Integrand size = 13, antiderivative size = 175 \[ \int \left (b x+c x^2\right )^{5/2} \, dx=\frac {5 b^5 \sqrt {b x+c x^2}}{512 c^3}-\frac {5 b^4 x \sqrt {b x+c x^2}}{768 c^2}+\frac {b^3 x^2 \sqrt {b x+c x^2}}{192 c}+\frac {9}{32} b^2 x^3 \sqrt {b x+c x^2}+\frac {5}{12} b c x^4 \sqrt {b x+c x^2}+\frac {1}{6} c^2 x^5 \sqrt {b x+c x^2}-\frac {5 b^6 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{7/2}} \] Output:
5/512*b^5*(c*x^2+b*x)^(1/2)/c^3-5/768*b^4*x*(c*x^2+b*x)^(1/2)/c^2+1/192*b^ 3*x^2*(c*x^2+b*x)^(1/2)/c+9/32*b^2*x^3*(c*x^2+b*x)^(1/2)+5/12*b*c*x^4*(c*x ^2+b*x)^(1/2)+1/6*c^2*x^5*(c*x^2+b*x)^(1/2)-5/512*b^6*arctanh(c^(1/2)*x/(c *x^2+b*x)^(1/2))/c^(7/2)
Time = 0.61 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.74 \[ \int \left (b x+c x^2\right )^{5/2} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (15 b^5-10 b^4 c x+8 b^3 c^2 x^2+432 b^2 c^3 x^3+640 b c^4 x^4+256 c^5 x^5\right )+\frac {30 b^6 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{1536 c^{7/2}} \] Input:
Integrate[(b*x + c*x^2)^(5/2),x]
Output:
(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^5 - 10*b^4*c*x + 8*b^3*c^2*x^2 + 432*b^2 *c^3*x^3 + 640*b*c^4*x^4 + 256*c^5*x^5) + (30*b^6*ArcTanh[(Sqrt[c]*Sqrt[x] )/(Sqrt[b] - Sqrt[b + c*x])])/(Sqrt[x]*Sqrt[b + c*x])))/(1536*c^(7/2))
Time = 0.43 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1087, 1087, 1087, 1091, 219}
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.
\(\displaystyle \int \left (b x+c x^2\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {5 b^2 \int \left (c x^2+b x\right )^{3/2}dx}{24 c}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {5 b^2 \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^2+b x}dx}{16 c}\right )}{24 c}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {5 b^2 \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{16 c}\right )}{24 c}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {5 b^2 \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{16 c}\right )}{24 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {5 b^2 \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{16 c}\right )}{24 c}\) |
Input:
Int[(b*x + c*x^2)^(5/2),x]
Output:
((b + 2*c*x)*(b*x + c*x^2)^(5/2))/(12*c) - (5*b^2*(((b + 2*c*x)*(b*x + c*x ^2)^(3/2))/(8*c) - (3*b^2*(((b + 2*c*x)*Sqrt[b*x + c*x^2])/(4*c) - (b^2*Ar cTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2))))/(16*c)))/(24*c)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Time = 0.38 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.61
method | result | size |
risch | \(\frac {\left (256 c^{5} x^{5}+640 b \,x^{4} c^{4}+432 b^{2} c^{3} x^{3}+8 c^{2} x^{2} b^{3}-10 b^{4} c x +15 b^{5}\right ) x \left (c x +b \right )}{1536 c^{3} \sqrt {x \left (c x +b \right )}}-\frac {5 b^{6} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{1024 c^{\frac {7}{2}}}\) | \(106\) |
default | \(\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{12 c}-\frac {5 b^{2} \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\) | \(118\) |
Input:
int((c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/1536*(256*c^5*x^5+640*b*c^4*x^4+432*b^2*c^3*x^3+8*b^3*c^2*x^2-10*b^4*c*x +15*b^5)*x*(c*x+b)/c^3/(x*(c*x+b))^(1/2)-5/1024*b^6/c^(7/2)*ln((1/2*b+c*x) /c^(1/2)+(c*x^2+b*x)^(1/2))
Time = 0.08 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.22 \[ \int \left (b x+c x^2\right )^{5/2} \, dx=\left [\frac {15 \, b^{6} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (256 \, c^{6} x^{5} + 640 \, b c^{5} x^{4} + 432 \, b^{2} c^{4} x^{3} + 8 \, b^{3} c^{3} x^{2} - 10 \, b^{4} c^{2} x + 15 \, b^{5} c\right )} \sqrt {c x^{2} + b x}}{3072 \, c^{4}}, \frac {15 \, b^{6} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) + {\left (256 \, c^{6} x^{5} + 640 \, b c^{5} x^{4} + 432 \, b^{2} c^{4} x^{3} + 8 \, b^{3} c^{3} x^{2} - 10 \, b^{4} c^{2} x + 15 \, b^{5} c\right )} \sqrt {c x^{2} + b x}}{1536 \, c^{4}}\right ] \] Input:
integrate((c*x^2+b*x)^(5/2),x, algorithm="fricas")
Output:
[1/3072*(15*b^6*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*( 256*c^6*x^5 + 640*b*c^5*x^4 + 432*b^2*c^4*x^3 + 8*b^3*c^3*x^2 - 10*b^4*c^2 *x + 15*b^5*c)*sqrt(c*x^2 + b*x))/c^4, 1/1536*(15*b^6*sqrt(-c)*arctan(sqrt (c*x^2 + b*x)*sqrt(-c)/(c*x + b)) + (256*c^6*x^5 + 640*b*c^5*x^4 + 432*b^2 *c^4*x^3 + 8*b^3*c^3*x^2 - 10*b^4*c^2*x + 15*b^5*c)*sqrt(c*x^2 + b*x))/c^4 ]
