\(\int \frac {(A+B x) (c^2-d^2 x^2)^{3/2}}{(c+d x)^2} \, dx\) [17]

Optimal result

Integrand size = 29, antiderivative size = 133 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^2} \, dx=-\frac {(2 B c-3 A d) x \sqrt {c^2-d^2 x^2}}{2 d}+\frac {B \left (c^2-d^2 x^2\right )^{3/2}}{3 d^2}-\frac {2 (B c-A d) \left (c^2-d^2 x^2\right )^{3/2}}{d^2 (c+d x)}-\frac {c^2 (2 B c-3 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^2} \] Output:

-1/2*(-3*A*d+2*B*c)*x*(-d^2*x^2+c^2)^(1/2)/d+1/3*B*(-d^2*x^2+c^2)^(3/2)/d^ 
2-2*(-A*d+B*c)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)-1/2*c^2*(-3*A*d+2*B*c)*arc 
tan(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^2} \, dx=\frac {d \sqrt {c^2-d^2 x^2} \left (-3 A d (-4 c+d x)-2 B \left (5 c^2-3 c d x+d^2 x^2\right )\right )+3 c^2 \sqrt {-d^2} (-2 B c+3 A d) \log \left (-\sqrt {-d^2} x+\sqrt {c^2-d^2 x^2}\right )}{6 d^3} \] Input:

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

Output:

(d*Sqrt[c^2 - d^2*x^2]*(-3*A*d*(-4*c + d*x) - 2*B*(5*c^2 - 3*c*d*x + d^2*x 
^2)) + 3*c^2*Sqrt[-d^2]*(-2*B*c + 3*A*d)*Log[-(Sqrt[-d^2]*x) + Sqrt[c^2 - 
d^2*x^2]])/(6*d^3)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {671, 466, 211, 224, 216}

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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

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 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 [A] (verified)

Time = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {\left (-2 B \,d^{2} x^{2}-3 A \,d^{2} x +6 B c d x +12 A c d -10 B \,c^{2}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{6 d^{2}}+\frac {c^{2} \left (3 A d -2 B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d \sqrt {d^{2}}}\)	\(99\)
default	\(\frac {B \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )\right )}{d^{2}}+\frac {\left (A d -B c \right ) \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}+\frac {3 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )\right )}{c}\right )}{d^{3}}\)	\(332\)

Input:

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

Output:

1/6*(-2*B*d^2*x^2-3*A*d^2*x+6*B*c*d*x+12*A*c*d-10*B*c^2)/d^2*(-d^2*x^2+c^2 
)^(1/2)+1/2*c^2/d*(3*A*d-2*B*c)/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2 
+c^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.76 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^2} \, dx=\frac {6 \, {\left (2 \, B c^{3} - 3 \, A c^{2} d\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (2 \, B d^{2} x^{2} + 10 \, B c^{2} - 12 \, A c d - 3 \, {\left (2 \, B c d - A d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{6 \, d^{2}} \] Input:

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

Output:

1/6*(6*(2*B*c^3 - 3*A*c^2*d)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) - ( 
2*B*d^2*x^2 + 10*B*c^2 - 12*A*c*d - 3*(2*B*c*d - A*d^2)*x)*sqrt(-d^2*x^2 + 
 c^2))/d^2
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.73 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^2} \, dx=-\frac {i \, B c^{3} \arcsin \left (\frac {d x}{c} + 2\right )}{2 \, d^{2}} - \frac {3 \, B c^{3} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{2}} + \frac {3 \, A c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c}{2 \, {\left (d^{3} x + c d^{2}\right )}} + \frac {\sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} B c x}{2 \, d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A}{2 \, {\left (d^{2} x + c d\right )}} + \frac {\sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} B c^{2}}{d^{2}} - \frac {3 \, \sqrt {-d^{2} x^{2} + c^{2}} B c^{2}}{2 \, d^{2}} + \frac {3 \, \sqrt {-d^{2} x^{2} + c^{2}} A c}{2 \, d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B}{3 \, d^{2}} \] Input:

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

Output:

-1/2*I*B*c^3*arcsin(d*x/c + 2)/d^2 - 3/2*B*c^3*arcsin(d*x/c)/d^2 + 3/2*A*c 
^2*arcsin(d*x/c)/d - 1/2*(-d^2*x^2 + c^2)^(3/2)*B*c/(d^3*x + c*d^2) + 1/2* 
sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*B*c*x/d + 1/2*(-d^2*x^2 + c^2)^(3/2)*A/(d^ 
2*x + c*d) + sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*B*c^2/d^2 - 3/2*sqrt(-d^2*x^2 
 + c^2)*B*c^2/d^2 + 3/2*sqrt(-d^2*x^2 + c^2)*A*c/d + 1/3*(-d^2*x^2 + c^2)^ 
(3/2)*B/d^2
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (119) = 238\).

Time = 0.15 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.09 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^2} \, dx=\frac {{\left (24 \, {\left (2 \, B c^{4} d^{4} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 3 \, A c^{3} d^{5} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )\right )} \arctan \left (\sqrt {\frac {2 \, c}{d x + c} - 1}\right ) - \frac {{\left (18 \, B c^{4} d^{4} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 15 \, A c^{3} d^{5} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 16 \, B c^{4} d^{4} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 24 \, A c^{3} d^{5} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 6 \, B c^{4} d^{4} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 9 \, A c^{3} d^{5} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )\right )} {\left (d x + c\right )}^{3}}{c^{3}}\right )} {\left | d \right |}}{24 \, c d^{7}} \] Input:

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

Output:

1/24*(24*(2*B*c^4*d^4*sgn(1/(d*x + c))*sgn(d) - 3*A*c^3*d^5*sgn(1/(d*x + c 
))*sgn(d))*arctan(sqrt(2*c/(d*x + c) - 1)) - (18*B*c^4*d^4*(2*c/(d*x + c) 
- 1)^(5/2)*sgn(1/(d*x + c))*sgn(d) - 15*A*c^3*d^5*(2*c/(d*x + c) - 1)^(5/2 
)*sgn(1/(d*x + c))*sgn(d) + 16*B*c^4*d^4*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/( 
d*x + c))*sgn(d) - 24*A*c^3*d^5*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c)) 
*sgn(d) + 6*B*c^4*d^4*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) - 9* 
A*c^3*d^5*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d))*(d*x + c)^3/c^3 
)*abs(d)/(c*d^7)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^2} \, dx=\frac {9 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d -6 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3}+12 \sqrt {-d^{2} x^{2}+c^{2}}\, a c d -3 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{2} x -10 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2}+6 \sqrt {-d^{2} x^{2}+c^{2}}\, b c d x -2 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{2} x^{2}-12 a \,c^{2} d +10 b \,c^{3}}{6 d^{2}} \] Input:

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

Output:

(9*asin((d*x)/c)*a*c**2*d - 6*asin((d*x)/c)*b*c**3 + 12*sqrt(c**2 - d**2*x 
**2)*a*c*d - 3*sqrt(c**2 - d**2*x**2)*a*d**2*x - 10*sqrt(c**2 - d**2*x**2) 
*b*c**2 + 6*sqrt(c**2 - d**2*x**2)*b*c*d*x - 2*sqrt(c**2 - d**2*x**2)*b*d* 
*2*x**2 - 12*a*c**2*d + 10*b*c**3)/(6*d**2)