\(\int \frac {A+B x}{(c+d x)^{5/2} \sqrt {c^2-d^2 x^2}} \, dx\) [98]

Optimal result

Integrand size = 31, antiderivative size = 151 \[ \int \frac {A+B x}{(c+d x)^{5/2} \sqrt {c^2-d^2 x^2}} \, dx=\frac {(B c-A d) \sqrt {c^2-d^2 x^2}}{4 c d^2 (c+d x)^{5/2}}-\frac {(5 B c+3 A d) \sqrt {c^2-d^2 x^2}}{16 c^2 d^2 (c+d x)^{3/2}}-\frac {(5 B c+3 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{16 \sqrt {2} c^{5/2} d^2} \] Output:

1/4*(-A*d+B*c)*(-d^2*x^2+c^2)^(1/2)/c/d^2/(d*x+c)^(5/2)-1/16*(3*A*d+5*B*c) 
*(-d^2*x^2+c^2)^(1/2)/c^2/d^2/(d*x+c)^(3/2)-1/32*(3*A*d+5*B*c)*arctanh(2^( 
1/2)*c^(1/2)*(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2))*2^(1/2)/c^(5/2)/d^2
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{(c+d x)^{5/2} \sqrt {c^2-d^2 x^2}} \, dx=\frac {-\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} (A d (7 c+3 d x)+B c (c+5 d x))}{(c+d x)^{5/2}}-\sqrt {2} (5 B c+3 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{32 c^{5/2} d^2} \] Input:

Integrate[(A + B*x)/((c + d*x)^(5/2)*Sqrt[c^2 - d^2*x^2]),x]
 

Output:

((-2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(A*d*(7*c + 3*d*x) + B*c*(c + 5*d*x)))/(c 
 + d*x)^(5/2) - Sqrt[2]*(5*B*c + 3*A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + 
d*x])/Sqrt[c^2 - d^2*x^2]])/(32*c^(5/2)*d^2)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {671, 470, 471, 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.

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {(3 A d+5 B c) \int \frac {1}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}dx}{8 c d}+\frac {\sqrt {c^2-d^2 x^2} (B c-A d)}{4 c d^2 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(3 A d+5 B c) \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{4 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c d}+\frac {\sqrt {c^2-d^2 x^2} (B c-A d)}{4 c d^2 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(3 A d+5 B c) \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{2 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c d}+\frac {\sqrt {c^2-d^2 x^2} (B c-A d)}{4 c d^2 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(3 A d+5 B c) \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{2 \sqrt {2} c^{3/2} d}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c d}+\frac {\sqrt {c^2-d^2 x^2} (B c-A d)}{4 c d^2 (c+d x)^{5/2}}\)

Input:

Int[(A + B*x)/((c + d*x)^(5/2)*Sqrt[c^2 - d^2*x^2]),x]
 

Output:

((B*c - A*d)*Sqrt[c^2 - d^2*x^2])/(4*c*d^2*(c + d*x)^(5/2)) + ((5*B*c + 3* 
A*d)*(-1/2*Sqrt[c^2 - d^2*x^2]/(c*d*(c + d*x)^(3/2)) - ArcTanh[Sqrt[c^2 - 
d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])]/(2*Sqrt[2]*c^(3/2)*d)))/(8*c*d)
 

Defintions of rubi rules used

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]
 

rule 470

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 
2*p + 2)/(2*c*(n + p + 1))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; 
 FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + 
 p + 1, 0] && IntegerQ[2*p]
 

rule 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m 
 + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m 
+ p + 1))   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, 
e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p 
 + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p 
 + 1, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(269\) vs. \(2(125)=250\).

