\(\int \frac {(a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx\) [364]

Optimal result

Integrand size = 19, antiderivative size = 128 \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=-\frac {2 (a+b x)^{5/2}}{3 d (c+d x)^{3/2}}-\frac {10 b (a+b x)^{3/2}}{3 d^2 \sqrt {c+d x}}+\frac {5 b^2 \sqrt {a+b x} \sqrt {c+d x}}{d^3}-\frac {5 b^{3/2} (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{7/2}} \] Output:

-2/3*(b*x+a)^(5/2)/d/(d*x+c)^(3/2)-10/3*b*(b*x+a)^(3/2)/d^2/(d*x+c)^(1/2)+ 
5*b^2*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^3-5*b^(3/2)*(-a*d+b*c)*arctanh(d^(1/2) 
*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(7/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (-2 a^2 d^2-2 a b d (5 c+7 d x)+b^2 \left (15 c^2+20 c d x+3 d^2 x^2\right )\right )}{3 d^3 (c+d x)^{3/2}}-\frac {5 b^{3/2} (b c-a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{d^{7/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*(-2*a^2*d^2 - 2*a*b*d*(5*c + 7*d*x) + b^2*(15*c^2 + 20*c*d* 
x + 3*d^2*x^2)))/(3*d^3*(c + d*x)^(3/2)) - (5*b^(3/2)*(b*c - a*d)*ArcTanh[ 
(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/d^(7/2)
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {57, 57, 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.

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, 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 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 [F]

\[\int \frac {\left (b x +a \right )^{\frac {5}{2}}}{\left (x d +c \right )^{\frac {5}{2}}}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (100) = 200\).

Time = 0.24 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.71 \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\left [-\frac {15 \, {\left (b^{2} c^{3} - a b c^{2} d + {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{2} + 2 \, {\left (b^{2} c^{2} d - a b c d^{2}\right )} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (3 \, b^{2} d^{2} x^{2} + 15 \, b^{2} c^{2} - 10 \, a b c d - 2 \, a^{2} d^{2} + 2 \, {\left (10 \, b^{2} c d - 7 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{12 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}}, \frac {15 \, {\left (b^{2} c^{3} - a b c^{2} d + {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{2} + 2 \, {\left (b^{2} c^{2} d - a b c d^{2}\right )} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, {\left (3 \, b^{2} d^{2} x^{2} + 15 \, b^{2} c^{2} - 10 \, a b c d - 2 \, a^{2} d^{2} + 2 \, {\left (10 \, b^{2} c d - 7 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}}\right ] \] Input:

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

Output:

[-1/12*(15*(b^2*c^3 - a*b*c^2*d + (b^2*c*d^2 - a*b*d^3)*x^2 + 2*(b^2*c^2*d 
 - a*b*c*d^2)*x)*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d 
^2 + 4*(2*b*d^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 
 8*(b^2*c*d + a*b*d^2)*x) - 4*(3*b^2*d^2*x^2 + 15*b^2*c^2 - 10*a*b*c*d - 2 
*a^2*d^2 + 2*(10*b^2*c*d - 7*a*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^5 
*x^2 + 2*c*d^4*x + c^2*d^3), 1/6*(15*(b^2*c^3 - a*b*c^2*d + (b^2*c*d^2 - a 
*b*d^3)*x^2 + 2*(b^2*c^2*d - a*b*c*d^2)*x)*sqrt(-b/d)*arctan(1/2*(2*b*d*x 
+ b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*x^2 + a*b*c + ( 
b^2*c + a*b*d)*x)) + 2*(3*b^2*d^2*x^2 + 15*b^2*c^2 - 10*a*b*c*d - 2*a^2*d^ 
2 + 2*(10*b^2*c*d - 7*a*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^5*x^2 + 
2*c*d^4*x + c^2*d^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((b*x+a)^(5/2)/(d*x+c)^(5/2),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. 276 vs. \(2 (100) = 200\).

