Integrand size = 31, antiderivative size = 190 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}} \, dx=\frac {(B c-A d) \sqrt {c^2-d^2 x^2}}{3 d^2 (c+d x)^{7/2}}-\frac {(13 B c-A d) \sqrt {c^2-d^2 x^2}}{24 c d^2 (c+d x)^{5/2}}+\frac {(3 B c+A d) \sqrt {c^2-d^2 x^2}}{32 c^2 d^2 (c+d x)^{3/2}}+\frac {(3 B c+A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{32 \sqrt {2} c^{5/2} d^2} \] Output:
1/3*(-A*d+B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(7/2)-1/24*(-A*d+13*B*c)*( -d^2*x^2+c^2)^(1/2)/c/d^2/(d*x+c)^(5/2)+1/32*(A*d+3*B*c)*(-d^2*x^2+c^2)^(1 /2)/c^2/d^2/(d*x+c)^(3/2)+1/64*(A*d+3*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
Time = 0.65 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.75 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}} \, dx=\frac {-\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (B c \left (11 c^2+34 c d x-9 d^2 x^2\right )+A d \left (25 c^2-10 c d x-3 d^2 x^2\right )\right )}{(c+d x)^{7/2}}+3 \sqrt {2} (3 B c+A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{192 c^{5/2} d^2} \] Input:
Integrate[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x)^(9/2),x]
Output:
((-2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(B*c*(11*c^2 + 34*c*d*x - 9*d^2*x^2) + A* d*(25*c^2 - 10*c*d*x - 3*d^2*x^2)))/(c + d*x)^(7/2) + 3*Sqrt[2]*(3*B*c + A *d)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/(192*c^( 5/2)*d^2)
Time = 0.50 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {671, 465, 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.
\(\displaystyle \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}} \, dx\) |
\(\Big \downarrow \) 671 |
\(\displaystyle \frac {(A d+3 B c) \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{7/2}}dx}{4 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{6 c d^2 (c+d x)^{9/2}}\) |
\(\Big \downarrow \) 465 |
\(\displaystyle \frac {(A d+3 B c) \left (-\frac {1}{4} \int \frac {1}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}dx-\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )}{4 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{6 c d^2 (c+d x)^{9/2}}\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {(A d+3 B c) \left (\frac {1}{4} \left (\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}-\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{4 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )}{4 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{6 c d^2 (c+d x)^{9/2}}\) |
\(\Big \downarrow \) 471 |
\(\displaystyle \frac {(A d+3 B c) \left (\frac {1}{4} \left (\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}-\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{2 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )}{4 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{6 c d^2 (c+d x)^{9/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(A d+3 B c) \left (\frac {1}{4} \left (\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{2 \sqrt {2} c^{3/2} d}+\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )-\frac {\sqrt {c^2-d^2 x^2}}{2 d (c+d x)^{5/2}}\right )}{4 c d}+\frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{6 c d^2 (c+d x)^{9/2}}\) |
Input:
Int[((A + B*x)*Sqrt[c^2 - d^2*x^2])/(c + d*x)^(9/2),x]
Output:
((B*c - A*d)*(c^2 - d^2*x^2)^(3/2))/(6*c*d^2*(c + d*x)^(9/2)) + ((3*B*c + A*d)*(-1/2*Sqrt[c^2 - d^2*x^2]/(d*(c + d*x)^(5/2)) + (Sqrt[c^2 - d^2*x^2]/ (2*c*d*(c + d*x)^(3/2)) + ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqr t[c + d*x])]/(2*Sqrt[2]*c^(3/2)*d))/4))/(4*c*d)
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]
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]
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]
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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(373\) vs. \(2(158)=316\).
