{3} x^{m}}{m + 3} & \text {for}\: m \neq -3 \\x^{3} x^{m} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((a+b*x**(1/2))**4*x**m,x)
Output:
a**4*Piecewise((x**(m + 1)/(m + 1), Ne(m, -1)), (log(x), True)) + 8*a**3*b *Piecewise((x**(3/2)*x**m/(2*m + 3), Ne(m, -3/2)), (x**(3/2)*x**m*log(sqrt (x)), True)) + 6*a**2*b**2*Piecewise((x**2*x**m/(m + 2), Ne(m, -2)), (x**2 *x**m*log(x), True)) + 8*a*b**3*Piecewise((x**(5/2)*x**m/(2*m + 5), Ne(m, -5/2)), (x**(5/2)*x**m*log(sqrt(x)), True)) + b**4*Piecewise((x**3*x**m/(m + 3), Ne(m, -3)), (x**3*x**m*log(x), True))
Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {b^{4} x^{m + 3}}{m + 3} + \frac {8 \, a b^{3} x^{m + \frac {5}{2}}}{2 \, m + 5} + \frac {6 \, a^{2} b^{2} x^{m + 2}}{m + 2} + \frac {8 \, a^{3} b x^{m + \frac {3}{2}}}{2 \, m + 3} + \frac {a^{4} x^{m + 1}}{m + 1} \] Input:
integrate((a+b*x^(1/2))^4*x^m,x, algorithm="maxima")
Output:
b^4*x^(m + 3)/(m + 3) + 8*a*b^3*x^(m + 5/2)/(2*m + 5) + 6*a^2*b^2*x^(m + 2 )/(m + 2) + 8*a^3*b*x^(m + 3/2)/(2*m + 3) + a^4*x^(m + 1)/(m + 1)
Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.22 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {b^{4} x^{3} \sqrt {x}^{2 \, m}}{m + 3} + \frac {8 \, a b^{3} x^{\frac {5}{2}} \sqrt {x}^{2 \, m}}{2 \, m + 5} + \frac {6 \, a^{2} b^{2} x^{2} \sqrt {x}^{2 \, m}}{m + 2} + \frac {8 \, a^{3} b x^{\frac {3}{2}} \sqrt {x}^{2 \, m}}{2 \, m + 3} + \frac {a^{4} x \sqrt {x}^{2 \, m}}{m + 1} \] Input:
integrate((a+b*x^(1/2))^4*x^m,x, algorithm="giac")
Output:
b^4*x^3*sqrt(x)^(2*m)/(m + 3) + 8*a*b^3*x^(5/2)*sqrt(x)^(2*m)/(2*m + 5) + 6*a^2*b^2*x^2*sqrt(x)^(2*m)/(m + 2) + 8*a^3*b*x^(3/2)*sqrt(x)^(2*m)/(2*m + 3) + a^4*x*sqrt(x)^(2*m)/(m + 1)
Time = 0.86 (sec) , antiderivative size = 292, normalized size of antiderivative = 3.36 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {b^4\,x^m\,x^3\,\left (4\,m^4+28\,m^3+71\,m^2+77\,m+30\right )}{4\,m^5+40\,m^4+155\,m^3+290\,m^2+261\,m+90}+\frac {a^4\,x\,x^m\,\left (4\,m^4+36\,m^3+119\,m^2+171\,m+90\right )}{4\,m^5+40\,m^4+155\,m^3+290\,m^2+261\,m+90}+\frac {8\,a\,b^3\,x^m\,x^{5/2}\,\left (2\,m^4+15\,m^3+40\,m^2+45\,m+18\right )}{4\,m^5+40\,m^4+155\,m^3+290\,m^2+261\,m+90}+\frac {8\,a^3\,b\,x^m\,x^{3/2}\,\left (2\,m^4+17\,m^3+52\,m^2+67\,m+30\right )}{4\,m^5+40\,m^4+155\,m^3+290\,m^2+261\,m+90}+\frac {6\,a^2\,b^2\,x^m\,x^2\,\left (4\,m^4+32\,m^3+91\,m^2+108\,m+45\right )}{4\,m^5+40\,m^4+155\,m^3+290\,m^2+261\,m+90} \] Input:
