\(\int (a+b \sqrt {x}+c x)^{5/2} \, dx\) [30]

Optimal result

Integrand size = 16, antiderivative size = 200 \[ \int \left (a+b \sqrt {x}+c x\right )^{5/2} \, dx=-\frac {5 b \left (b^2-4 a c\right )^2 \left (b+2 c \sqrt {x}\right ) \sqrt {a+b \sqrt {x}+c x}}{512 c^4}+\frac {5 b \left (b^2-4 a c\right ) \left (b+2 c \sqrt {x}\right ) \left (a+b \sqrt {x}+c x\right )^{3/2}}{192 c^3}-\frac {b \left (b+2 c \sqrt {x}\right ) \left (a+b \sqrt {x}+c x\right )^{5/2}}{12 c^2}+\frac {2 \left (a+b \sqrt {x}+c x\right )^{7/2}}{7 c}+\frac {5 b \left (b^2-4 a c\right )^3 \text {arctanh}\left (\frac {b+2 c \sqrt {x}}{2 \sqrt {c} \sqrt {a+b \sqrt {x}+c x}}\right )}{1024 c^{9/2}} \] Output:

-5/512*b*(-4*a*c+b^2)^2*(b+2*c*x^(1/2))*(a+b*x^(1/2)+c*x)^(1/2)/c^4+5/192* 
b*(-4*a*c+b^2)*(b+2*c*x^(1/2))*(a+b*x^(1/2)+c*x)^(3/2)/c^3-1/12*b*(b+2*c*x 
^(1/2))*(a+b*x^(1/2)+c*x)^(5/2)/c^2+2/7*(a+b*x^(1/2)+c*x)^(7/2)/c+5/1024*b 
*(-4*a*c+b^2)^3*arctanh(1/2*(b+2*c*x^(1/2))/c^(1/2)/(a+b*x^(1/2)+c*x)^(1/2 
))/c^(9/2)
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.98 \[ \int \left (a+b \sqrt {x}+c x\right )^{5/2} \, dx=\frac {\sqrt {a+b \sqrt {x}+c x} \left (-105 b^6+70 b^5 c \sqrt {x}-56 b^4 c (-20 a+c x)+48 b^3 c^2 \sqrt {x} (-14 a+c x)+3072 c^3 (a+c x)^3+32 b c^3 \sqrt {x} \left (57 a^2+394 a c x+232 c^2 x^2\right )+16 b^2 c^2 \left (-231 a^2+30 a c x+296 c^2 x^2\right )\right )}{10752 c^4}-\frac {5 b \left (b^2-4 a c\right )^3 \log \left (b+2 c \sqrt {x}-2 \sqrt {c} \sqrt {a+b \sqrt {x}+c x}\right )}{1024 c^{9/2}} \] Input:

Integrate[(a + b*Sqrt[x] + c*x)^(5/2),x]
 

Output:

(Sqrt[a + b*Sqrt[x] + c*x]*(-105*b^6 + 70*b^5*c*Sqrt[x] - 56*b^4*c*(-20*a 
+ c*x) + 48*b^3*c^2*Sqrt[x]*(-14*a + c*x) + 3072*c^3*(a + c*x)^3 + 32*b*c^ 
3*Sqrt[x]*(57*a^2 + 394*a*c*x + 232*c^2*x^2) + 16*b^2*c^2*(-231*a^2 + 30*a 
*c*x + 296*c^2*x^2)))/(10752*c^4) - (5*b*(b^2 - 4*a*c)^3*Log[b + 2*c*Sqrt[ 
x] - 2*Sqrt[c]*Sqrt[a + b*Sqrt[x] + c*x]])/(1024*c^(9/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1680, 1160, 1087, 1087, 1087, 1092, 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.

\(\displaystyle \int \left (a+b \sqrt {x}+c x\right )^{5/2} \, dx\)

\(\Big \downarrow \) 1680

\(\displaystyle 2 \int \sqrt {x} \left (a+c x+b \sqrt {x}\right )^{5/2}d\sqrt {x}\)

\(\Big \downarrow \) 1160

\(\displaystyle 2 \left (\frac {\left (a+b \sqrt {x}+c x\right )^{7/2}}{7 c}-\frac {b \int \left (a+c x+b \sqrt {x}\right )^{5/2}d\sqrt {x}}{2 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle 2 \left (\frac {\left (a+b \sqrt {x}+c x\right )^{7/2}}{7 c}-\frac {b \left (\frac {\left (b+2 c \sqrt {x}\right ) \left (a+b \sqrt {x}+c x\right )^{5/2}}{12 c}-\frac {5 \left (b^2-4 a c\right ) \int \left (a+c x+b \sqrt {x}\right )^{3/2}d\sqrt {x}}{24 c}\right )}{2 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle 2 \left (\frac {\left (a+b \sqrt {x}+c x\right )^{7/2}}{7 c}-\frac {b \left (\frac {\left (b+2 c \sqrt {x}\right ) \left (a+b \sqrt {x}+c x\right )^{5/2}}{12 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt {x}\right ) \left (a+b \sqrt {x}+c x\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {a+c x+b \sqrt {x}}d\sqrt {x}}{16 c}\right )}{24 c}\right )}{2 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle 2 \left (\frac {\left (a+b \sqrt {x}+c x\right )^{7/2}}{7 c}-\frac {b \left (\frac {\left (b+2 c \sqrt {x}\right ) \left (a+b \sqrt {x}+c x\right )^{5/2}}{12 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt {x}\right ) \left (a+b \sqrt {x}+c x\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt {x}\right ) \sqrt {a+b \sqrt {x}+c x}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {a+c x+b \sqrt {x}}}d\sqrt {x}}{8 c}\right )}{16 c}\right )}{24 c}\right )}{2 c}\right )\)