Time = 0.35 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.97
method | result | size |
default | \(\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (3 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{4} x^{3}+9 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{3} x^{3}+9 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{3} x^{2}+27 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x^{2}+9 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x +6 A \,d^{3} x^{2} \sqrt {-d x +c}\, \sqrt {c}+27 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d x +18 B \,c^{\frac {3}{2}} d^{2} x^{2} \sqrt {-d x +c}+3 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d +20 A \,c^{\frac {3}{2}} d^{2} x \sqrt {-d x +c}+9 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}-68 B \,c^{\frac {5}{2}} d x \sqrt {-d x +c}-50 A \sqrt {-d x +c}\, c^{\frac {5}{2}} d -22 B \sqrt {-d x +c}\, c^{\frac {7}{2}}\right )}{192 c^{\frac {5}{2}} \left (d x +c \right )^{\frac {7}{2}} \sqrt {-d x +c}\, d^{2}}\) | \(374\) |
Input:
int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(9/2),x,method=_RETURNVERBOSE)
Output:
1/192*(-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^4*x^3+9*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c ^(1/2))*c*d^3*x^3+9*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))* c*d^3*x^2+27*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^2 *x^2+9*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^2*x+6*A *d^3*x^2*(-d*x+c)^(1/2)*c^(1/2)+27*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^ (1/2)/c^(1/2))*c^3*d*x+18*B*c^(3/2)*d^2*x^2*(-d*x+c)^(1/2)+3*A*2^(1/2)*arc tanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d+20*A*c^(3/2)*d^2*x*(-d*x+c) ^(1/2)+9*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4-68*B*c^ (5/2)*d*x*(-d*x+c)^(1/2)-50*A*(-d*x+c)^(1/2)*c^(5/2)*d-22*B*(-d*x+c)^(1/2) *c^(7/2))/(d*x+c)^(7/2)/(-d*x+c)^(1/2)/d^2
Time = 0.11 (sec) , antiderivative size = 579, normalized size of antiderivative = 3.05 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (3 \, B c^{5} + A c^{4} d + {\left (3 \, B c d^{4} + A d^{5}\right )} x^{4} + 4 \, {\left (3 \, B c^{2} d^{3} + A c d^{4}\right )} x^{3} + 6 \, {\left (3 \, B c^{3} d^{2} + A c^{2} d^{3}\right )} x^{2} + 4 \, {\left (3 \, B c^{4} d + A c^{3} d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x - 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 4 \, {\left (11 \, B c^{4} + 25 \, A c^{3} d - 3 \, {\left (3 \, B c^{2} d^{2} + A c d^{3}\right )} x^{2} + 2 \, {\left (17 \, B c^{3} d - 5 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{384 \, {\left (c^{3} d^{6} x^{4} + 4 \, c^{4} d^{5} x^{3} + 6 \, c^{5} d^{4} x^{2} + 4 \, c^{6} d^{3} x + c^{7} d^{2}\right )}}, -\frac {3 \, \sqrt {2} {\left (3 \, B c^{5} + A c^{4} d + {\left (3 \, B c d^{4} + A d^{5}\right )} x^{4} + 4 \, {\left (3 \, B c^{2} d^{3} + A c d^{4}\right )} x^{3} + 6 \, {\left (3 \, B c^{3} d^{2} + A c^{2} d^{3}\right )} x^{2} + 4 \, {\left (3 \, B c^{4} d + A c^{3} d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + 2 \, {\left (11 \, B c^{4} + 25 \, A c^{3} d - 3 \, {\left (3 \, B c^{2} d^{2} + A c d^{3}\right )} x^{2} + 2 \, {\left (17 \, B c^{3} d - 5 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{192 \, {\left (c^{3} d^{6} x^{4} + 4 \, c^{4} d^{5} x^{3} + 6 \, c^{5} d^{4} x^{2} + 4 \, c^{6} d^{3} x + c^{7} d^{2}\right )}}\right ] \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(9/2),x, algorithm="fricas" )
Output:
[1/384*(3*sqrt(2)*(3*B*c^5 + A*c^4*d + (3*B*c*d^4 + A*d^5)*x^4 + 4*(3*B*c^ 2*d^3 + A*c*d^4)*x^3 + 6*(3*B*c^3*d^2 + A*c^2*d^3)*x^2 + 4*(3*B*c^4*d + A* c^3*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*(11*B*c^4 + 25*A*c^3*d - 3*(3*B*c^2*d^2 + A*c*d^3)*x^2 + 2*(17*B*c^3*d - 5*A*c^2*d^ 2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^3*d^6*x^4 + 4*c^4*d^5*x^3 + 6 *c^5*d^4*x^2 + 4*c^6*d^3*x + c^7*d^2), -1/192*(3*sqrt(2)*(3*B*c^5 + A*c^4* d + (3*B*c*d^4 + A*d^5)*x^4 + 4*(3*B*c^2*d^3 + A*c*d^4)*x^3 + 6*(3*B*c^3*d ^2 + A*c^2*d^3)*x^2 + 4*(3*B*c^4*d + A*c^3*d^2)*x)*sqrt(-c)*arctan(1/2*sqr t(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + c^2)) + 2*(11*B* c^4 + 25*A*c^3*d - 3*(3*B*c^2*d^2 + A*c*d^3)*x^2 + 2*(17*B*c^3*d - 5*A*c^2 *d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^3*d^6*x^4 + 4*c^4*d^5*x^3 + 6*c^5*d^4*x^2 + 4*c^6*d^3*x + c^7*d^2)]
