\(\int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx\) [343]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (33 a^2 d^2+2 a b d (-20 c+13 d x)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )}{24 d^3}-\frac {5 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 \sqrt {b} d^{7/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*Sqrt[c + d*x]*(33*a^2*d^2 + 2*a*b*d*(-20*c + 13*d*x) + b^2* 
(15*c^2 - 10*c*d*x + 8*d^2*x^2)))/(24*d^3) - (5*(b*c - a*d)^3*ArcTanh[(Sqr 
t[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(8*Sqrt[b]*d^(7/2))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05, 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.

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*d) - (5*(b*c - a*d)*(((a + b*x)^(3/2)*S 
qrt[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*d)
 

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 = 0.14 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.17

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

Input:

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

Output:

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

integrate((b*x+a)^(5/2)/(d*x+c)^(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 + 15*b^3*c^2*d - 40*a*b^2*c*d^2 + 33*a^2*b*d^3 - 2*(5*b^3*c*d^2 
 - 13*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^4), 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 + 15*b^3*c^2*d - 40*a*b^2* 
c*d^2 + 33*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 13*a*b^2*d^3)*x)*sqrt(b*x + a)*sqr 
t(d*x + c))/(b*d^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((b*x+a)^(5/2)/(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 [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.34 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b d} - \frac {5 \, {\left (b c d^{3} - a d^{4}\right )}}{b d^{5}}\right )} + \frac {15 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}}{b d^{5}}\right )} + \frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{3}}\right )} b}{24 \, {\left | b \right |}} \] Input:

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

Output:

1/24*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b 
*x + a)/(b*d) - 5*(b*c*d^3 - a*d^4)/(b*d^5)) + 15*(b^2*c^2*d^2 - 2*a*b*c*d 
^3 + a^2*d^4)/(b*d^5)) + 15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3 
*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^3))*b/abs(b)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.16 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx=\frac {33 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b \,d^{3}-40 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} c \,d^{2}+26 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} d^{3} x +15 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c^{2} d -10 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c \,d^{2} x +8 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} d^{3} x^{2}+15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} d^{3}-45 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b c \,d^{2}+45 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{2} c^{2} d -15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{3} c^{3}}{24 b \,d^{4}} \] Input:

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

Output:

(33*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 + 26*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*d**3*x + 15*sqrt( 
c + d*x)*sqrt(a + b*x)*b**3*c**2*d - 10*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*d**4)