\(\int (a+b x)^{3/2} \sqrt {c+d x} \, dx\) [312]

Optimal result

Integrand size = 19, antiderivative size = 154 \[ \int (a+b x)^{3/2} \sqrt {c+d x} \, dx=-\frac {(b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 b d^2}+\frac {(b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}+\frac {(b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{5/2}} \] Output:

-1/8*(-a*d+b*c)^2*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b/d^2+1/12*(-a*d+b*c)*(b*x+a 
)^(3/2)*(d*x+c)^(1/2)/b/d+1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b+1/8*(-a*d+b*c) 
^3*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(3/2)/d^(5/2)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.83 \[ \int (a+b x)^{3/2} \sqrt {c+d x} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 a^2 d^2+2 a b d (4 c+7 d x)+b^2 \left (-3 c^2+2 c d x+8 d^2 x^2\right )\right )}{24 b d^2}+\frac {(b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{3/2} d^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {60, 60, 60, 66, 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.

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {(b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\)

Input:

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

Output:

((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*b) + ((b*c - a*d)*(((a + b*x)^(3/2)*Sqr 
t[c + d*x])/(2*d) - (3*(b*c - a*d)*((Sqrt[a + b*x]*Sqrt[c + d*x])/d - ((b* 
c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b 
]*d^(3/2))))/(4*d)))/(6*b)
 

Defintions of rubi rules used

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 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

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]
 

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.12

method	result	size
default	\(\frac {\left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}}}{3 d}-\frac {\left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \left (x d +c \right )^{\frac {3}{2}}}{2 d}-\frac {\left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {x d +c}}{b}-\frac {\left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {d b}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 b \sqrt {x d +c}\, \sqrt {b x +a}\, \sqrt {d b}}\right )}{4 d}\right )}{2 d}\)	\(173\)

Input:

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

Output:

1/3/d*(b*x+a)^(3/2)*(d*x+c)^(3/2)-1/2*(-a*d+b*c)/d*(1/2/d*(b*x+a)^(1/2)*(d 
*x+c)^(3/2)-1/4*(-a*d+b*c)/d*((b*x+a)^(1/2)*(d*x+c)^(1/2)/b-1/2*(a*d-b*c)/ 
b*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*ln((1/2*a*d+1/2*b*c+ 
b*d*x)/(d*b)^(1/2)+(b*d*x^2+(a*d+b*c)*x+a*c)^(1/2))/(d*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.66 \[ \int (a+b x)^{3/2} \sqrt {c+d x} \, dx=\left [-\frac {3 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d + 8 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} + 7 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{2} d^{3}}, -\frac {3 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d + 8 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} + 7 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{2} d^{3}}\right ] \] Input:

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

Output:

[-1/96*(3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(b*d)*lo 
g(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b*d*x + b*c + a*d)* 
sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(8*b^ 
3*d^3*x^2 - 3*b^3*c^2*d + 8*a*b^2*c*d^2 + 3*a^2*b*d^3 + 2*(b^3*c*d^2 + 7*a 
*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^3), -1/48*(3*(b^3*c^3 - 3 
*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b 
*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + 
(b^2*c*d + a*b*d^2)*x)) - 2*(8*b^3*d^3*x^2 - 3*b^3*c^2*d + 8*a*b^2*c*d^2 + 
 3*a^2*b*d^3 + 2*(b^3*c*d^2 + 7*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c)) 
/(b^2*d^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int (a+b x)^{3/2} \sqrt {c+d x} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (122) = 244\).

Time = 0.19 (sec) , antiderivative size = 423, normalized size of antiderivative = 2.75 \[ \int (a+b x)^{3/2} \sqrt {c+d x} \, dx=-\frac {\frac {24 \, {\left (\frac {{\left (b^{2} c - a b d\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d}} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a}\right )} a^{2} {\left | b \right |}}{b^{2}} - \frac {12 \, {\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d}\right )} a {\left | b \right |}}{b^{2}} - \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (4 \, b x + 4 \, a + \frac {b c d^{3} - 13 \, a d^{4}}{d^{4}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (b^{2} c^{2} d^{2} + 2 \, a b c d^{3} - 11 \, a^{2} d^{4}\right )}}{d^{4}}\right )} \sqrt {b x + a} - \frac {3 \, {\left (b^{4} c^{3} + a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - 5 \, a^{3} b d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{2}}\right )} {\left | b \right |}}{b^{2}}}{24 \, b} \] Input:

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

Output:

-1/24*(24*((b^2*c - a*b*d)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + 
 (b*x + a)*b*d - a*b*d)))/sqrt(b*d) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)* 
sqrt(b*x + a))*a^2*abs(b)/b^2 - 12*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2 
*b*x + 2*a + (b*c*d - 5*a*d^2)/d^2)*sqrt(b*x + a) + (b^3*c^2 + 2*a*b^2*c*d 
 - 3*a^2*b*d^2)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)* 
b*d - a*b*d)))/(sqrt(b*d)*d))*a*abs(b)/b^2 - (sqrt(b^2*c + (b*x + a)*b*d - 
 a*b*d)*(2*(4*b*x + 4*a + (b*c*d^3 - 13*a*d^4)/d^4)*(b*x + a) - 3*(b^2*c^2 
*d^2 + 2*a*b*c*d^3 - 11*a^2*d^4)/d^4)*sqrt(b*x + a) - 3*(b^4*c^3 + a*b^3*c 
^2*d + 3*a^2*b^2*c*d^2 - 5*a^3*b*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + s 
qrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d^2))*abs(b)/b^2)/b
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.08 \[ \int (a+b x)^{3/2} \sqrt {c+d x} \, dx=\frac {3 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b \,d^{3}+8 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} c \,d^{2}+14 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} d^{3} x -3 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c^{2} d +2 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c \,d^{2} x +8 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} d^{3} x^{2}-3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} d^{3}+9 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b c \,d^{2}-9 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{2} c^{2} d +3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{3} c^{3}}{24 b^{2} d^{3}} \] Input:

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

Output:

(3*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*d**3 + 8*sqrt(c + d*x)*sqrt(a + b*x) 
*a*b**2*c*d**2 + 14*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*d**3*x - 3*sqrt(c + 
 d*x)*sqrt(a + b*x)*b**3*c**2*d + 2*sqrt(c + d*x)*sqrt(a + b*x)*b**3*c*d** 
2*x + 8*sqrt(c + d*x)*sqrt(a + b*x)*b**3*d**3*x**2 - 3*sqrt(d)*sqrt(b)*log 
((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*d** 
3 + 9*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/ 
sqrt(a*d - b*c))*a**2*b*c*d**2 - 9*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b 
*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b**2*c**2*d + 3*sqrt(d)*sq 
rt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c)) 
*b**3*c**3)/(24*b**2*d**3)