\(\int \frac {(c+d x)^{5/2}}{\sqrt {a+b x}} \, dx\) [334]

Optimal result

Integrand size = 19, antiderivative size = 148 \[ \int \frac {(c+d x)^{5/2}}{\sqrt {a+b x}} \, dx=\frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 b^3}+\frac {5 (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^2}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b}+\frac {5 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} \sqrt {d}} \] Output:

5/8*(-a*d+b*c)^2*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3+5/12*(-a*d+b*c)*(b*x+a)^( 
1/2)*(d*x+c)^(3/2)/b^2+1/3*(b*x+a)^(1/2)*(d*x+c)^(5/2)/b+5/8*(-a*d+b*c)^3* 
arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(7/2)/d^(1/2)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.84 \[ \int \frac {(c+d x)^{5/2}}{\sqrt {a+b x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-10 a b d (4 c+d x)+b^2 \left (33 c^2+26 c d x+8 d^2 x^2\right )\right )}{24 b^3}+\frac {5 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{7/2} \sqrt {d}} \] Input:

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

Output:

(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^2*d^2 - 10*a*b*d*(4*c + d*x) + b^2*(33* 
c^2 + 26*c*d*x + 8*d^2*x^2)))/(24*b^3) + (5*(b*c - a*d)^3*ArcTanh[(Sqrt[b] 
*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(8*b^(7/2)*Sqrt[d])
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.04, 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.

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*b) + (5*(b*c - a*d)*((Sqrt[a + b*x]*(c 
+ d*x)^(3/2))/(2*b) + (3*(b*c - a*d)*((Sqrt[a + b*x]*Sqrt[c + d*x])/b + (( 
b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(b^(3 
/2)*Sqrt[d])))/(4*b)))/(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 = 0.22 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.17

method	result	size
default	\(\frac {\sqrt {b x +a}\, \left (x d +c \right )^{\frac {5}{2}}}{3 b}-\frac {5 \left (a d -b c \right ) \left (\frac {\sqrt {b x +a}\, \left (x d +c \right )^{\frac {3}{2}}}{2 b}-\frac {3 \left (a d -b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {x d +c}}{b}-\frac {\left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {d b}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 b \sqrt {x d +c}\, \sqrt {b x +a}\, \sqrt {d b}}\right )}{4 b}\right )}{6 b}\)	\(173\)

Input:

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

Output:

1/3*(b*x+a)^(1/2)*(d*x+c)^(5/2)/b-5/6*(a*d-b*c)/b*(1/2*(b*x+a)^(1/2)*(d*x+ 
c)^(3/2)/b-3/4*(a*d-b*c)/b*((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 = 0.11 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.78 \[ \int \frac {(c+d x)^{5/2}}{\sqrt {a+b x}} \, dx=\left [-\frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (8 \, b^{3} d^{3} x^{2} + 33 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (13 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{4} d}, -\frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (8 \, b^{3} d^{3} x^{2} + 33 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (13 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{4} d}\right ] \] Input:

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

Output:

[-1/96*(15*(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)*l 
og(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 + 33*b^3*c^2*d - 40*a*b^2*c*d^2 + 15*a^2*b*d^3 + 2*(13*b^3*c*d^ 
2 - 5*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*d), -1/48*(15*(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 + 33*b^3*c^2*d - 40*a*b^2 
*c*d^2 + 15*a^2*b*d^3 + 2*(13*b^3*c*d^2 - 5*a*b^2*d^3)*x)*sqrt(b*x + a)*sq 
rt(d*x + c))/(b^4*d)]
 

Sympy [F]

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

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

Output:

Integral((c + d*x)**(5/2)/sqrt(a + b*x), x)
 

Maxima [F(-2)]

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

integrate((d*x+c)^(5/2)/(b*x+a)^(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. 427 vs. \(2 (116) = 232\).

Time = 0.18 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.89 \[ \int \frac {(c+d x)^{5/2}}{\sqrt {a+b x}} \, dx=-\frac {\frac {24 \, {\left (\frac {{\left (b^{2} c - a b d\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d}} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a}\right )} c^{2} {\left | b \right |}}{b^{2}} - \frac {12 \, {\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d}\right )} c d {\left | b \right |}}{b^{3}} - \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (4 \, b x + 4 \, a + \frac {b c d^{3} - 13 \, a d^{4}}{d^{4}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (b^{2} c^{2} d^{2} + 2 \, a b c d^{3} - 11 \, a^{2} d^{4}\right )}}{d^{4}}\right )} \sqrt {b x + a} - \frac {3 \, {\left (b^{4} c^{3} + a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - 5 \, a^{3} b d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{2}}\right )} d^{2} {\left | b \right |}}{b^{4}}}{24 \, b} \] Input:

integrate((d*x+c)^(5/2)/(b*x+a)^(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))*c^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))*c*d*abs(b)/b^3 - (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) + 
 sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d^2))*d^2*abs(b)/b^4)/b
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.16 \[ \int \frac {(c+d x)^{5/2}}{\sqrt {a+b x}} \, dx=\frac {15 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b \,d^{3}-40 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} c \,d^{2}-10 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} d^{3} x +33 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c^{2} d +26 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c \,d^{2} x +8 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} d^{3} x^{2}-15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} d^{3}+45 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b c \,d^{2}-45 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{2} c^{2} d +15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{3} c^{3}}{24 b^{4} d} \] Input:

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

Output:

(15*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*d**3 - 40*sqrt(c + d*x)*sqrt(a + b* 
x)*a*b**2*c*d**2 - 10*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*d**3*x + 33*sqrt( 
c + d*x)*sqrt(a + b*x)*b**3*c**2*d + 26*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 - 15*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 + 45*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 - 45*sqrt(d)*sqrt(b)*log((sqrt(d)*sqr 
t(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b**2*c**2*d + 15*sq 
rt(d)*sqrt(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**4*d)