\(\int \frac {(A+B x) (c^2-d^2 x^2)^{3/2}}{(c+d x)^{5/2}} \, dx\) [77]

Optimal result

Integrand size = 31, antiderivative size = 173 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}} \, dx=-\frac {4 c (B c-A d) \sqrt {c^2-d^2 x^2}}{d^2 \sqrt {c+d x}}-\frac {2 (B c-A d) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^2 (c+d x)^{3/2}}-\frac {2 B \left (c^2-d^2 x^2\right )^{5/2}}{5 d^2 (c+d x)^{5/2}}+\frac {4 \sqrt {2} c^{3/2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{d^2} \] Output:

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

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.72 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {2 \left (-\frac {\sqrt {c^2-d^2 x^2} \left (5 A d (-7 c+d x)+B \left (38 c^2-11 c d x+3 d^2 x^2\right )\right )}{\sqrt {c+d x}}+30 \sqrt {2} c^{3/2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )\right )}{15 d^2} \] Input:

Integrate[((A + B*x)*(c^2 - d^2*x^2)^(3/2))/(c + d*x)^(5/2),x]
 

Output:

(2*(-((Sqrt[c^2 - d^2*x^2]*(5*A*d*(-7*c + d*x) + B*(38*c^2 - 11*c*d*x + 3* 
d^2*x^2)))/Sqrt[c + d*x]) + 30*Sqrt[2]*c^(3/2)*(B*c - A*d)*ArcTanh[(Sqrt[2 
]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]]))/(15*d^2)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {672, 466, 466, 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.

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

\(\Big \downarrow \) 672

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 471

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

\(\Big \downarrow \) 221

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

Input:

Int[((A + B*x)*(c^2 - d^2*x^2)^(3/2))/(c + d*x)^(5/2),x]
 

Output:

(-2*B*(c^2 - d^2*x^2)^(5/2))/(5*d^2*(c + d*x)^(5/2)) - ((B*c - A*d)*((2*(c 
^2 - d^2*x^2)^(3/2))/(3*d*(c + d*x)^(3/2)) + 2*c*((2*Sqrt[c^2 - d^2*x^2])/ 
(d*Sqrt[c + d*x]) - (2*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2 
]*Sqrt[c]*Sqrt[c + d*x])])/d)))/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 466

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 
2*(n + 2*p + 1)))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr 
eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 
] || EqQ[n + p + 1, 0]) && NeQ[n + 2*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 672

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), 
 x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2))   Int[(d + e*x 
)^m*(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] && NeQ[m + 2*p + 2, 0]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (30 A \,c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d -30 B \,c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )+3 B \,d^{2} x^{2} \sqrt {-d x +c}+5 A \,d^{2} x \sqrt {-d x +c}-11 B c d x \sqrt {-d x +c}-35 A c d \sqrt {-d x +c}+38 B \,c^{2} \sqrt {-d x +c}\right )}{15 \sqrt {d x +c}\, \sqrt {-d x +c}\, d^{2}}\)	\(162\)
risch	\(\frac {2 \left (-3 B \,d^{2} x^{2}-5 A \,d^{2} x +11 B c d x +35 A c d -38 B \,c^{2}\right ) \sqrt {-d x +c}\, \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{15 d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}-\frac {4 c^{\frac {3}{2}} \left (A d -B c \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(171\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/15*(-d^2*x^2+c^2)^(1/2)*(30*A*c^(3/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2 
)*2^(1/2)/c^(1/2))*d-30*B*c^(5/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/ 
2)/c^(1/2))+3*B*d^2*x^2*(-d*x+c)^(1/2)+5*A*d^2*x*(-d*x+c)^(1/2)-11*B*c*d*x 
*(-d*x+c)^(1/2)-35*A*c*d*(-d*x+c)^(1/2)+38*B*c^2*(-d*x+c)^(1/2))/(d*x+c)^( 
1/2)/(-d*x+c)^(1/2)/d^2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.94 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}} \, dx=\left [-\frac {2 \, {\left (15 \, \sqrt {2} {\left (B c^{3} - A c^{2} d + {\left (B c^{2} d - A c d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + {\left (3 \, B d^{2} x^{2} + 38 \, B c^{2} - 35 \, A c d - {\left (11 \, B c d - 5 \, A d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}\right )}}{15 \, {\left (d^{3} x + c d^{2}\right )}}, -\frac {2 \, {\left (30 \, \sqrt {2} {\left (B c^{3} - A c^{2} d + {\left (B c^{2} d - A c d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + {\left (3 \, B d^{2} x^{2} + 38 \, B c^{2} - 35 \, A c d - {\left (11 \, B c d - 5 \, A d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}\right )}}{15 \, {\left (d^{3} x + c d^{2}\right )}}\right ] \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas" 
)
 

