\(\int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{11/2}} \, dx\) [72]

Optimal result

Integrand size = 31, antiderivative size = 236 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{11/2}} \, dx=\frac {(B c-A d) \sqrt {c^2-d^2 x^2}}{4 d^2 (c+d x)^{9/2}}-\frac {(17 B c-A d) \sqrt {c^2-d^2 x^2}}{48 c d^2 (c+d x)^{7/2}}+\frac {(11 B c+5 A d) \sqrt {c^2-d^2 x^2}}{384 c^2 d^2 (c+d x)^{5/2}}+\frac {(11 B c+5 A d) \sqrt {c^2-d^2 x^2}}{512 c^3 d^2 (c+d x)^{3/2}}+\frac {(11 B c+5 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{512 \sqrt {2} c^{7/2} d^2} \] Output:

1/4*(-A*d+B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(9/2)-1/48*(-A*d+17*B*c)*( 
-d^2*x^2+c^2)^(1/2)/c/d^2/(d*x+c)^(7/2)+1/384*(5*A*d+11*B*c)*(-d^2*x^2+c^2 
)^(1/2)/c^2/d^2/(d*x+c)^(5/2)+1/512*(5*A*d+11*B*c)*(-d^2*x^2+c^2)^(1/2)/c^ 
3/d^2/(d*x+c)^(3/2)+1/1024*(5*A*d+11*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^(7/2)/d^2
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.70 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{11/2}} \, dx=\frac {-\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (B c \left (83 c^3+357 c^2 d x-143 c d^2 x^2-33 d^3 x^3\right )-A d \left (-317 c^3+117 c^2 d x+65 c d^2 x^2+15 d^3 x^3\right )\right )}{(c+d x)^{9/2}}+3 \sqrt {2} (11 B c+5 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{3072 c^{7/2} d^2} \] Input:

Integrate[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x)^(11/2),x]
 

Output:

((-2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(B*c*(83*c^3 + 357*c^2*d*x - 143*c*d^2*x^ 
2 - 33*d^3*x^3) - A*d*(-317*c^3 + 117*c^2*d*x + 65*c*d^2*x^2 + 15*d^3*x^3) 
))/(c + d*x)^(9/2) + 3*Sqrt[2]*(11*B*c + 5*A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]*S 
qrt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/(3072*c^(7/2)*d^2)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {671, 465, 470, 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.

\(\displaystyle \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{11/2}} \, dx\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {(5 A d+11 B c) \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}}dx}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{8 c d^2 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(5 A d+11 B c) \left (-\frac {1}{6} \int \frac {1}{(c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}dx-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{8 c d^2 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(5 A d+11 B c) \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \int \frac {1}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}dx}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{8 c d^2 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(5 A d+11 B c) \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{4 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{8 c d^2 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(5 A d+11 B c) \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{2 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{8 c d^2 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(5 A d+11 B c) \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{2 \sqrt {2} c^{3/2} d}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{8 c d^2 (c+d x)^{11/2}}\)

Input:

Int[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x)^(11/2),x]
 

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(3/2))/(8*c*d^2*(c + d*x)^(11/2)) + ((11*B*c 
+ 5*A*d)*(-1/3*Sqrt[c^2 - d^2*x^2]/(d*(c + d*x)^(7/2)) + (Sqrt[c^2 - d^2*x 
^2]/(4*c*d*(c + d*x)^(5/2)) - (3*(-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*S 
qrt[2]*c^(3/2)*d)))/(8*c))/6))/(16*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 465

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 + p + 1))), x] - Simp[b*(p/(d^2*(n + 
 p + 1)))   Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, 
b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LtQ[n, -2] || EqQ[n 
+ 2*p + 1, 0]) && NeQ[n + p + 1, 0] && IntegerQ[2*p]
 

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. \(477\) vs. \(2(198)=396\).

Time = 0.38 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.03

method	result	size
default	\(\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (15 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{5} x^{4}+33 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{4} x^{4}+60 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{4} x^{3}+132 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{3} x^{3}+90 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{3} x^{2}+30 A \,d^{4} x^{3} \sqrt {-d x +c}\, \sqrt {c}+198 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{2} x^{2}+66 B \,c^{\frac {3}{2}} d^{3} x^{3} \sqrt {-d x +c}+60 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{2} x +130 A \,c^{\frac {3}{2}} d^{3} x^{2} \sqrt {-d x +c}+132 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d x +286 B \,c^{\frac {5}{2}} d^{2} x^{2} \sqrt {-d x +c}+15 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d +234 A \,c^{\frac {5}{2}} d^{2} x \sqrt {-d x +c}+33 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5}-714 B \,c^{\frac {7}{2}} d x \sqrt {-d x +c}-634 A \,c^{\frac {7}{2}} \sqrt {-d x +c}\, d -166 B \,c^{\frac {9}{2}} \sqrt {-d x +c}\right )}{3072 c^{\frac {7}{2}} \left (d x +c \right )^{\frac {9}{2}} \sqrt {-d x +c}\, d^{2}}\)	\(478\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(11/2),x,method=_RETURNVERBOSE)
 

Output:

1/3072*(-d^2*x^2+c^2)^(1/2)/c^(7/2)*(15*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/ 
2)*2^(1/2)/c^(1/2))*d^5*x^4+33*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2 
)/c^(1/2))*c*d^4*x^4+60*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/ 
2))*c*d^4*x^3+132*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^ 
2*d^3*x^3+90*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^3 
*x^2+30*A*d^4*x^3*(-d*x+c)^(1/2)*c^(1/2)+198*B*2^(1/2)*arctanh(1/2*(-d*x+c 
)^(1/2)*2^(1/2)/c^(1/2))*c^3*d^2*x^2+66*B*c^(3/2)*d^3*x^3*(-d*x+c)^(1/2)+6 
0*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d^2*x+130*A*c^ 
(3/2)*d^3*x^2*(-d*x+c)^(1/2)+132*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1 
/2)/c^(1/2))*c^4*d*x+286*B*c^(5/2)*d^2*x^2*(-d*x+c)^(1/2)+15*A*2^(1/2)*arc 
tanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4*d+234*A*c^(5/2)*d^2*x*(-d*x+c 
)^(1/2)+33*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^5-714*B 
*c^(7/2)*d*x*(-d*x+c)^(1/2)-634*A*c^(7/2)*(-d*x+c)^(1/2)*d-166*B*c^(9/2)*( 
-d*x+c)^(1/2))/(d*x+c)^(9/2)/(-d*x+c)^(1/2)/d^2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 709, normalized size of antiderivative = 3.00 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{11/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (11 \, B c^{6} + 5 \, A c^{5} d + {\left (11 \, B c d^{5} + 5 \, A d^{6}\right )} x^{5} + 5 \, {\left (11 \, B c^{2} d^{4} + 5 \, A c d^{5}\right )} x^{4} + 10 \, {\left (11 \, B c^{3} d^{3} + 5 \, A c^{2} d^{4}\right )} x^{3} + 10 \, {\left (11 \, B c^{4} d^{2} + 5 \, A c^{3} d^{3}\right )} x^{2} + 5 \, {\left (11 \, B c^{5} d + 5 \, A c^{4} 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 (83 \, B c^{5} + 317 \, A c^{4} d - 3 \, {\left (11 \, B c^{2} d^{3} + 5 \, A c d^{4}\right )} x^{3} - 13 \, {\left (11 \, B c^{3} d^{2} + 5 \, A c^{2} d^{3}\right )} x^{2} + 3 \, {\left (119 \, B c^{4} d - 39 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{6144 \, {\left (c^{4} d^{7} x^{5} + 5 \, c^{5} d^{6} x^{4} + 10 \, c^{6} d^{5} x^{3} + 10 \, c^{7} d^{4} x^{2} + 5 \, c^{8} d^{3} x + c^{9} d^{2}\right )}}, -\frac {3 \, \sqrt {2} {\left (11 \, B c^{6} + 5 \, A c^{5} d + {\left (11 \, B c d^{5} + 5 \, A d^{6}\right )} x^{5} + 5 \, {\left (11 \, B c^{2} d^{4} + 5 \, A c d^{5}\right )} x^{4} + 10 \, {\left (11 \, B c^{3} d^{3} + 5 \, A c^{2} d^{4}\right )} x^{3} + 10 \, {\left (11 \, B c^{4} d^{2} + 5 \, A c^{3} d^{3}\right )} x^{2} + 5 \, {\left (11 \, B c^{5} d + 5 \, A c^{4} 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 (83 \, B c^{5} + 317 \, A c^{4} d - 3 \, {\left (11 \, B c^{2} d^{3} + 5 \, A c d^{4}\right )} x^{3} - 13 \, {\left (11 \, B c^{3} d^{2} + 5 \, A c^{2} d^{3}\right )} x^{2} + 3 \, {\left (119 \, B c^{4} d - 39 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{3072 \, {\left (c^{4} d^{7} x^{5} + 5 \, c^{5} d^{6} x^{4} + 10 \, c^{6} d^{5} x^{3} + 10 \, c^{7} d^{4} x^{2} + 5 \, c^{8} d^{3} x + c^{9} d^{2}\right )}}\right ] \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(11/2),x, algorithm="fricas 
")
 

Output:

