Integrand size = 41, antiderivative size = 184 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \left (49 c^2 C d+35 B c d^2+105 A d^3+27 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{105 d^4 \sqrt {c+d x}}+\frac {2 \left (14 c C d-35 B d^2-33 c^2 D\right ) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{105 d^4}-\frac {2 (7 C d-9 c D) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{35 d^4}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4} \] Output:
-2/105*(105*A*d^3+35*B*c*d^2+49*C*c^2*d+27*D*c^3)*(-d^2*x^2+c^2)^(1/2)/d^4 /(d*x+c)^(1/2)+2/105*(-35*B*d^2+14*C*c*d-33*D*c^2)*(d*x+c)^(1/2)*(-d^2*x^2 +c^2)^(1/2)/d^4-2/35*(7*C*d-9*D*c)*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2)/d^4- 2/7*D*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(1/2)/d^4
Time = 0.45 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.53 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \sqrt {c^2-d^2 x^2} \left (48 c^3 D+8 c^2 d (7 C+3 D x)+2 c d^2 (35 B+x (14 C+9 D x))+d^3 (105 A+x (35 B+3 x (7 C+5 D x)))\right )}{105 d^4 \sqrt {c+d x}} \] Input:
Integrate[(Sqrt[c + d*x]*(A + B*x + C*x^2 + D*x^3))/Sqrt[c^2 - d^2*x^2],x]
Output:
(-2*Sqrt[c^2 - d^2*x^2]*(48*c^3*D + 8*c^2*d*(7*C + 3*D*x) + 2*c*d^2*(35*B + x*(14*C + 9*D*x)) + d^3*(105*A + x*(35*B + 3*x*(7*C + 5*D*x)))))/(105*d^ 4*Sqrt[c + d*x])
Time = 1.00 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {2170, 27, 2170, 27, 672, 458}
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.
\(\displaystyle \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx\) |
\(\Big \downarrow \) 2170 |
\(\displaystyle -\frac {2 \int -\frac {\sqrt {c+d x} \left ((7 C d-9 c D) x^2 d^4+\left (3 D c^2+7 B d^2\right ) x d^3+\left (5 D c^3+7 A d^3\right ) d^2\right )}{2 \sqrt {c^2-d^2 x^2}}dx}{7 d^5}-\frac {2 D \sqrt {c^2-d^2 x^2} (c+d x)^{5/2}}{7 d^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {c+d x} \left ((7 C d-9 c D) x^2 d^4+\left (3 D c^2+7 B d^2\right ) x d^3+\left (5 D c^3+7 A d^3\right ) d^2\right )}{\sqrt {c^2-d^2 x^2}}dx}{7 d^5}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}\) |
\(\Big \downarrow \) 2170 |
\(\displaystyle \frac {-\frac {2 \int -\frac {d^6 \sqrt {c+d x} \left (-2 D c^3+21 C d c^2+35 A d^3-d \left (-33 D c^2+14 C d c-35 B d^2\right ) x\right )}{2 \sqrt {c^2-d^2 x^2}}dx}{5 d^4}-\frac {2}{5} d \sqrt {c^2-d^2 x^2} (c+d x)^{3/2} (7 C d-9 c D)}{7 d^5}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{5} d^2 \int \frac {\sqrt {c+d x} \left (-2 D c^3+21 C d c^2+35 A d^3-d \left (-33 D c^2+14 C d c-35 B d^2\right ) x\right )}{\sqrt {c^2-d^2 x^2}}dx-\frac {2}{5} d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} (7 C d-9 c D)}{7 d^5}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}\) |
\(\Big \downarrow \) 672 |