Time = 0.36 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.79

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (3 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{3} x^{2}+5 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{2} x^{2}+6 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{2} x +10 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d x +3 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d +6 A \,d^{2} x \sqrt {-d x +c}\, \sqrt {c}+5 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}+10 B \,c^{\frac {3}{2}} d x \sqrt {-d x +c}+14 A \sqrt {-d x +c}\, c^{\frac {3}{2}} d +2 B \sqrt {-d x +c}\, c^{\frac {5}{2}}\right )}{32 c^{\frac {5}{2}} \left (d x +c \right )^{\frac {5}{2}} \sqrt {-d x +c}\, d^{2}}\)	\(270\)

Input:

int((B*x+A)/(d*x+c)^(5/2)/(-d^2*x^2+c^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/32*(-d^2*x^2+c^2)^(1/2)/c^(5/2)*(3*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2) 
*2^(1/2)/c^(1/2))*d^3*x^2+5*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c 
^(1/2))*c*d^2*x^2+6*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))* 
c*d^2*x+10*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d*x+3 
*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d+6*A*d^2*x*(-d 
*x+c)^(1/2)*c^(1/2)+5*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2) 
)*c^3+10*B*c^(3/2)*d*x*(-d*x+c)^(1/2)+14*A*(-d*x+c)^(1/2)*c^(3/2)*d+2*B*(- 
d*x+c)^(1/2)*c^(5/2))/(d*x+c)^(5/2)/(-d*x+c)^(1/2)/d^2
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 467, normalized size of antiderivative = 3.09 \[ \int \frac {A+B x}{(c+d x)^{5/2} \sqrt {c^2-d^2 x^2}} \, dx=\left [\frac {\sqrt {2} {\left (5 \, B c^{4} + 3 \, A c^{3} d + {\left (5 \, B c d^{3} + 3 \, A d^{4}\right )} x^{3} + 3 \, {\left (5 \, B c^{2} d^{2} + 3 \, A c d^{3}\right )} x^{2} + 3 \, {\left (5 \, B c^{3} d + 3 \, A c^{2} d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 4 \, {\left (B c^{3} + 7 \, A c^{2} d + {\left (5 \, B c^{2} d + 3 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{64 \, {\left (c^{3} d^{5} x^{3} + 3 \, c^{4} d^{4} x^{2} + 3 \, c^{5} d^{3} x + c^{6} d^{2}\right )}}, \frac {\sqrt {2} {\left (5 \, B c^{4} + 3 \, A c^{3} d + {\left (5 \, B c d^{3} + 3 \, A d^{4}\right )} x^{3} + 3 \, {\left (5 \, B c^{2} d^{2} + 3 \, A c d^{3}\right )} x^{2} + 3 \, {\left (5 \, B c^{3} d + 3 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) - 2 \, {\left (B c^{3} + 7 \, A c^{2} d + {\left (5 \, B c^{2} d + 3 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{32 \, {\left (c^{3} d^{5} x^{3} + 3 \, c^{4} d^{4} x^{2} + 3 \, c^{5} d^{3} x + c^{6} d^{2}\right )}}\right ] \] Input:

integrate((B*x+A)/(d*x+c)^(5/2)/(-d^2*x^2+c^2)^(1/2),x, algorithm="fricas" 
)
 

Output:

[1/64*(sqrt(2)*(5*B*c^4 + 3*A*c^3*d + (5*B*c*d^3 + 3*A*d^4)*x^3 + 3*(5*B*c 
^2*d^2 + 3*A*c*d^3)*x^2 + 3*(5*B*c^3*d + 3*A*c^2*d^2)*x)*sqrt(c)*log(-(d^2 
*x^2 - 2*c*d*x + 2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(c) - 3* 
c^2)/(d^2*x^2 + 2*c*d*x + c^2)) - 4*(B*c^3 + 7*A*c^2*d + (5*B*c^2*d + 3*A* 
c*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^3*d^5*x^3 + 3*c^4*d^4*x^2 
 + 3*c^5*d^3*x + c^6*d^2), 1/32*(sqrt(2)*(5*B*c^4 + 3*A*c^3*d + (5*B*c*d^3 
 + 3*A*d^4)*x^3 + 3*(5*B*c^2*d^2 + 3*A*c*d^3)*x^2 + 3*(5*B*c^3*d + 3*A*c^2 
*d^2)*x)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sq 
rt(-c)/(c*d*x + c^2)) - 2*(B*c^3 + 7*A*c^2*d + (5*B*c^2*d + 3*A*c*d^2)*x)* 
sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^3*d^5*x^3 + 3*c^4*d^4*x^2 + 3*c^5*d 
^3*x + c^6*d^2)]
 

Sympy [F]