[1/6144*(3*sqrt(2)*(11*B*c^6 + 5*A*c^5*d + (11*B*c*d^5 + 5*A*d^6)*x^5 + 5* 
(11*B*c^2*d^4 + 5*A*c*d^5)*x^4 + 10*(11*B*c^3*d^3 + 5*A*c^2*d^4)*x^3 + 10* 
(11*B*c^4*d^2 + 5*A*c^3*d^3)*x^2 + 5*(11*B*c^5*d + 5*A*c^4*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)*sq 
rt(c) - 3*c^2)/(d^2*x^2 + 2*c*d*x + c^2)) - 4*(83*B*c^5 + 317*A*c^4*d - 3* 
(11*B*c^2*d^3 + 5*A*c*d^4)*x^3 - 13*(11*B*c^3*d^2 + 5*A*c^2*d^3)*x^2 + 3*( 
119*B*c^4*d - 39*A*c^3*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^4*d^ 
7*x^5 + 5*c^5*d^6*x^4 + 10*c^6*d^5*x^3 + 10*c^7*d^4*x^2 + 5*c^8*d^3*x + c^ 
9*d^2), -1/3072*(3*sqrt(2)*(11*B*c^6 + 5*A*c^5*d + (11*B*c*d^5 + 5*A*d^6)* 
x^5 + 5*(11*B*c^2*d^4 + 5*A*c*d^5)*x^4 + 10*(11*B*c^3*d^3 + 5*A*c^2*d^4)*x 
^3 + 10*(11*B*c^4*d^2 + 5*A*c^3*d^3)*x^2 + 5*(11*B*c^5*d + 5*A*c^4*d^2)*x) 
*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/( 
c*d*x + c^2)) + 2*(83*B*c^5 + 317*A*c^4*d - 3*(11*B*c^2*d^3 + 5*A*c*d^4)*x 
^3 - 13*(11*B*c^3*d^2 + 5*A*c^2*d^3)*x^2 + 3*(119*B*c^4*d - 39*A*c^3*d^2)* 
x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^4*d^7*x^5 + 5*c^5*d^6*x^4 + 10*c 
^6*d^5*x^3 + 10*c^7*d^4*x^2 + 5*c^8*d^3*x + c^9*d^2)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{11/2}} \, dx=\int \frac {\sqrt {- \left (- c + d x\right ) \left (c + d x\right )} \left (A + B x\right )}{\left (c + d x\right )^{\frac {11}{2}}}\, dx \] Input:

integrate((B*x+A)*(-d**2*x**2+c**2)**(1/2)/(d*x+c)**(11/2),x)
 

Output:

Integral(sqrt(-(-c + d*x)*(c + d*x))*(A + B*x)/(c + d*x)**(11/2), x)
 

Maxima [F]

\[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{11/2}} \, dx=\int { \frac {\sqrt {-d^{2} x^{2} + c^{2}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {11}{2}}} \,d x } \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(11/2),x, algorithm="maxima 
")
 

Output:

integrate(sqrt(-d^2*x^2 + c^2)*(B*x + A)/(d*x + c)^(11/2), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{11/2}} \, dx=-\frac {\frac {3 \, \sqrt {2} {\left (11 \, B c + 5 \, A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{3}} - \frac {2 \, {\left (33 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} B c + 242 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B c^{2} - 28 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c^{3} - 264 \, \sqrt {-d x + c} B c^{4} + 15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} A d + 110 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} A c d - 292 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c^{2} d - 120 \, \sqrt {-d x + c} A c^{3} d\right )}}{{\left (d x + c\right )}^{4} c^{3}}}{3072 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(11/2),x, algorithm="giac")
 

Output:

-1/3072*(3*sqrt(2)*(11*B*c + 5*A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt 
(-c))/(sqrt(-c)*c^3) - 2*(33*(d*x - c)^3*sqrt(-d*x + c)*B*c + 242*(d*x - c 
)^2*sqrt(-d*x + c)*B*c^2 - 28*(-d*x + c)^(3/2)*B*c^3 - 264*sqrt(-d*x + c)* 
B*c^4 + 15*(d*x - c)^3*sqrt(-d*x + c)*A*d + 110*(d*x - c)^2*sqrt(-d*x + c) 
*A*c*d - 292*(-d*x + c)^(3/2)*A*c^2*d - 120*sqrt(-d*x + c)*A*c^3*d)/((d*x 
+ c)^4*c^3))/d^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{11/2}} \, dx=\int \frac {\sqrt {c^2-d^2\,x^2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^{11/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x)^(11/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x)^(11/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.19 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{11/2}} \, dx =\text {Too large to display} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(11/2),x)
 

Output:

( - 634*sqrt(c - d*x)*a*c**4*d + 234*sqrt(c - d*x)*a*c**3*d**2*x + 130*sqr 
t(c - d*x)*a*c**2*d**3*x**2 + 30*sqrt(c - d*x)*a*c*d**4*x**3 - 166*sqrt(c 
- d*x)*b*c**5 - 714*sqrt(c - d*x)*b*c**4*d*x + 286*sqrt(c - d*x)*b*c**3*d* 
*2*x**2 + 66*sqrt(c - d*x)*b*c**2*d**3*x**3 - 15*sqrt(c)*sqrt(2)*log(tan(a 
sin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**4*d - 60*sqrt(c)*sqrt(2)*log 
(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**3*d**2*x - 90*sqrt(c)* 
sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**2*d**3*x**2 
 - 60*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a* 
c*d**4*x**3 - 15*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt( 
2)))/2))*a*d**5*x**4 - 33*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt 
(c)*sqrt(2)))/2))*b*c**5 - 132*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/ 
(sqrt(c)*sqrt(2)))/2))*b*c**4*d*x - 198*sqrt(c)*sqrt(2)*log(tan(asin(sqrt( 
c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**3*d**2*x**2 - 132*sqrt(c)*sqrt(2)*log 
(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**2*d**3*x**3 - 33*sqrt( 
c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c*d**4*x**4 
)/(3072*c**4*d**2*(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + 
d**4*x**4))