Time = 0.20 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.16 \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {{\left ({\left (b x + a\right )} {\left (\frac {3 \, {\left (b^{6} c d^{4} - a b^{5} d^{5}\right )} {\left (b x + a\right )}}{b^{2} c d^{5} {\left | b \right |} - a b d^{6} {\left | b \right |}} + \frac {20 \, {\left (b^{7} c^{2} d^{3} - 2 \, a b^{6} c d^{4} + a^{2} b^{5} d^{5}\right )}}{b^{2} c d^{5} {\left | b \right |} - a b d^{6} {\left | b \right |}}\right )} + \frac {15 \, {\left (b^{8} c^{3} d^{2} - 3 \, a b^{7} c^{2} d^{3} + 3 \, a^{2} b^{6} c d^{4} - a^{3} b^{5} d^{5}\right )}}{b^{2} c d^{5} {\left | b \right |} - a b d^{6} {\left | b \right |}}\right )} \sqrt {b x + a}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} + \frac {5 \, {\left (b^{4} c - a b^{3} d\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{3} {\left | b \right |}} \] Input:

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

Output:

1/3*((b*x + a)*(3*(b^6*c*d^4 - a*b^5*d^5)*(b*x + a)/(b^2*c*d^5*abs(b) - a* 
b*d^6*abs(b)) + 20*(b^7*c^2*d^3 - 2*a*b^6*c*d^4 + a^2*b^5*d^5)/(b^2*c*d^5* 
abs(b) - a*b*d^6*abs(b))) + 15*(b^8*c^3*d^2 - 3*a*b^7*c^2*d^3 + 3*a^2*b^6* 
c*d^4 - a^3*b^5*d^5)/(b^2*c*d^5*abs(b) - a*b*d^6*abs(b)))*sqrt(b*x + a)/(b 
^2*c + (b*x + a)*b*d - a*b*d)^(3/2) + 5*(b^4*c - a*b^3*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)*d^3*ab 
s(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 504, normalized size of antiderivative = 3.94 \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {-4 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} d^{3}-20 \sqrt {d x +c}\, \sqrt {b x +a}\, a b c \,d^{2}-28 \sqrt {d x +c}\, \sqrt {b x +a}\, a b \,d^{3} x +30 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{2} c^{2} d +40 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{2} c \,d^{2} x +6 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{2} d^{3} x^{2}+30 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a b \,c^{2} d +60 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a b c \,d^{2} x +30 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a b \,d^{3} x^{2}-30 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{2} c^{3}-60 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{2} c^{2} d x -30 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{2} c \,d^{2} x^{2}+5 \sqrt {d}\, \sqrt {b}\, a b \,c^{2} d +10 \sqrt {d}\, \sqrt {b}\, a b c \,d^{2} x +5 \sqrt {d}\, \sqrt {b}\, a b \,d^{3} x^{2}-5 \sqrt {d}\, \sqrt {b}\, b^{2} c^{3}-10 \sqrt {d}\, \sqrt {b}\, b^{2} c^{2} d x -5 \sqrt {d}\, \sqrt {b}\, b^{2} c \,d^{2} x^{2}}{6 d^{4} \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:

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

Output:

( - 4*sqrt(c + d*x)*sqrt(a + b*x)*a**2*d**3 - 20*sqrt(c + d*x)*sqrt(a + b* 
x)*a*b*c*d**2 - 28*sqrt(c + d*x)*sqrt(a + b*x)*a*b*d**3*x + 30*sqrt(c + d* 
x)*sqrt(a + b*x)*b**2*c**2*d + 40*sqrt(c + d*x)*sqrt(a + b*x)*b**2*c*d**2* 
x + 6*sqrt(c + d*x)*sqrt(a + b*x)*b**2*d**3*x**2 + 30*sqrt(d)*sqrt(b)*log( 
(sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b*c**2* 
d + 60*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x)) 
/sqrt(a*d - b*c))*a*b*c*d**2*x + 30*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + 
b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b*d**3*x**2 - 30*sqrt(d)* 
sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c 
))*b**2*c**3 - 60*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqr 
t(c + d*x))/sqrt(a*d - b*c))*b**2*c**2*d*x - 30*sqrt(d)*sqrt(b)*log((sqrt( 
d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**2*c*d**2*x** 
2 + 5*sqrt(d)*sqrt(b)*a*b*c**2*d + 10*sqrt(d)*sqrt(b)*a*b*c*d**2*x + 5*sqr 
t(d)*sqrt(b)*a*b*d**3*x**2 - 5*sqrt(d)*sqrt(b)*b**2*c**3 - 10*sqrt(d)*sqrt 
(b)*b**2*c**2*d*x - 5*sqrt(d)*sqrt(b)*b**2*c*d**2*x**2)/(6*d**4*(c**2 + 2* 
c*d*x + d**2*x**2))