\[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}} \, dx=\int \frac {\sqrt {- \left (- c + d x\right ) \left (c + d x\right )} \left (A + B x\right )}{\left (c + d x\right )^{\frac {9}{2}}}\, dx \] Input:
integrate((B*x+A)*(-d**2*x**2+c**2)**(1/2)/(d*x+c)**(9/2),x)
Output:
Integral(sqrt(-(-c + d*x)*(c + d*x))*(A + B*x)/(c + d*x)**(9/2), x)
\[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}} \, dx=\int { \frac {\sqrt {-d^{2} x^{2} + c^{2}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {9}{2}}} \,d x } \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(9/2),x, algorithm="maxima" )
Output:
integrate(sqrt(-d^2*x^2 + c^2)*(B*x + A)/(d*x + c)^(9/2), x)
Time = 0.12 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}} \, dx=-\frac {\frac {3 \, \sqrt {2} {\left (3 \, B c + A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {2 \, {\left (9 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B c + 16 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c^{2} - 36 \, \sqrt {-d x + c} B c^{3} + 3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} A d - 16 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c d - 12 \, \sqrt {-d x + c} A c^{2} d\right )}}{{\left (d x + c\right )}^{3} c^{2}}}{192 \, d^{2}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(9/2),x, algorithm="giac")
Output:
-1/192*(3*sqrt(2)*(3*B*c + A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c) )/(sqrt(-c)*c^2) - 2*(9*(d*x - c)^2*sqrt(-d*x + c)*B*c + 16*(-d*x + c)^(3/ 2)*B*c^2 - 36*sqrt(-d*x + c)*B*c^3 + 3*(d*x - c)^2*sqrt(-d*x + c)*A*d - 16 *(-d*x + c)^(3/2)*A*c*d - 12*sqrt(-d*x + c)*A*c^2*d)/((d*x + c)^3*c^2))/d^ 2
Timed out. \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}} \, dx=\int \frac {\sqrt {c^2-d^2\,x^2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^{9/2}} \,d x \] Input:
int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x)^(9/2),x)
Output:
int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x)^(9/2), x)
Time = 0.26 (sec) , antiderivative size = 397, normalized size of antiderivative = 2.09 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}} \, dx=\frac {-50 \sqrt {-d x +c}\, a \,c^{3} d +20 \sqrt {-d x +c}\, a \,c^{2} d^{2} x +6 \sqrt {-d x +c}\, a c \,d^{3} x^{2}-22 \sqrt {-d x +c}\, b \,c^{4}-68 \sqrt {-d x +c}\, b \,c^{3} d x +18 \sqrt {-d x +c}\, b \,c^{2} d^{2} x^{2}-3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a \,c^{3} d -9 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a \,c^{2} d^{2} x -9 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a c \,d^{3} x^{2}-3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a \,d^{4} x^{3}-9 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{4}-27 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{3} d x -27 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{2} d^{2} x^{2}-9 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b c \,d^{3} x^{3}}{192 c^{3} d^{2} \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )} \] Input:
int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(9/2),x)
Output:
( - 50*sqrt(c - d*x)*a*c**3*d + 20*sqrt(c - d*x)*a*c**2*d**2*x + 6*sqrt(c - d*x)*a*c*d**3*x**2 - 22*sqrt(c - d*x)*b*c**4 - 68*sqrt(c - d*x)*b*c**3*d *x + 18*sqrt(c - d*x)*b*c**2*d**2*x**2 - 3*sqrt(c)*sqrt(2)*log(tan(asin(sq rt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**3*d - 9*sqrt(c)*sqrt(2)*log(tan(as in(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**2*d**2*x - 9*sqrt(c)*sqrt(2)* log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c*d**3*x**2 - 3*sqrt(c )*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*d**4*x**3 - 9*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**4 - 27*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b* c**3*d*x - 27*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)) )/2))*b*c**2*d**2*x**2 - 9*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqr t(c)*sqrt(2)))/2))*b*c*d**3*x**3)/(192*c**3*d**2*(c**3 + 3*c**2*d*x + 3*c* d**2*x**2 + d**3*x**3))