Integrand size = 31, antiderivative size = 117 \[ \int \frac {(A+B x) (c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {8 c (3 B c+5 A d) \sqrt {c^2-d^2 x^2}}{15 d^2 \sqrt {c+d x}}-\frac {2 (3 B c+5 A d) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{15 d^2}-\frac {2 B (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^2} \] Output:
-8/15*c*(5*A*d+3*B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(1/2)-2/15*(5*A*d+3 *B*c)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)/d^2-2/5*B*(d*x+c)^(3/2)*(-d^2*x^2 +c^2)^(1/2)/d^2
Time = 0.22 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.56 \[ \int \frac {(A+B x) (c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \sqrt {c^2-d^2 x^2} \left (5 A d (5 c+d x)+3 B \left (6 c^2+3 c d x+d^2 x^2\right )\right )}{15 d^2 \sqrt {c+d x}} \] Input:
Integrate[((A + B*x)*(c + d*x)^(3/2))/Sqrt[c^2 - d^2*x^2],x]
Output:
(-2*Sqrt[c^2 - d^2*x^2]*(5*A*d*(5*c + d*x) + 3*B*(6*c^2 + 3*c*d*x + d^2*x^ 2)))/(15*d^2*Sqrt[c + d*x])
Time = 0.38 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {672, 459, 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.
\(\displaystyle \int \frac {(A+B x) (c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}} \, dx\) |
\(\Big \downarrow \) 672 |
\(\displaystyle \frac {(5 A d+3 B c) \int \frac {(c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}}dx}{5 d}-\frac {2 B (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^2}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {(5 A d+3 B c) \left (\frac {4}{3} c \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right )}{5 d}-\frac {2 B (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^2}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {\left (-\frac {8 c \sqrt {c^2-d^2 x^2}}{3 d \sqrt {c+d x}}-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right ) (5 A d+3 B c)}{5 d}-\frac {2 B (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^2}\) |
Input:
Int[((A + B*x)*(c + d*x)^(3/2))/Sqrt[c^2 - d^2*x^2],x]
Output:
(-2*B*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2])/(5*d^2) + ((3*B*c + 5*A*d)*((-8 *c*Sqrt[c^2 - d^2*x^2])/(3*d*Sqrt[c + d*x]) - (2*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2])/(3*d)))/(5*d)
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[((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*(n + 2*p + 1))), x] + Simp[2*c* (Simplify[n + p]/(n + 2*p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[Simplif y[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]
Time = 0.37 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.52
method | result | size |
default | \(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (3 B \,d^{2} x^{2}+5 A \,d^{2} x +9 B c d x +25 A c d +18 B \,c^{2}\right )}{15 \sqrt {d x +c}\, d^{2}}\) | \(61\) |
gosper | \(-\frac {2 \left (-d x +c \right ) \left (3 B \,d^{2} x^{2}+5 A \,d^{2} x +9 B c d x +25 A c d +18 B \,c^{2}\right ) \sqrt {d x +c}}{15 d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\) | \(67\) |
orering | \(-\frac {2 \left (-d x +c \right ) \left (3 B \,d^{2} x^{2}+5 A \,d^{2} x +9 B c d x +25 A c d +18 B \,c^{2}\right ) \sqrt {d x +c}}{15 d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\) | \(67\) |
risch | \(-\frac {2 \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}\, \left (3 B \,d^{2} x^{2}+5 A \,d^{2} x +9 B c d x +25 A c d +18 B \,c^{2}\right ) \sqrt {-d x +c}}{15 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}\) | \(91\) |