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

integrate((B*x+A)/(d*x+c)**(5/2)/(-d**2*x**2+c**2)**(1/2),x)
 

Output:

Integral((A + B*x)/(sqrt(-(-c + d*x)*(c + d*x))*(c + d*x)**(5/2)), x)
 

Maxima [F]

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

integrate((B*x+A)/(d*x+c)^(5/2)/(-d^2*x^2+c^2)^(1/2),x, algorithm="maxima" 
)
 

Output:

integrate((B*x + A)/(sqrt(-d^2*x^2 + c^2)*(d*x + c)^(5/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.73 \[ \int \frac {A+B x}{(c+d x)^{5/2} \sqrt {c^2-d^2 x^2}} \, dx=\frac {\frac {\sqrt {2} {\left (5 \, B c + 3 \, A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{2}} + \frac {2 \, {\left (5 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c - 6 \, \sqrt {-d x + c} B c^{2} + 3 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d - 10 \, \sqrt {-d x + c} A c d\right )}}{{\left (d x + c\right )}^{2} c^{2}}}{32 \, d^{2}} \] Input:

integrate((B*x+A)/(d*x+c)^(5/2)/(-d^2*x^2+c^2)^(1/2),x, algorithm="giac")
 

Output:

1/32*(sqrt(2)*(5*B*c + 3*A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c))/ 
(sqrt(-c)*c^2) + 2*(5*(-d*x + c)^(3/2)*B*c - 6*sqrt(-d*x + c)*B*c^2 + 3*(- 
d*x + c)^(3/2)*A*d - 10*sqrt(-d*x + c)*A*c*d)/((d*x + c)^2*c^2))/d^2
 

Mupad [F(-1)]

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

int((A + B*x)/((c^2 - d^2*x^2)^(1/2)*(c + d*x)^(5/2)),x)
 

Output:

int((A + B*x)/((c^2 - d^2*x^2)^(1/2)*(c + d*x)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 402, normalized size of antiderivative = 2.66 \[ \int \frac {A+B x}{(c+d x)^{5/2} \sqrt {c^2-d^2 x^2}} \, dx=\frac {-28 \sqrt {-d x +c}\, a \,c^{2} d -12 \sqrt {-d x +c}\, a c \,d^{2} x -4 \sqrt {-d x +c}\, b \,c^{3}-20 \sqrt {-d x +c}\, b \,c^{2} d x +3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,c^{2} d +6 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a c \,d^{2} x +3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,d^{3} x^{2}+5 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b \,c^{3}+10 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b \,c^{2} d x +5 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c \,d^{2} x^{2}-3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,c^{2} d -6 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a c \,d^{2} x -3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,d^{3} x^{2}-5 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b \,c^{3}-10 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b \,c^{2} d x -5 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c \,d^{2} x^{2}}{64 c^{3} d^{2} \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:

int((B*x+A)/(d*x+c)^(5/2)/(-d^2*x^2+c^2)^(1/2),x)
 

Output:

( - 28*sqrt(c - d*x)*a*c**2*d - 12*sqrt(c - d*x)*a*c*d**2*x - 4*sqrt(c - d 
*x)*b*c**3 - 20*sqrt(c - d*x)*b*c**2*d*x + 3*sqrt(c)*sqrt(2)*log(sqrt(c - 
d*x) - sqrt(c)*sqrt(2))*a*c**2*d + 6*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - s 
qrt(c)*sqrt(2))*a*c*d**2*x + 3*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c) 
*sqrt(2))*a*d**3*x**2 + 5*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt 
(2))*b*c**3 + 10*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c* 
*2*d*x + 5*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c*d**2*x 
**2 - 3*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*c**2*d - 6* 
sqrt(c)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*c*d**2*x - 3*sqrt(c 
)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*d**3*x**2 - 5*sqrt(c)*sqr 
t(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**3 - 10*sqrt(c)*sqrt(2)*log( 
sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**2*d*x - 5*sqrt(c)*sqrt(2)*log(sqrt(c 
 - d*x) + sqrt(c)*sqrt(2))*b*c*d**2*x**2)/(64*c**3*d**2*(c**2 + 2*c*d*x + 
d**2*x**2))