Time = 0.49 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.62 \[ \int \left (b x+c x^2\right )^{5/2} \, dx =\text {Too large to display} \] Input:
integrate((c*x**2+b*x)**(5/2),x)
Output:
b**2*Piecewise((-5*b**4*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/ (2*c) + x)**2), True))/(128*c**3) + sqrt(b*x + c*x**2)*(5*b**3/(64*c**3) - 5*b**2*x/(96*c**2) + b*x**2/(24*c) + x**3/4), Ne(c, 0)), (2*(b*x)**(7/2)/ (7*b**3), Ne(b, 0)), (0, True)) + 2*b*c*Piecewise((7*b**5*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c ) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(256*c**4) + sqrt (b*x + c*x**2)*(-7*b**4/(128*c**4) + 7*b**3*x/(192*c**3) - 7*b**2*x**2/(24 0*c**2) + b*x**3/(40*c) + x**4/5), Ne(c, 0)), (2*(b*x)**(9/2)/(9*b**4), Ne (b, 0)), (0, True)) + c**2*Piecewise((-21*b**6*Piecewise((log(b + 2*sqrt(c )*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x)*log( b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(1024*c**5) + sqrt(b*x + c*x **2)*(21*b**5/(512*c**5) - 7*b**4*x/(256*c**4) + 7*b**3*x**2/(320*c**3) - 3*b**2*x**3/(160*c**2) + b*x**4/(60*c) + x**5/6), Ne(c, 0)), (2*(b*x)**(11 /2)/(11*b**5), Ne(b, 0)), (0, True))
Time = 0.03 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.81 \[ \int \left (b x+c x^2\right )^{5/2} \, dx=\frac {1}{6} \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} x + \frac {5 \, \sqrt {c x^{2} + b x} b^{4} x}{256 \, c^{2}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} x}{96 \, c} - \frac {5 \, b^{6} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} b^{5}}{512 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3}}{192 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} b}{12 \, c} \] Input:
integrate((c*x^2+b*x)^(5/2),x, algorithm="maxima")
Output:
1/6*(c*x^2 + b*x)^(5/2)*x + 5/256*sqrt(c*x^2 + b*x)*b^4*x/c^2 - 5/96*(c*x^ 2 + b*x)^(3/2)*b^2*x/c - 5/1024*b^6*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sq rt(c))/c^(7/2) + 5/512*sqrt(c*x^2 + b*x)*b^5/c^3 - 5/192*(c*x^2 + b*x)^(3/ 2)*b^3/c^2 + 1/12*(c*x^2 + b*x)^(5/2)*b/c
Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.60 \[ \int \left (b x+c x^2\right )^{5/2} \, dx=\frac {5 \, b^{6} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{1024 \, c^{\frac {7}{2}}} + \frac {1}{1536} \, \sqrt {c x^{2} + b x} {\left (\frac {15 \, b^{5}}{c^{3}} - 2 \, {\left (\frac {5 \, b^{4}}{c^{2}} - 4 \, {\left (\frac {b^{3}}{c} + 2 \, {\left (27 \, b^{2} + 8 \, {\left (2 \, c^{2} x + 5 \, b c\right )} x\right )} x\right )} x\right )} x\right )} \] Input:
integrate((c*x^2+b*x)^(5/2),x, algorithm="giac")
Output:
5/1024*b^6*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(7/2) + 1/1536*sqrt(c*x^2 + b*x)*(15*b^5/c^3 - 2*(5*b^4/c^2 - 4*(b^3/c + 2*(27* b^2 + 8*(2*c^2*x + 5*b*c)*x)*x)*x)*x)
Time = 9.22 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.68 \[ \int \left (b x+c x^2\right )^{5/2} \, dx=\frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (\frac {b}{2}+c\,x\right )}{6\,c}-\frac {5\,b^2\,\left (\frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (\frac {b}{2}+c\,x\right )}{4\,c}-\frac {3\,b^2\,\left (\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c}\right )}{24\,c} \] Input:
int((b*x + c*x^2)^(5/2),x)
Output:
((b*x + c*x^2)^(5/2)*(b/2 + c*x))/(6*c) - (5*b^2*(((b*x + c*x^2)^(3/2)*(b/ 2 + c*x))/(4*c) - (3*b^2*((b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) - (b^2*log(( b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/(8*c^(3/2))))/(16*c)))/(24*c)
Time = 0.22 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.76 \[ \int \left (b x+c x^2\right )^{5/2} \, dx=\frac {15 \sqrt {x}\, \sqrt {c x +b}\, b^{5} c -10 \sqrt {x}\, \sqrt {c x +b}\, b^{4} c^{2} x +8 \sqrt {x}\, \sqrt {c x +b}\, b^{3} c^{3} x^{2}+432 \sqrt {x}\, \sqrt {c x +b}\, b^{2} c^{4} x^{3}+640 \sqrt {x}\, \sqrt {c x +b}\, b \,c^{5} x^{4}+256 \sqrt {x}\, \sqrt {c x +b}\, c^{6} x^{5}-15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{6}}{1536 c^{4}} \] Input:
int((c*x^2+b*x)^(5/2),x)
Output:
(15*sqrt(x)*sqrt(b + c*x)*b**5*c - 10*sqrt(x)*sqrt(b + c*x)*b**4*c**2*x + 8*sqrt(x)*sqrt(b + c*x)*b**3*c**3*x**2 + 432*sqrt(x)*sqrt(b + c*x)*b**2*c* *4*x**3 + 640*sqrt(x)*sqrt(b + c*x)*b*c**5*x**4 + 256*sqrt(x)*sqrt(b + c*x )*c**6*x**5 - 15*sqrt(c)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b))*b* *6)/(1536*c**4)