\(\int \frac {(d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [373]

Optimal result

Integrand size = 30, antiderivative size = 161 \[ \int \frac {(d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (b d-a e) (a+b x) \sqrt {d+e x}}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (d+e x)^{3/2}}{3 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (b d-a e)^{3/2} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

2*(-a*e+b*d)*(b*x+a)*(e*x+d)^(1/2)/b^2/((b*x+a)^2)^(1/2)+2/3*(b*x+a)*(e*x+ 
d)^(3/2)/b/((b*x+a)^2)^(1/2)-2*(-a*e+b*d)^(3/2)*(b*x+a)*arctanh(b^(1/2)*(e 
*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(5/2)/((b*x+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.60 \[ \int \frac {(d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\sqrt {b} \sqrt {d+e x} (4 b d-3 a e+b e x)+3 (-b d+a e)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{3 b^{5/2} \sqrt {(a+b x)^2}} \] Input:

Integrate[(d + e*x)^(3/2)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
 

Output:

(2*(a + b*x)*(Sqrt[b]*Sqrt[d + e*x]*(4*b*d - 3*a*e + b*e*x) + 3*(-(b*d) + 
a*e)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]]))/(3*b^(5/2) 
*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.73, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1102, 27, 60, 60, 73, 221}

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.

\(\displaystyle \int \frac {(d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

\(\Big \downarrow \) 1102

\(\displaystyle \frac {b (a+b x) \int \frac {(d+e x)^{3/2}}{b (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^{3/2}}{a+b x}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(a+b x) \left (\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(a+b x) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a+b x) \left (\frac {(b d-a e) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(a+b x) \left (\frac {(b d-a e) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

Input:

Int[(d + e*x)^(3/2)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
 

Output:

((a + b*x)*((2*(d + e*x)^(3/2))/(3*b) + ((b*d - a*e)*((2*Sqrt[d + e*x])/b 
- (2*Sqrt[b*d - a*e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^( 
3/2)))/b))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 1102

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F 
racPart[p]))   Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, 
 d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
 

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {2 \left (-b e x +3 a e -4 b d \right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{3 b^{2} \left (b x +a \right )}+\frac {2 \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{2} \sqrt {b \left (a e -b d \right )}\, \left (b x +a \right )}\)	\(120\)
default	\(\frac {2 \left (b x +a \right ) \left (\left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, b +3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) e^{2} a^{2}-6 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a b d e +3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) b^{2} d^{2}-3 \sqrt {e x +d}\, a e \sqrt {b \left (a e -b d \right )}+3 \sqrt {e x +d}\, d b \sqrt {b \left (a e -b d \right )}\right )}{3 \sqrt {\left (b x +a \right )^{2}}\, b^{2} \sqrt {b \left (a e -b d \right )}}\)	\(188\)

Input:

int((e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3*(-b*e*x+3*a*e-4*b*d)*(e*x+d)^(1/2)/b^2*((b*x+a)^2)^(1/2)/(b*x+a)+2*(a 
^2*e^2-2*a*b*d*e+b^2*d^2)/b^2/(b*(a*e-b*d))^(1/2)*arctan(b*(e*x+d)^(1/2)/( 
b*(a*e-b*d))^(1/2))*((b*x+a)^2)^(1/2)/(b*x+a)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.17 \[ \int \frac {(d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {3 \, {\left (b d - a e\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (b e x + 4 \, b d - 3 \, a e\right )} \sqrt {e x + d}}{3 \, b^{2}}, -\frac {2 \, {\left (3 \, {\left (b d - a e\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (b e x + 4 \, b d - 3 \, a e\right )} \sqrt {e x + d}\right )}}{3 \, b^{2}}\right ] \] Input:

integrate((e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/3*(3*(b*d - a*e)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e + 2*sqrt 
(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) - 2*(b*e*x + 4*b*d - 3*a*e)*sq 
rt(e*x + d))/b^2, -2/3*(3*(b*d - a*e)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e* 
x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (b*e*x + 4*b*d - 3*a*e)*sqrt( 
e*x + d))/b^2]
 

Sympy [F]

\[ \int \frac {(d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {3}{2}}}{\sqrt {\left (a + b x\right )^{2}}}\, dx \] Input:

integrate((e*x+d)**(3/2)/((b*x+a)**2)**(1/2),x)
 

Output:

Integral((d + e*x)**(3/2)/sqrt((a + b*x)**2), x)
 

Maxima [F]

\[ \int \frac {(d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {3}{2}}}{\sqrt {{\left (b x + a\right )}^{2}}} \,d x } \] Input:

integrate((e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate((e*x + d)^(3/2)/sqrt((b*x + a)^2), x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \, {\left (b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b d e \mathrm {sgn}\left (b x + a\right ) + a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{2}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} b^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, \sqrt {e x + d} b^{2} d \mathrm {sgn}\left (b x + a\right ) - 3 \, \sqrt {e x + d} a b e \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, b^{3}} \] Input:

integrate((e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")
 

Output:

2*(b^2*d^2*sgn(b*x + a) - 2*a*b*d*e*sgn(b*x + a) + a^2*e^2*sgn(b*x + a))*a 
rctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^2) + 2 
/3*((e*x + d)^(3/2)*b^2*sgn(b*x + a) + 3*sqrt(e*x + d)*b^2*d*sgn(b*x + a) 
- 3*sqrt(e*x + d)*a*b*e*sgn(b*x + a))/b^3
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^{3/2}}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \] Input:

int((d + e*x)^(3/2)/((a + b*x)^2)^(1/2),x)
 

Output:

int((d + e*x)^(3/2)/((a + b*x)^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.74 \[ \int \frac {(d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a e -2 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b d -2 \sqrt {e x +d}\, a b e +\frac {8 \sqrt {e x +d}\, b^{2} d}{3}+\frac {2 \sqrt {e x +d}\, b^{2} e x}{3}}{b^{3}} \] Input:

int((e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x)
 

Output:

(2*(3*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b 
*d)))*a*e - 3*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt 
(a*e - b*d)))*b*d - 3*sqrt(d + e*x)*a*b*e + 4*sqrt(d + e*x)*b**2*d + sqrt( 
d + e*x)*b**2*e*x))/(3*b**3)