Output:

[-2/15*(15*sqrt(2)*(B*c^3 - A*c^2*d + (B*c^2*d - A*c*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)) + (3*B*d^2*x^2 + 38*B*c^2 - 35*A*c*d - 
 (11*B*c*d - 5*A*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(d^3*x + c*d^ 
2), -2/15*(30*sqrt(2)*(B*c^3 - A*c^2*d + (B*c^2*d - A*c*d^2)*x)*sqrt(-c)*a 
rctan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + c^2 
)) + (3*B*d^2*x^2 + 38*B*c^2 - 35*A*c*d - (11*B*c*d - 5*A*d^2)*x)*sqrt(-d^ 
2*x^2 + c^2)*sqrt(d*x + c))/(d^3*x + c*d^2)]
 

Sympy [F]

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

integrate((B*x+A)*(-d**2*x**2+c**2)**(3/2)/(d*x+c)**(5/2),x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(3/2)*(A + B*x)/(c + d*x)**(5/2), x)
 

Maxima [F]

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

integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(5/2),x, algorithm="maxima" 
)
 

Output:

integrate((-d^2*x^2 + c^2)^(3/2)*(B*x + A)/(d*x + c)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B + 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c + 30 \, \sqrt {-d x + c} B c^{2} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d - 30 \, \sqrt {-d x + c} A c d + \frac {30 \, \sqrt {2} {\left (B c^{3} - A c^{2} d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}}\right )}}{15 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

-2/15*(3*(d*x - c)^2*sqrt(-d*x + c)*B + 5*(-d*x + c)^(3/2)*B*c + 30*sqrt(- 
d*x + c)*B*c^2 - 5*(-d*x + c)^(3/2)*A*d - 30*sqrt(-d*x + c)*A*c*d + 30*sqr 
t(2)*(B*c^3 - A*c^2*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c))/sqrt(-c 
))/d^2
 

Mupad [F(-1)]

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

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x))/(c + d*x)^(5/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x))/(c + d*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {\frac {14 \sqrt {-d x +c}\, a c d}{3}-\frac {2 \sqrt {-d x +c}\, a \,d^{2} x}{3}-\frac {76 \sqrt {-d x +c}\, b \,c^{2}}{15}+\frac {22 \sqrt {-d x +c}\, b c d x}{15}-\frac {2 \sqrt {-d x +c}\, b \,d^{2} x^{2}}{5}+4 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a c d -4 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{2}-\frac {16 \sqrt {c}\, \sqrt {2}\, a c d}{3}+\frac {104 \sqrt {c}\, \sqrt {2}\, b \,c^{2}}{15}}{d^{2}} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(5/2),x)
 

Output:

(2*(35*sqrt(c - d*x)*a*c*d - 5*sqrt(c - d*x)*a*d**2*x - 38*sqrt(c - d*x)*b 
*c**2 + 11*sqrt(c - d*x)*b*c*d*x - 3*sqrt(c - d*x)*b*d**2*x**2 + 30*sqrt(c 
)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c*d - 30*sqr 
t(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**2 - 40 
*sqrt(c)*sqrt(2)*a*c*d + 52*sqrt(c)*sqrt(2)*b*c**2))/(15*d**2)