\(\int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx\) [177]

Optimal result

Integrand size = 24, antiderivative size = 111 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=-\frac {2 (B d-A e) (a+b x)^{3/2}}{3 e (b d-a e) (d+e x)^{3/2}}-\frac {2 B \sqrt {a+b x}}{e^2 \sqrt {d+e x}}+\frac {2 \sqrt {b} B \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{5/2}} \] Output:

-2/3*(-A*e+B*d)*(b*x+a)^(3/2)/e/(-a*e+b*d)/(e*x+d)^(3/2)-2*B*(b*x+a)^(1/2) 
/e^2/(e*x+d)^(1/2)+2*b^(1/2)*B*arctanh(e^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(e*x+ 
d)^(1/2))/e^(5/2)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=-\frac {2 \sqrt {a+b x} \left (A b e^2 x-b B d (3 d+4 e x)+a e (2 B d+A e+3 B e x)\right )}{3 e^2 (-b d+a e) (d+e x)^{3/2}}+\frac {2 \sqrt {b} B \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{5/2}} \] Input:

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

Output:

(-2*Sqrt[a + b*x]*(A*b*e^2*x - b*B*d*(3*d + 4*e*x) + a*e*(2*B*d + A*e + 3* 
B*e*x)))/(3*e^2*(-(b*d) + a*e)*(d + e*x)^(3/2)) + (2*Sqrt[b]*B*ArcTanh[(Sq 
rt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/e^(5/2)
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 57, 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.

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-2*(B*d - A*e)*(a + b*x)^(3/2))/(3*e*(b*d - a*e)*(d + e*x)^(3/2)) + (B*(( 
-2*Sqrt[a + b*x])/(e*Sqrt[d + e*x]) + (2*Sqrt[b]*ArcTanh[(Sqrt[e]*Sqrt[a + 
 b*x])/(Sqrt[b]*Sqrt[d + e*x])])/e^(3/2)))/e
 

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 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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(502\) vs. \(2(89)=178\).

Time = 0.27 (sec) , antiderivative size = 503, normalized size of antiderivative = 4.53

method	result	size
default	\(-\frac {\left (-3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a b \,e^{3} x^{2}+3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b^{2} d \,e^{2} x^{2}-6 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a b d \,e^{2} x +6 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b^{2} d^{2} e x +2 A b \,e^{2} x \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}-3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a b \,d^{2} e +3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b^{2} d^{3}+6 B a \,e^{2} x \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}-8 B b d e x \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+2 A a \,e^{2} \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+4 B a d e \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}-6 B b \,d^{2} \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\right ) \sqrt {b x +a}}{3 \sqrt {b e}\, \left (a e -d b \right ) \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, e^{2} \left (e x +d \right )^{\frac {3}{2}}}\)	\(503\)

Input:

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

Output:

-1/3*(-3*B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/ 
(b*e)^(1/2))*a*b*e^3*x^2+3*B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b* 
e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*b^2*d*e^2*x^2-6*B*ln(1/2*(2*b*e*x+2*((e*x+d 
)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*a*b*d*e^2*x+6*B*ln(1/2* 
(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*b^2*d 
^2*e*x+2*A*b*e^2*x*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)-3*B*ln(1/2*(2*b*e*x 
+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*a*b*d^2*e+3*B 
*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2 
))*b^2*d^3+6*B*a*e^2*x*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)-8*B*b*d*e*x*((e 
*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+2*A*a*e^2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^( 
1/2)+4*B*a*d*e*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)-6*B*b*d^2*((e*x+d)*(b*x 
+a))^(1/2)*(b*e)^(1/2))*(b*x+a)^(1/2)/(b*e)^(1/2)/(a*e-b*d)/((e*x+d)*(b*x+ 
a))^(1/2)/e^2/(e*x+d)^(3/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (89) = 178\).

Time = 0.67 (sec) , antiderivative size = 515, normalized size of antiderivative = 4.64 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=\left [\frac {3 \, {\left (B b d^{3} - B a d^{2} e + {\left (B b d e^{2} - B a e^{3}\right )} x^{2} + 2 \, {\left (B b d^{2} e - B a d e^{2}\right )} x\right )} \sqrt {\frac {b}{e}} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e^{2} x + b d e + a e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {\frac {b}{e}} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \, {\left (3 \, B b d^{2} - 2 \, B a d e - A a e^{2} + {\left (4 \, B b d e - {\left (3 \, B a + A b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{6 \, {\left (b d^{3} e^{2} - a d^{2} e^{3} + {\left (b d e^{4} - a e^{5}\right )} x^{2} + 2 \, {\left (b d^{2} e^{3} - a d e^{4}\right )} x\right )}}, -\frac {3 \, {\left (B b d^{3} - B a d^{2} e + {\left (B b d e^{2} - B a e^{3}\right )} x^{2} + 2 \, {\left (B b d^{2} e - B a d e^{2}\right )} x\right )} \sqrt {-\frac {b}{e}} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {-\frac {b}{e}}}{2 \, {\left (b^{2} e x^{2} + a b d + {\left (b^{2} d + a b e\right )} x\right )}}\right ) + 2 \, {\left (3 \, B b d^{2} - 2 \, B a d e - A a e^{2} + {\left (4 \, B b d e - {\left (3 \, B a + A b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{3 \, {\left (b d^{3} e^{2} - a d^{2} e^{3} + {\left (b d e^{4} - a e^{5}\right )} x^{2} + 2 \, {\left (b d^{2} e^{3} - a d e^{4}\right )} x\right )}}\right ] \] Input:

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

Output:

[1/6*(3*(B*b*d^3 - B*a*d^2*e + (B*b*d*e^2 - B*a*e^3)*x^2 + 2*(B*b*d^2*e - 
B*a*d*e^2)*x)*sqrt(b/e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 
+ 4*(2*b*e^2*x + b*d*e + a*e^2)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(b/e) + 8* 
(b^2*d*e + a*b*e^2)*x) - 4*(3*B*b*d^2 - 2*B*a*d*e - A*a*e^2 + (4*B*b*d*e - 
 (3*B*a + A*b)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b*d^3*e^2 - a*d^2*e^3 
 + (b*d*e^4 - a*e^5)*x^2 + 2*(b*d^2*e^3 - a*d*e^4)*x), -1/3*(3*(B*b*d^3 - 
B*a*d^2*e + (B*b*d*e^2 - B*a*e^3)*x^2 + 2*(B*b*d^2*e - B*a*d*e^2)*x)*sqrt( 
-b/e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(-b 
/e)/(b^2*e*x^2 + a*b*d + (b^2*d + a*b*e)*x)) + 2*(3*B*b*d^2 - 2*B*a*d*e - 
A*a*e^2 + (4*B*b*d*e - (3*B*a + A*b)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d))/ 
(b*d^3*e^2 - a*d^2*e^3 + (b*d*e^4 - a*e^5)*x^2 + 2*(b*d^2*e^3 - a*d*e^4)*x 
)]
 

Sympy [F]

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

integrate((b*x+a)**(1/2)*(B*x+A)/(e*x+d)**(5/2),x)
 

Output:

Integral((A + B*x)*sqrt(a + b*x)/(d + e*x)**(5/2), x)
 

Maxima [F(-2)]

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

integrate((b*x+a)^(1/2)*(B*x+A)/(e*x+d)^(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(e*(a*e-b*d)>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (89) = 178\).

Time = 0.18 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=-\frac {2 \, B {\left | b \right |} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} e^{2}} - \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (4 \, B b^{4} d e^{2} {\left | b \right |} - 3 \, B a b^{3} e^{3} {\left | b \right |} - A b^{4} e^{3} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{3} d e^{3} - a b^{2} e^{4}} + \frac {3 \, {\left (B b^{5} d^{2} e {\left | b \right |} - 2 \, B a b^{4} d e^{2} {\left | b \right |} + B a^{2} b^{3} e^{3} {\left | b \right |}\right )}}{b^{3} d e^{3} - a b^{2} e^{4}}\right )}}{3 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {3}{2}}} \] Input:

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

Output:

-2*B*abs(b)*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e 
- a*b*e)))/(sqrt(b*e)*e^2) - 2/3*sqrt(b*x + a)*((4*B*b^4*d*e^2*abs(b) - 3* 
B*a*b^3*e^3*abs(b) - A*b^4*e^3*abs(b))*(b*x + a)/(b^3*d*e^3 - a*b^2*e^4) + 
 3*(B*b^5*d^2*e*abs(b) - 2*B*a*b^4*d*e^2*abs(b) + B*a^2*b^3*e^3*abs(b))/(b 
^3*d*e^3 - a*b^2*e^4))/(b^2*d + (b*x + a)*b*e - a*b*e)^(3/2)
 

Mupad [F(-1)]

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

int(((A + B*x)*(a + b*x)^(1/2))/(d + e*x)^(5/2),x)
 

Output:

int(((A + B*x)*(a + b*x)^(1/2))/(d + e*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.86 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=\frac {-\frac {2 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,e^{2}}{3}-2 \sqrt {e x +d}\, \sqrt {b x +a}\, b d e -\frac {8 \sqrt {e x +d}\, \sqrt {b x +a}\, b \,e^{2} x}{3}+2 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) b \,d^{2}+4 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) b d e x +2 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) b \,e^{2} x^{2}}{e^{3} \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:

int((b*x+a)^(1/2)*(B*x+A)/(e*x+d)^(5/2),x)
 

Output:

(2*( - sqrt(d + e*x)*sqrt(a + b*x)*a*e**2 - 3*sqrt(d + e*x)*sqrt(a + b*x)* 
b*d*e - 4*sqrt(d + e*x)*sqrt(a + b*x)*b*e**2*x + 3*sqrt(e)*sqrt(b)*log((sq 
rt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*b*d**2 + 6*s 
qrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a* 
e - b*d))*b*d*e*x + 3*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b) 
*sqrt(d + e*x))/sqrt(a*e - b*d))*b*e**2*x**2))/(3*e**3*(d**2 + 2*d*e*x + e 
**2*x**2))