\(\Big \downarrow \) 1092

\(\displaystyle 2 \left (\frac {\left (a+b \sqrt {x}+c x\right )^{7/2}}{7 c}-\frac {b \left (\frac {\left (b+2 c \sqrt {x}\right ) \left (a+b \sqrt {x}+c x\right )^{5/2}}{12 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt {x}\right ) \left (a+b \sqrt {x}+c x\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt {x}\right ) \sqrt {a+b \sqrt {x}+c x}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-x}d\frac {b+2 c \sqrt {x}}{\sqrt {a+c x+b \sqrt {x}}}}{4 c}\right )}{16 c}\right )}{24 c}\right )}{2 c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 2 \left (\frac {\left (a+b \sqrt {x}+c x\right )^{7/2}}{7 c}-\frac {b \left (\frac {\left (b+2 c \sqrt {x}\right ) \left (a+b \sqrt {x}+c x\right )^{5/2}}{12 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt {x}\right ) \left (a+b \sqrt {x}+c x\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c \sqrt {x}\right ) \sqrt {a+b \sqrt {x}+c x}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c \sqrt {x}}{2 \sqrt {c} \sqrt {a+b \sqrt {x}+c x}}\right )}{8 c^{3/2}}\right )}{16 c}\right )}{24 c}\right )}{2 c}\right )\)

Input:

Int[(a + b*Sqrt[x] + c*x)^(5/2),x]
 

Output:

2*((a + b*Sqrt[x] + c*x)^(7/2)/(7*c) - (b*(((b + 2*c*Sqrt[x])*(a + b*Sqrt[ 
x] + c*x)^(5/2))/(12*c) - (5*(b^2 - 4*a*c)*(((b + 2*c*Sqrt[x])*(a + b*Sqrt 
[x] + c*x)^(3/2))/(8*c) - (3*(b^2 - 4*a*c)*(((b + 2*c*Sqrt[x])*Sqrt[a + b* 
Sqrt[x] + c*x])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*Sqrt[x])/(2*Sqrt[c 
]*Sqrt[a + b*Sqrt[x] + c*x])])/(8*c^(3/2))))/(16*c)))/(24*c)))/(2*c))
 

Defintions of rubi rules used

rule 219

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])
 

rule 1087

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])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

rule 1680

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k 
 = Denominator[n]}, Simp[k   Subst[Int[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k* 
n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && Fr 
actionQ[n]
 

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {2 \left (a +b \sqrt {x}+c x \right )^{\frac {7}{2}}}{7 c}-\frac {b \left (\frac {\left (b +2 c \sqrt {x}\right ) \left (a +b \sqrt {x}+c x \right )^{\frac {5}{2}}}{12 c}+\frac {5 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \sqrt {x}\right ) \left (a +b \sqrt {x}+c x \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \sqrt {x}\right ) \sqrt {a +b \sqrt {x}+c x}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \sqrt {x}}{\sqrt {c}}+\sqrt {a +b \sqrt {x}+c x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\right )}{c}\)	\(175\)
default	\(\frac {2 \left (a +b \sqrt {x}+c x \right )^{\frac {7}{2}}}{7 c}-\frac {b \left (\frac {\left (b +2 c \sqrt {x}\right ) \left (a +b \sqrt {x}+c x \right )^{\frac {5}{2}}}{12 c}+\frac {5 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \sqrt {x}\right ) \left (a +b \sqrt {x}+c x \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (b +2 c \sqrt {x}\right ) \sqrt {a +b \sqrt {x}+c x}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \sqrt {x}}{\sqrt {c}}+\sqrt {a +b \sqrt {x}+c x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\right )}{c}\)	\(175\)

Input:

int((a+b*x^(1/2)+c*x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/7*(a+b*x^(1/2)+c*x)^(7/2)/c-b/c*(1/12*(b+2*c*x^(1/2))/c*(a+b*x^(1/2)+c*x 
)^(5/2)+5/24*(4*a*c-b^2)/c*(1/8*(b+2*c*x^(1/2))/c*(a+b*x^(1/2)+c*x)^(3/2)+ 
3/16*(4*a*c-b^2)/c*(1/4*(b+2*c*x^(1/2))/c*(a+b*x^(1/2)+c*x)^(1/2)+1/8*(4*a 
*c-b^2)/c^(3/2)*ln((1/2*b+c*x^(1/2))/c^(1/2)+(a+b*x^(1/2)+c*x)^(1/2)))))
 

