Integrand size = 41, antiderivative size = 230 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) (c+d x)^{3/2}}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\left (3 c^2 C d+B c d^2-A d^3+5 c^3 D\right ) \sqrt {c+d x}}{2 c^2 d^4 \sqrt {c^2-d^2 x^2}}-\frac {2 D \sqrt {c^2-d^2 x^2}}{d^4 \sqrt {c+d x}}-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{2 \sqrt {2} c^{5/2} d^4} \] Output:
1/3*(A*d^3+B*c*d^2+C*c^2*d+D*c^3)*(d*x+c)^(3/2)/c/d^4/(-d^2*x^2+c^2)^(3/2) -1/2*(-A*d^3+B*c*d^2+3*C*c^2*d+5*D*c^3)*(d*x+c)^(1/2)/c^2/d^4/(-d^2*x^2+c^ 2)^(1/2)-2*D*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)^(1/2)-1/4*(A*d^3-B*c*d^2+C*c ^2*d-D*c^3)*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^4
Time = 3.27 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.80 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {-\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (25 c^4 D+3 A d^4 x-c d^3 (5 A+3 B x)+c^3 d (7 C-39 D x)+c^2 d^2 (B+3 x (-3 C+4 D x))\right )}{(c-d x)^2 \sqrt {c+d x}}+3 \sqrt {2} \left (-c^2 C d+B c d^2-A d^3+c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{12 c^{5/2} d^4} \] Input:
Integrate[((c + d*x)^(3/2)*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(5/2 ),x]
Output:
((-2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(25*c^4*D + 3*A*d^4*x - c*d^3*(5*A + 3*B* x) + c^3*d*(7*C - 39*D*x) + c^2*d^2*(B + 3*x*(-3*C + 4*D*x))))/((c - d*x)^ 2*Sqrt[c + d*x]) + 3*Sqrt[2]*(-(c^2*C*d) + B*c*d^2 - A*d^3 + c^3*D)*ArcTan h[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/(12*c^(5/2)*d^4)
Time = 1.13 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2166, 27, 2166, 27, 600, 458, 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.
\(\displaystyle \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\int \frac {3 \sqrt {c+d x} \left (\frac {2 c D x^2}{d}+\frac {2 c (C d+c D) x}{d^2}+\frac {D c^3+C d c^2+B d^2 c-A d^3}{d^3}\right )}{2 \left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\int \frac {\sqrt {c+d x} \left (\frac {2 c D x^2}{d}+\frac {2 c (C d+c D) x}{d^2}+\frac {D c^3+C d c^2+B d^2 c-A d^3}{d^3}\right )}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{2 c}\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {\sqrt {c+d x} \left (-A d^3+B c d^2+5 c^3 D+3 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\int \frac {3 D c^3+C d c^2+4 d D x c^2-B d^2 c+A d^3}{2 d^3 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{c}}{2 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {\sqrt {c+d x} \left (-A d^3+B c d^2+5 c^3 D+3 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\int \frac {3 D c^3+C d c^2+4 d D x c^2-B d^2 c+A d^3}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{2 c d^3}}{2 c}\) |
\(\Big \downarrow \) 600 |
\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {\sqrt {c+d x} \left (-A d^3+B c d^2+5 c^3 D+3 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx+4 c^2 D \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx}{2 c d^3}}{2 c}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {\sqrt {c+d x} \left (-A d^3+B c d^2+5 c^3 D+3 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx-\frac {8 c^2 D \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}}{2 c d^3}}{2 c}\) |
\(\Big \downarrow \) 471 |
\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {\sqrt {c+d x} \left (-A d^3+B c d^2+5 c^3 D+3 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {2 d \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}-\frac {8 c^2 D \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}}{2 c d^3}}{2 c}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {\sqrt {c+d x} \left (-A d^3+B c d^2+5 c^3 D+3 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right ) \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{\sqrt {c} d}-\frac {8 c^2 D \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}}{2 c d^3}}{2 c}\) |
Input:
Int[((c + d*x)^(3/2)*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(5/2),x]
Output:
((c^2*C*d + B*c*d^2 + A*d^3 + c^3*D)*(c + d*x)^(3/2))/(3*c*d^4*(c^2 - d^2* x^2)^(3/2)) - (((3*c^2*C*d + B*c*d^2 - A*d^3 + 5*c^3*D)*Sqrt[c + d*x])/(c* d^4*Sqrt[c^2 - d^2*x^2]) - ((-8*c^2*D*Sqrt[c^2 - d^2*x^2])/(d*Sqrt[c + d*x ]) - (Sqrt[2]*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*ArcTanh[Sqrt[c^2 - d^2*x ^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])])/(Sqrt[c]*d))/(2*c*d^3))/(2*c)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c , d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + p, 0]
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[((A_.) + (B_.)*(x_))/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2] ), x_Symbol] :> Simp[B/d Int[Sqrt[c + d*x]/Sqrt[a + b*x^2], x], x] - Simp [(B*c - A*d)/d Int[1/(Sqrt[c + d*x]*Sqrt[a + b*x^2]), x], x] /; FreeQ[{a, b, c, d, A, B}, x] && NegQ[b/a]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e *(p + 1))), x] + Simp[d/(2*a*(p + 1)) Int[(d + e*x)^(m - 1)*(a + b*x^2)^( p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ [{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 0] && GtQ[m, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(405\) vs. \(2(200)=400\).
