\(\int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx\) [378]

Optimal result

Integrand size = 19, antiderivative size = 138 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {15 d (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^3}+\frac {5 d \sqrt {a+b x} (c+d x)^{3/2}}{2 b^2}-\frac {2 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {15 \sqrt {d} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{7/2}} \] Output:

15/4*d*(-a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3+5/2*d*(b*x+a)^(1/2)*(d*x 
+c)^(3/2)/b^2-2*(d*x+c)^(5/2)/b/(b*x+a)^(1/2)+15/4*d^(1/2)*(-a*d+b*c)^2*ar 
ctanh(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 = 124, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {c+d x} \left (-15 a^2 d^2-5 a b d (-5 c+d x)+b^2 \left (-8 c^2+9 c d x+2 d^2 x^2\right )\right )}{4 b^3 \sqrt {a+b x}}+\frac {15 \sqrt {d} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 b^{7/2}} \] Input:

Integrate[(c + d*x)^(5/2)/(a + b*x)^(3/2),x]
 

Output:

(Sqrt[c + d*x]*(-15*a^2*d^2 - 5*a*b*d*(-5*c + d*x) + b^2*(-8*c^2 + 9*c*d*x 
 + 2*d^2*x^2)))/(4*b^3*Sqrt[a + b*x]) + (15*Sqrt[d]*(b*c - a*d)^2*ArcTanh[ 
(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(4*b^(7/2))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 143, 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 = {57, 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.

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

Int[(c + d*x)^(5/2)/(a + b*x)^(3/2),x]
 

Output:

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

\[\int \frac {\left (x d +c \right )^{\frac {5}{2}}}{\left (b x +a \right )^{\frac {3}{2}}}d x\]

Input:

int((d*x+c)^(5/2)/(b*x+a)^(3/2),x)
 

Output:

int((d*x+c)^(5/2)/(b*x+a)^(3/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 439, normalized size of antiderivative = 3.18 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\left [\frac {15 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} + {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x\right )} \sqrt {\frac {d}{b}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d^{2} x^{2} - 8 \, b^{2} c^{2} + 25 \, a b c d - 15 \, a^{2} d^{2} + {\left (9 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (b^{4} x + a b^{3}\right )}}, -\frac {15 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} + {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{2} - 8 \, b^{2} c^{2} + 25 \, a b c d - 15 \, a^{2} d^{2} + {\left (9 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \] Input:

integrate((d*x+c)^(5/2)/(b*x+a)^(3/2),x, algorithm="fricas")
 

Output:

[1/16*(15*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2 + (b^3*c^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*(2*b^2*d^2*x^2 - 8*b^2*c^2 + 25*a*b*c*d - 15*a^2* 
d^2 + (9*b^2*c*d - 5*a*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*x + a*b 
^3), -1/8*(15*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2 + (b^3*c^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*(2*b 
^2*d^2*x^2 - 8*b^2*c^2 + 25*a*b*c*d - 15*a^2*d^2 + (9*b^2*c*d - 5*a*b*d^2) 
*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*x + a*b^3)]
 

Sympy [F]

\[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((d*x+c)**(5/2)/(b*x+a)**(3/2),x)
 

Output:

Integral((c + d*x)**(5/2)/(a + b*x)**(3/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x+c)^(5/2)/(b*x+a)^(3/2),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. 287 vs. \(2 (108) = 216\).

Time = 0.22 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.08 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {1}{4} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} d^{2} {\left | b \right |}}{b^{5}} + \frac {9 \, {\left (b^{10} c d^{3} {\left | b \right |} - a b^{9} d^{4} {\left | b \right |}\right )}}{b^{14} d^{2}}\right )} - \frac {15 \, {\left (\sqrt {b d} b^{2} c^{2} {\left | b \right |} - 2 \, \sqrt {b d} a b c d {\left | b \right |} + \sqrt {b d} a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{8 \, b^{5}} - \frac {4 \, {\left (\sqrt {b d} b^{3} c^{3} {\left | b \right |} - 3 \, \sqrt {b d} a b^{2} c^{2} d {\left | b \right |} + 3 \, \sqrt {b d} a^{2} b c d^{2} {\left | b \right |} - \sqrt {b d} a^{3} d^{3} {\left | b \right |}\right )}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} b^{4}} \] Input:

integrate((d*x+c)^(5/2)/(b*x+a)^(3/2),x, algorithm="giac")
 

Output:

1/4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*d^2*abs 
(b)/b^5 + 9*(b^10*c*d^3*abs(b) - a*b^9*d^4*abs(b))/(b^14*d^2)) - 15/8*(sqr 
t(b*d)*b^2*c^2*abs(b) - 2*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b*d)*a^2*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*(sqrt(b*d)*b^3*c^3*abs(b) - 3*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 3*s 
qrt(b*d)*a^2*b*c*d^2*abs(b) - sqrt(b*d)*a^3*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)*b^4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/2}}{{\left (a+b\,x\right )}^{3/2}} \,d x \] Input:

int((c + d*x)^(5/2)/(a + b*x)^(3/2),x)
 

Output:

int((c + d*x)^(5/2)/(a + b*x)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.20 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {15 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} d^{2}-30 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a b c d +15 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{2} c^{2}-10 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a^{2} d^{2}+20 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, a b c d -10 \sqrt {d}\, \sqrt {b}\, \sqrt {b x +a}\, b^{2} c^{2}-15 \sqrt {d x +c}\, a^{2} b \,d^{2}+25 \sqrt {d x +c}\, a \,b^{2} c d -5 \sqrt {d x +c}\, a \,b^{2} d^{2} x -8 \sqrt {d x +c}\, b^{3} c^{2}+9 \sqrt {d x +c}\, b^{3} c d x +2 \sqrt {d x +c}\, b^{3} d^{2} x^{2}}{4 \sqrt {b x +a}\, b^{4}} \] Input:

int((d*x+c)^(5/2)/(b*x+a)^(3/2),x)
 

Output:

(15*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqr 
t(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 + 15*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**2 - 10*sqrt(d)*sqrt(b)*sqrt(a + b 
*x)*a**2*d**2 + 20*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b*c*d - 10*sqrt(d)*sqrt 
(b)*sqrt(a + b*x)*b**2*c**2 - 15*sqrt(c + d*x)*a**2*b*d**2 + 25*sqrt(c + d 
*x)*a*b**2*c*d - 5*sqrt(c + d*x)*a*b**2*d**2*x - 8*sqrt(c + d*x)*b**3*c**2 
 + 9*sqrt(c + d*x)*b**3*c*d*x + 2*sqrt(c + d*x)*b**3*d**2*x**2)/(4*sqrt(a 
+ b*x)*b**4)