\(\int \frac {(c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx\) [404]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.93 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx=\frac {\sqrt {c+d x} \left (15 a^2 d^2-10 a b d (c-2 d x)+b^2 \left (-2 c^2-14 c d x+3 d^2 x^2\right )\right )}{3 b^3 (a+b x)^{3/2}}+\frac {5 d^{3/2} (b c-a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{7/2}} \] Input:

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

Output:

(Sqrt[c + d*x]*(15*a^2*d^2 - 10*a*b*d*(c - 2*d*x) + b^2*(-2*c^2 - 14*c*d*x 
 + 3*d^2*x^2)))/(3*b^3*(a + b*x)^(3/2)) + (5*d^(3/2)*(b*c - a*d)*ArcTanh[( 
Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/b^(7/2)
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.06, 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.

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

int((d*x+c)^(5/2)/(b*x+a)^(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 = 0.20 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.71 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx=\left [-\frac {15 \, {\left (a^{2} b c d - a^{3} d^{2} + {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{2} c d - a^{2} b d^{2}\right )} x\right )} \sqrt {\frac {d}{b}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (3 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2} - 2 \, {\left (7 \, b^{2} c d - 10 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{12 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (a^{2} b c d - a^{3} d^{2} + {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{2} c d - a^{2} b d^{2}\right )} x\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) - 2 \, {\left (3 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2} - 2 \, {\left (7 \, b^{2} c d - 10 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((d*x+c)^(5/2)/(b*x+a)^(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. 650 vs. \(2 (100) = 200\).

Time = 0.29 (sec) , antiderivative size = 650, normalized size of antiderivative = 5.08 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx=\frac {\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} d^{2} {\left | b \right |}}{b^{5}} - \frac {5 \, {\left (\sqrt {b d} b c d {\left | b \right |} - \sqrt {b d} a d^{2} {\left | b \right |}\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 )}^{2}\right )}{2 \, b^{5}} - \frac {4 \, {\left (7 \, \sqrt {b d} b^{6} c^{4} d {\left | b \right |} - 28 \, \sqrt {b d} a b^{5} c^{3} d^{2} {\left | b \right |} + 42 \, \sqrt {b d} a^{2} b^{4} c^{2} d^{3} {\left | b \right |} - 28 \, \sqrt {b d} a^{3} b^{3} c d^{4} {\left | b \right |} + 7 \, \sqrt {b d} a^{4} b^{2} d^{5} {\left | b \right |} - 12 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{4} c^{3} d {\left | b \right |} + 36 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{3} c^{2} d^{2} {\left | b \right |} - 36 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{2} c d^{3} {\left | b \right |} + 12 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{3} b d^{4} {\left | b \right |} + 9 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} b^{2} c^{2} d {\left | b \right |} - 18 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a b c d^{2} {\left | b \right |} + 9 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a^{2} d^{3} {\left | b \right |}\right )}}{3 \, {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{3} b^{4}} \] Input:

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

Output:

sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*d^2*abs(b)/b^5 - 5/2*(sq 
rt(b*d)*b*c*d*abs(b) - sqrt(b*d)*a*d^2*abs(b))*log((sqrt(b*d)*sqrt(b*x + a 
) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/b^5 - 4/3*(7*sqrt(b*d)*b^6*c^4 
*d*abs(b) - 28*sqrt(b*d)*a*b^5*c^3*d^2*abs(b) + 42*sqrt(b*d)*a^2*b^4*c^2*d 
^3*abs(b) - 28*sqrt(b*d)*a^3*b^3*c*d^4*abs(b) + 7*sqrt(b*d)*a^4*b^2*d^5*ab 
s(b) - 12*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d 
- a*b*d))^2*b^4*c^3*d*abs(b) + 36*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr 
t(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^3*c^2*d^2*abs(b) - 36*sqrt(b*d)*(s 
qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^2*c* 
d^3*abs(b) + 12*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a 
)*b*d - a*b*d))^2*a^3*b*d^4*abs(b) + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) 
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^2*c^2*d*abs(b) - 18*sqrt(b*d)*( 
sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b*c*d^2 
*abs(b) + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b* 
d - a*b*d))^4*a^2*d^3*abs(b))/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - 
 sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)^3*b^4)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.96 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx=\frac {-30 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} d^{2}+30 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a b c d -30 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a b \,d^{2} x +30 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{2} c d x -5 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{2} d^{2}+5 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a b c d -5 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a b \,d^{2} x +5 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b^{2} c d x +30 \sqrt {d x +c}\, a^{2} b \,d^{2}-20 \sqrt {d x +c}\, a \,b^{2} c d +40 \sqrt {d x +c}\, a \,b^{2} d^{2} x -4 \sqrt {d x +c}\, b^{3} c^{2}-28 \sqrt {d x +c}\, b^{3} c d x +6 \sqrt {d x +c}\, b^{3} d^{2} x^{2}}{6 \sqrt {b x +a}\, b^{4} \left (b x +a \right )} \] Input:

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

Output:

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