Time = 0.37 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.77
method | result | size |
default | \(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (-3 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{4} x +3 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{3} x -3 C \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x +3 D \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d x +3 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{3}-3 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2}+3 C \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d -3 D \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}+24 D c^{\frac {5}{2}} d^{2} x^{2}+6 A \sqrt {c}\, d^{4} x -6 B \,c^{\frac {3}{2}} d^{3} x -18 C \,c^{\frac {5}{2}} d^{2} x -78 D c^{\frac {7}{2}} d x -10 A \,c^{\frac {3}{2}} d^{3}+2 B \,c^{\frac {5}{2}} d^{2}+14 C \,c^{\frac {7}{2}} d +50 D c^{\frac {9}{2}}\right )}{12 c^{\frac {5}{2}} \sqrt {d x +c}\, \left (-d x +c \right )^{2} d^{4}}\) | \(406\) |
Input:
int((d*x+c)^(3/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x,method=_RETUR NVERBOSE)
Output:
-1/12*(-d^2*x^2+c^2)^(1/2)/c^(5/2)*(-3*A*(-d*x+c)^(1/2)*2^(1/2)*arctanh(1/ 2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*d^4*x+3*B*(-d*x+c)^(1/2)*2^(1/2)*arctanh (1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^3*x-3*C*(-d*x+c)^(1/2)*2^(1/2)*ar ctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^2*x+3*D*(-d*x+c)^(1/2)*2^( 1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d*x+3*A*(-d*x+c)^(1/2 )*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^3-3*B*(-d*x+c)^( 1/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^2+3*C*(-d*x +c)^(1/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d-3*D*(- d*x+c)^(1/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4+24*D* c^(5/2)*d^2*x^2+6*A*c^(1/2)*d^4*x-6*B*c^(3/2)*d^3*x-18*C*c^(5/2)*d^2*x-78* D*c^(7/2)*d*x-10*A*c^(3/2)*d^3+2*B*c^(5/2)*d^2+14*C*c^(7/2)*d+50*D*c^(9/2) )/(d*x+c)^(1/2)/(-d*x+c)^2/d^4
Time = 0.11 (sec) , antiderivative size = 693, normalized size of antiderivative = 3.01 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, \sqrt {2} {\left (D c^{6} - C c^{5} d + B c^{4} d^{2} - A c^{3} d^{3} + {\left (D c^{3} d^{3} - C c^{2} d^{4} + B c d^{5} - A d^{6}\right )} x^{3} - {\left (D c^{4} d^{2} - C c^{3} d^{3} + B c^{2} d^{4} - A c d^{5}\right )} x^{2} - {\left (D c^{5} d - C c^{4} d^{2} + B c^{3} d^{3} - A c^{2} d^{4}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 4 \, {\left (12 \, D c^{3} d^{2} x^{2} + 25 \, D c^{5} + 7 \, C c^{4} d + B c^{3} d^{2} - 5 \, A c^{2} d^{3} - 3 \, {\left (13 \, D c^{4} d + 3 \, C c^{3} d^{2} + B c^{2} d^{3} - A c d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{24 \, {\left (c^{3} d^{7} x^{3} - c^{4} d^{6} x^{2} - c^{5} d^{5} x + c^{6} d^{4}\right )}}, -\frac {3 \, \sqrt {2} {\left (D c^{6} - C c^{5} d + B c^{4} d^{2} - A c^{3} d^{3} + {\left (D c^{3} d^{3} - C c^{2} d^{4} + B c d^{5} - A d^{6}\right )} x^{3} - {\left (D c^{4} d^{2} - C c^{3} d^{3} + B c^{2} d^{4} - A c d^{5}\right )} x^{2} - {\left (D c^{5} d - C c^{4} d^{2} + B c^{3} d^{3} - A c^{2} d^{4}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + 2 \, {\left (12 \, D c^{3} d^{2} x^{2} + 25 \, D c^{5} + 7 \, C c^{4} d + B c^{3} d^{2} - 5 \, A c^{2} d^{3} - 3 \, {\left (13 \, D c^{4} d + 3 \, C c^{3} d^{2} + B c^{2} d^{3} - A c d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{12 \, {\left (c^{3} d^{7} x^{3} - c^{4} d^{6} x^{2} - c^{5} d^{5} x + c^{6} d^{4}\right )}}\right ] \] Input:
integrate((d*x+c)^(3/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algori thm="fricas")
Output:
[-1/24*(3*sqrt(2)*(D*c^6 - C*c^5*d + B*c^4*d^2 - A*c^3*d^3 + (D*c^3*d^3 - C*c^2*d^4 + B*c*d^5 - A*d^6)*x^3 - (D*c^4*d^2 - C*c^3*d^3 + B*c^2*d^4 - A* c*d^5)*x^2 - (D*c^5*d - C*c^4*d^2 + B*c^3*d^3 - A*c^2*d^4)*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*(12*D*c^3*d^2*x^2 + 25*D*c^5 + 7* C*c^4*d + B*c^3*d^2 - 5*A*c^2*d^3 - 3*(13*D*c^4*d + 3*C*c^3*d^2 + B*c^2*d^ 3 - A*c*d^4)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^3*d^7*x^3 - c^4*d^6 *x^2 - c^5*d^5*x + c^6*d^4), -1/12*(3*sqrt(2)*(D*c^6 - C*c^5*d + B*c^4*d^2 - A*c^3*d^3 + (D*c^3*d^3 - C*c^2*d^4 + B*c*d^5 - A*d^6)*x^3 - (D*c^4*d^2 - C*c^3*d^3 + B*c^2*d^4 - A*c*d^5)*x^2 - (D*c^5*d - C*c^4*d^2 + B*c^3*d^3 - A*c^2*d^4)*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*(12*D*c^3*d^2*x^2 + 25*D*c^5 + 7*C*c^4*d + B*c^3*d^2 - 5*A*c^2*d^3 - 3*(13*D*c^4*d + 3*C*c^3*d^2 + B*c^2*d^3 - A*c* d^4)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^3*d^7*x^3 - c^4*d^6*x^2 - c ^5*d^5*x + c^6*d^4)]
\[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {3}{2}} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((d*x+c)**(3/2)*(D*x**3+C*x**2+B*x+A)/(-d**2*x**2+c**2)**(5/2),x)
Output:
Integral((c + d*x)**(3/2)*(A + B*x + C*x**2 + D*x**3)/(-(-c + d*x)*(c + d* x))**(5/2), x)
\[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{\frac {3}{2}}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((d*x+c)^(3/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algori thm="maxima")
Output:
integrate((D*x^3 + C*x^2 + B*x + A)*(d*x + c)^(3/2)/(-d^2*x^2 + c^2)^(5/2) , x)
Time = 0.13 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {\frac {24 \, \sqrt {-d x + c} D}{d^{3}} + \frac {3 \, \sqrt {2} {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{2} d^{3}} + \frac {2 \, {\left (15 \, {\left (d x - c\right )} D c^{3} + 2 \, D c^{4} + 9 \, {\left (d x - c\right )} C c^{2} d + 2 \, C c^{3} d + 3 \, {\left (d x - c\right )} B c d^{2} + 2 \, B c^{2} d^{2} - 3 \, {\left (d x - c\right )} A d^{3} + 2 \, A c d^{3}\right )}}{{\left (d x - c\right )} \sqrt {-d x + c} c^{2} d^{3}}}{12 \, d} \] Input:
integrate((d*x+c)^(3/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algori thm="giac")
Output:
-1/12*(24*sqrt(-d*x + c)*D/d^3 + 3*sqrt(2)*(D*c^3 - C*c^2*d + B*c*d^2 - A* d^3)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c))/(sqrt(-c)*c^2*d^3) + 2*(1 5*(d*x - c)*D*c^3 + 2*D*c^4 + 9*(d*x - c)*C*c^2*d + 2*C*c^3*d + 3*(d*x - c )*B*c*d^2 + 2*B*c^2*d^2 - 3*(d*x - c)*A*d^3 + 2*A*c*d^3)/((d*x - c)*sqrt(- d*x + c)*c^2*d^3))/d
Timed out. \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{3/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c^2-d^2\,x^2\right )}^{5/2}} \,d x \] Input:
int(((c + d*x)^(3/2)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(5/2),x)
Output:
int(((c + d*x)^(3/2)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(5/2), x)
Time = 0.21 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.50 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a c \,d^{2}-3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,d^{3} x -3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b \,c^{2} d +3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c \,d^{2} x -3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a c \,d^{2}+3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,d^{3} x +3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b \,c^{2} d -3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c \,d^{2} x +20 a \,c^{2} d^{2}-12 a c \,d^{3} x -4 b \,c^{3} d +12 b \,c^{2} d^{2} x -128 c^{5}+192 c^{4} d x -48 c^{3} d^{2} x^{2}}{24 \sqrt {-d x +c}\, c^{3} d^{3} \left (-d x +c \right )} \] Input:
int((d*x+c)^(3/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x)
Output:
(3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*a*c* d**2 - 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2) )*a*d**3*x - 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*s qrt(2))*b*c**2*d + 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqr t(c)*sqrt(2))*b*c*d**2*x - 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d* x) + sqrt(c)*sqrt(2))*a*c*d**2 + 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt( c - d*x) + sqrt(c)*sqrt(2))*a*d**3*x + 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log (sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**2*d - 3*sqrt(c)*sqrt(c - d*x)*sqrt( 2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c*d**2*x + 20*a*c**2*d**2 - 12*a *c*d**3*x - 4*b*c**3*d + 12*b*c**2*d**2*x - 128*c**5 + 192*c**4*d*x - 48*c **3*d**2*x**2)/(24*sqrt(c - d*x)*c**3*d**3*(c - d*x))