Fricas [F(-1)]

Timed out. \[ \int \left (a+b \sqrt {x}+c x\right )^{5/2} \, dx=\text {Timed out} \] Input:

integrate((a+b*x^(1/2)+c*x)^(5/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [A] (verification not implemented)

Time = 1.74 (sec) , antiderivative size = 1076, normalized size of antiderivative = 5.38 \[ \int \left (a+b \sqrt {x}+c x\right )^{5/2} \, dx=\text {Too large to display} \] Input:

integrate((a+b*x**(1/2)+c*x)**(5/2),x)
 

Output:

2*Piecewise(((-a*(3*a**2*b - 3*a*(359*a*b*c/84 + b**3 - 9*b*(15*a*c**2/7 + 
 185*b**2*c/168)/(10*c))/(4*c) - 5*b*(3*a**2*c + 3*a*b**2 - 4*a*(15*a*c**2 
/7 + 185*b**2*c/168)/(5*c) - 7*b*(359*a*b*c/84 + b**3 - 9*b*(15*a*c**2/7 + 
 185*b**2*c/168)/(10*c))/(8*c))/(6*c))/(2*c) - b*(a**3 - 2*a*(3*a**2*c + 3 
*a*b**2 - 4*a*(15*a*c**2/7 + 185*b**2*c/168)/(5*c) - 7*b*(359*a*b*c/84 + b 
**3 - 9*b*(15*a*c**2/7 + 185*b**2*c/168)/(10*c))/(8*c))/(3*c) - 3*b*(3*a** 
2*b - 3*a*(359*a*b*c/84 + b**3 - 9*b*(15*a*c**2/7 + 185*b**2*c/168)/(10*c) 
)/(4*c) - 5*b*(3*a**2*c + 3*a*b**2 - 4*a*(15*a*c**2/7 + 185*b**2*c/168)/(5 
*c) - 7*b*(359*a*b*c/84 + b**3 - 9*b*(15*a*c**2/7 + 185*b**2*c/168)/(10*c) 
)/(8*c))/(6*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*sqrt 
(x) + c*x) + 2*c*sqrt(x))/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + sqr 
t(x))*log(b/(2*c) + sqrt(x))/sqrt(c*(b/(2*c) + sqrt(x))**2), True)) + sqrt 
(a + b*sqrt(x) + c*x)*(29*b*c*x**(5/2)/84 + c**2*x**3/7 + x**(3/2)*(359*a* 
b*c/84 + b**3 - 9*b*(15*a*c**2/7 + 185*b**2*c/168)/(10*c))/(4*c) + sqrt(x) 
*(3*a**2*b - 3*a*(359*a*b*c/84 + b**3 - 9*b*(15*a*c**2/7 + 185*b**2*c/168) 
/(10*c))/(4*c) - 5*b*(3*a**2*c + 3*a*b**2 - 4*a*(15*a*c**2/7 + 185*b**2*c/ 
168)/(5*c) - 7*b*(359*a*b*c/84 + b**3 - 9*b*(15*a*c**2/7 + 185*b**2*c/168) 
/(10*c))/(8*c))/(6*c))/(2*c) + x**2*(15*a*c**2/7 + 185*b**2*c/168)/(5*c) + 
 x*(3*a**2*c + 3*a*b**2 - 4*a*(15*a*c**2/7 + 185*b**2*c/168)/(5*c) - 7*b*( 
359*a*b*c/84 + b**3 - 9*b*(15*a*c**2/7 + 185*b**2*c/168)/(10*c))/(8*c))...
 

Maxima [F]

\[ \int \left (a+b \sqrt {x}+c x\right )^{5/2} \, dx=\int { {\left (c x + b \sqrt {x} + a\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a+b*x^(1/2)+c*x)^(5/2),x, algorithm="maxima")
 

Output:

integrate((c*x + b*sqrt(x) + a)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \left (a+b \sqrt {x}+c x\right )^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*x^(1/2)+c*x)^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage3:=type(sage2):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m 
& i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(const gen & 
 e,const
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sqrt {x}+c x\right )^{5/2} \, dx=\int {\left (a+c\,x+b\,\sqrt {x}\right )}^{5/2} \,d x \] Input:

int((a + c*x + b*x^(1/2))^(5/2),x)
 

Output:

int((a + c*x + b*x^(1/2))^(5/2), x)
 

Reduce [F]

\[ \int \left (a+b \sqrt {x}+c x\right )^{5/2} \, dx=\int \left (a +b \sqrt {x}+c x \right )^{\frac {5}{2}}d x \] Input:

int((a+b*x^(1/2)+c*x)^(5/2),x)
 

Output:

int((a+b*x^(1/2)+c*x)^(5/2),x)