Input:
int((B*x+A)*(d*x+c)^(3/2)/(-d^2*x^2+c^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/15/(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)*(3*B*d^2*x^2+5*A*d^2*x+9*B*c*d*x+ 25*A*c*d+18*B*c^2)/d^2
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.61 \[ \int \frac {(A+B x) (c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \, {\left (3 \, B d^{2} x^{2} + 18 \, B c^{2} + 25 \, A c d + {\left (9 \, B c d + 5 \, A d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{15 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:
integrate((B*x+A)*(d*x+c)^(3/2)/(-d^2*x^2+c^2)^(1/2),x, algorithm="fricas" )
Output:
-2/15*(3*B*d^2*x^2 + 18*B*c^2 + 25*A*c*d + (9*B*c*d + 5*A*d^2)*x)*sqrt(-d^ 2*x^2 + c^2)*sqrt(d*x + c)/(d^3*x + c*d^2)
\[ \int \frac {(A+B x) (c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}} \, dx=\int \frac {\left (A + B x\right ) \left (c + d x\right )^{\frac {3}{2}}}{\sqrt {- \left (- c + d x\right ) \left (c + d x\right )}}\, dx \] Input:
integrate((B*x+A)*(d*x+c)**(3/2)/(-d**2*x**2+c**2)**(1/2),x)
Output:
Integral((A + B*x)*(c + d*x)**(3/2)/sqrt(-(-c + d*x)*(c + d*x)), x)
Time = 0.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.65 \[ \int \frac {(A+B x) (c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \, {\left (d^{2} x^{2} + 4 \, c d x - 5 \, c^{2}\right )} A}{3 \, \sqrt {-d x + c} d} + \frac {2 \, {\left (d^{3} x^{3} + 2 \, c d^{2} x^{2} + 3 \, c^{2} d x - 6 \, c^{3}\right )} B}{5 \, \sqrt {-d x + c} d^{2}} \] Input:
integrate((B*x+A)*(d*x+c)^(3/2)/(-d^2*x^2+c^2)^(1/2),x, algorithm="maxima" )
Output:
2/3*(d^2*x^2 + 4*c*d*x - 5*c^2)*A/(sqrt(-d*x + c)*d) + 2/5*(d^3*x^3 + 2*c* d^2*x^2 + 3*c^2*d*x - 6*c^3)*B/(sqrt(-d*x + c)*d^2)
Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.66 \[ \int \frac {(A+B x) (c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B - 15 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c + 30 \, \sqrt {-d x + c} B c^{2} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d + 30 \, \sqrt {-d x + c} A c d\right )}}{15 \, d^{2}} \] Input:
integrate((B*x+A)*(d*x+c)^(3/2)/(-d^2*x^2+c^2)^(1/2),x, algorithm="giac")
Output:
-2/15*(3*(d*x - c)^2*sqrt(-d*x + c)*B - 15*(-d*x + c)^(3/2)*B*c + 30*sqrt( -d*x + c)*B*c^2 - 5*(-d*x + c)^(3/2)*A*d + 30*sqrt(-d*x + c)*A*c*d)/d^2
Time = 9.80 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B x) (c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {\left (36\,B\,c^2+50\,A\,d\,c\right )\,\sqrt {c+d\,x}}{15\,d^3}+\frac {x\,\left (10\,A\,d^2+18\,B\,c\,d\right )\,\sqrt {c+d\,x}}{15\,d^3}+\frac {2\,B\,x^2\,\sqrt {c+d\,x}}{5\,d}\right )}{x+\frac {c}{d}} \] Input:
int(((A + B*x)*(c + d*x)^(3/2))/(c^2 - d^2*x^2)^(1/2),x)
Output:
-((c^2 - d^2*x^2)^(1/2)*(((36*B*c^2 + 50*A*c*d)*(c + d*x)^(1/2))/(15*d^3) + (x*(10*A*d^2 + 18*B*c*d)*(c + d*x)^(1/2))/(15*d^3) + (2*B*x^2*(c + d*x)^ (1/2))/(5*d)))/(x + c/d)
Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.39 \[ \int \frac {(A+B x) (c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \sqrt {-d x +c}\, \left (-3 b \,d^{2} x^{2}-5 a \,d^{2} x -9 b c d x -25 a c d -18 b \,c^{2}\right )}{15 d^{2}} \] Input:
int((B*x+A)*(d*x+c)^(3/2)/(-d^2*x^2+c^2)^(1/2),x)
Output:
(2*sqrt(c - d*x)*( - 25*a*c*d - 5*a*d**2*x - 18*b*c**2 - 9*b*c*d*x - 3*b*d **2*x**2))/(15*d**2)