int(x^m*(a + b*x^(1/2))^4,x)
Output:
(b^4*x^m*x^3*(77*m + 71*m^2 + 28*m^3 + 4*m^4 + 30))/(261*m + 290*m^2 + 155 *m^3 + 40*m^4 + 4*m^5 + 90) + (a^4*x*x^m*(171*m + 119*m^2 + 36*m^3 + 4*m^4 + 90))/(261*m + 290*m^2 + 155*m^3 + 40*m^4 + 4*m^5 + 90) + (8*a*b^3*x^m*x ^(5/2)*(45*m + 40*m^2 + 15*m^3 + 2*m^4 + 18))/(261*m + 290*m^2 + 155*m^3 + 40*m^4 + 4*m^5 + 90) + (8*a^3*b*x^m*x^(3/2)*(67*m + 52*m^2 + 17*m^3 + 2*m ^4 + 30))/(261*m + 290*m^2 + 155*m^3 + 40*m^4 + 4*m^5 + 90) + (6*a^2*b^2*x ^m*x^2*(108*m + 91*m^2 + 32*m^3 + 4*m^4 + 45))/(261*m + 290*m^2 + 155*m^3 + 40*m^4 + 4*m^5 + 90)
Time = 0.21 (sec) , antiderivative size = 278, normalized size of antiderivative = 3.20 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {x^{m} x \left (16 \sqrt {x}\, a^{3} b \,m^{4}+136 \sqrt {x}\, a^{3} b \,m^{3}+416 \sqrt {x}\, a^{3} b \,m^{2}+536 \sqrt {x}\, a^{3} b m +4 b^{4} m^{4} x^{2}+28 b^{4} m^{3} x^{2}+71 b^{4} m^{2} x^{2}+77 b^{4} m \,x^{2}+240 \sqrt {x}\, a^{3} b +270 a^{2} b^{2} x +144 \sqrt {x}\, a \,b^{3} x +90 a^{4}+16 \sqrt {x}\, a \,b^{3} m^{4} x +120 \sqrt {x}\, a \,b^{3} m^{3} x +320 \sqrt {x}\, a \,b^{3} m^{2} x +360 \sqrt {x}\, a \,b^{3} m x +4 a^{4} m^{4}+36 a^{4} m^{3}+119 a^{4} m^{2}+171 a^{4} m +30 b^{4} x^{2}+24 a^{2} b^{2} m^{4} x +192 a^{2} b^{2} m^{3} x +546 a^{2} b^{2} m^{2} x +648 a^{2} b^{2} m x \right )}{4 m^{5}+40 m^{4}+155 m^{3}+290 m^{2}+261 m +90} \] Input:
int((a+b*x^(1/2))^4*x^m,x)
Output:
(x**m*x*(16*sqrt(x)*a**3*b*m**4 + 136*sqrt(x)*a**3*b*m**3 + 416*sqrt(x)*a* *3*b*m**2 + 536*sqrt(x)*a**3*b*m + 240*sqrt(x)*a**3*b + 16*sqrt(x)*a*b**3* m**4*x + 120*sqrt(x)*a*b**3*m**3*x + 320*sqrt(x)*a*b**3*m**2*x + 360*sqrt( x)*a*b**3*m*x + 144*sqrt(x)*a*b**3*x + 4*a**4*m**4 + 36*a**4*m**3 + 119*a* *4*m**2 + 171*a**4*m + 90*a**4 + 24*a**2*b**2*m**4*x + 192*a**2*b**2*m**3* x + 546*a**2*b**2*m**2*x + 648*a**2*b**2*m*x + 270*a**2*b**2*x + 4*b**4*m* *4*x**2 + 28*b**4*m**3*x**2 + 71*b**4*m**2*x**2 + 77*b**4*m*x**2 + 30*b**4 *x**2))/(4*m**5 + 40*m**4 + 155*m**3 + 290*m**2 + 261*m + 90)