\(\displaystyle \frac {\frac {1}{5} d^2 \left (\frac {1}{3} \left (105 A d^3+35 B c d^2+27 c^3 D+49 c^2 C d\right ) \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx+\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \left (-35 B d^2-33 c^2 D+14 c C d\right )}{3 d}\right )-\frac {2}{5} d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} (7 C d-9 c D)}{7 d^5}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {\frac {1}{5} d^2 \left (\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \left (-35 B d^2-33 c^2 D+14 c C d\right )}{3 d}-\frac {2 \sqrt {c^2-d^2 x^2} \left (105 A d^3+35 B c d^2+27 c^3 D+49 c^2 C d\right )}{3 d \sqrt {c+d x}}\right )-\frac {2}{5} d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} (7 C d-9 c D)}{7 d^5}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}\) |
Input:
Int[(Sqrt[c + d*x]*(A + B*x + C*x^2 + D*x^3))/Sqrt[c^2 - d^2*x^2],x]
Output:
(-2*D*(c + d*x)^(5/2)*Sqrt[c^2 - d^2*x^2])/(7*d^4) + ((-2*d*(7*C*d - 9*c*D )*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2])/5 + (d^2*((-2*(49*c^2*C*d + 35*B*c* d^2 + 105*A*d^3 + 27*c^3*D)*Sqrt[c^2 - d^2*x^2])/(3*d*Sqrt[c + d*x]) + (2* (14*c*C*d - 35*B*d^2 - 33*c^2*D)*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2])/(3*d)) )/5)/(7*d^5)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((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]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) ^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si mp[1/(b*e^q*(m + q + 2*p + 1)) Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - b*d*x), x], x], x] /; NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && !IGtQ[m, 0]
Time = 0.36 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.57
method | result | size |
default | \(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (15 D x^{3} d^{3}+21 C \,d^{3} x^{2}+18 D c \,d^{2} x^{2}+35 B \,d^{3} x +28 C c \,d^{2} x +24 D c^{2} d x +105 A \,d^{3}+70 B c \,d^{2}+56 C \,c^{2} d +48 D c^{3}\right )}{105 \sqrt {d x +c}\, d^{4}}\) | \(105\) |
gosper | \(-\frac {2 \left (-d x +c \right ) \left (15 D x^{3} d^{3}+21 C \,d^{3} x^{2}+18 D c \,d^{2} x^{2}+35 B \,d^{3} x +28 C c \,d^{2} x +24 D c^{2} d x +105 A \,d^{3}+70 B c \,d^{2}+56 C \,c^{2} d +48 D c^{3}\right ) \sqrt {d x +c}}{105 d^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\) | \(111\) |
orering | \(-\frac {2 \left (-d x +c \right ) \left (15 D x^{3} d^{3}+21 C \,d^{3} x^{2}+18 D c \,d^{2} x^{2}+35 B \,d^{3} x +28 C c \,d^{2} x +24 D c^{2} d x +105 A \,d^{3}+70 B c \,d^{2}+56 C \,c^{2} d +48 D c^{3}\right ) \sqrt {d x +c}}{105 d^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\) | \(111\) |
Input:
int((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(1/2),x,method=_RETUR NVERBOSE)
Output:
-2/105/(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)*(15*D*d^3*x^3+21*C*d^3*x^2+18*D* c*d^2*x^2+35*B*d^3*x+28*C*c*d^2*x+24*D*c^2*d*x+105*A*d^3+70*B*c*d^2+56*C*c ^2*d+48*D*c^3)/d^4
Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \, {\left (15 \, D d^{3} x^{3} + 48 \, D c^{3} + 56 \, C c^{2} d + 70 \, B c d^{2} + 105 \, A d^{3} + 3 \, {\left (6 \, D c d^{2} + 7 \, C d^{3}\right )} x^{2} + {\left (24 \, D c^{2} d + 28 \, C c d^{2} + 35 \, B d^{3}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{105 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:
integrate((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(1/2),x, algori thm="fricas")
Output:
-2/105*(15*D*d^3*x^3 + 48*D*c^3 + 56*C*c^2*d + 70*B*c*d^2 + 105*A*d^3 + 3* (6*D*c*d^2 + 7*C*d^3)*x^2 + (24*D*c^2*d + 28*C*c*d^2 + 35*B*d^3)*x)*sqrt(- d^2*x^2 + c^2)*sqrt(d*x + c)/(d^5*x + c*d^4)
\[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=\int \frac {\sqrt {c + d x} \left (A + B x + C x^{2} + D x^{3}\right )}{\sqrt {- \left (- c + d x\right ) \left (c + d x\right )}}\, dx \] Input:
integrate((d*x+c)**(1/2)*(D*x**3+C*x**2+B*x+A)/(-d**2*x**2+c**2)**(1/2),x)
Output:
Integral(sqrt(c + d*x)*(A + B*x + C*x**2 + D*x**3)/sqrt(-(-c + d*x)*(c + d *x)), x)
Time = 0.06 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \, {\left (d x - c\right )} A}{\sqrt {-d x + c} d} + \frac {2 \, {\left (d^{2} x^{2} + c d x - 2 \, c^{2}\right )} B}{3 \, \sqrt {-d x + c} d^{2}} + \frac {2 \, {\left (3 \, d^{3} x^{3} + c d^{2} x^{2} + 4 \, c^{2} d x - 8 \, c^{3}\right )} C}{15 \, \sqrt {-d x + c} d^{3}} + \frac {2 \, {\left (5 \, d^{4} x^{4} + c d^{3} x^{3} + 2 \, c^{2} d^{2} x^{2} + 8 \, c^{3} d x - 16 \, c^{4}\right )} D}{35 \, \sqrt {-d x + c} d^{4}} \] Input:
integrate((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(1/2),x, algori thm="maxima")
Output:
2*(d*x - c)*A/(sqrt(-d*x + c)*d) + 2/3*(d^2*x^2 + c*d*x - 2*c^2)*B/(sqrt(- d*x + c)*d^2) + 2/15*(3*d^3*x^3 + c*d^2*x^2 + 4*c^2*d*x - 8*c^3)*C/(sqrt(- d*x + c)*d^3) + 2/35*(5*d^4*x^4 + c*d^3*x^3 + 2*c^2*d^2*x^2 + 8*c^3*d*x - 16*c^4)*D/(sqrt(-d*x + c)*d^4)
Time = 0.13 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} D + 63 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D c - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c^{2} + 105 \, \sqrt {-d x + c} D c^{3} + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} C d - 70 \, {\left (-d x + c\right )}^{\frac {3}{2}} C c d + 105 \, \sqrt {-d x + c} C c^{2} d - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} B d^{2} + 105 \, \sqrt {-d x + c} B c d^{2} + 105 \, \sqrt {-d x + c} A d^{3}\right )}}{105 \, d^{4}} \] Input:
integrate((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(1/2),x, algori thm="giac")
Output:
-2/105*(15*(d*x - c)^3*sqrt(-d*x + c)*D + 63*(d*x - c)^2*sqrt(-d*x + c)*D* c - 105*(-d*x + c)^(3/2)*D*c^2 + 105*sqrt(-d*x + c)*D*c^3 + 21*(d*x - c)^2 *sqrt(-d*x + c)*C*d - 70*(-d*x + c)^(3/2)*C*c*d + 105*sqrt(-d*x + c)*C*c^2 *d - 35*(-d*x + c)^(3/2)*B*d^2 + 105*sqrt(-d*x + c)*B*c*d^2 + 105*sqrt(-d* x + c)*A*d^3)/d^4
Timed out. \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=\int \frac {\sqrt {c+d\,x}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{\sqrt {c^2-d^2\,x^2}} \,d x \] Input:
int(((c + d*x)^(1/2)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(1/2),x)
Output:
int(((c + d*x)^(1/2)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(1/2), x)
Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.33 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \sqrt {-d x +c}\, \left (-15 d^{3} x^{3}-39 c \,d^{2} x^{2}-35 b \,d^{2} x -52 c^{2} d x -105 a \,d^{2}-70 b c d -104 c^{3}\right )}{105 d^{3}} \] Input:
int((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(1/2),x)
Output:
(2*sqrt(c - d*x)*( - 105*a*d**2 - 70*b*c*d - 35*b*d**2*x - 104*c**3 - 52*c **2*d*x - 39*c*d**2*x**2 - 15*d**3*x**3))/(105*d**3)