Integrand size = 31, antiderivative size = 110 \[ \int \frac {(A+B x) (c+d x)^{7/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {4 c (B c+A d) (c+d x)^{3/2}}{3 d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {2 (3 B c+A d) \sqrt {c+d x}}{d^2 \sqrt {c^2-d^2 x^2}}-\frac {2 B \sqrt {c^2-d^2 x^2}}{d^2 \sqrt {c+d x}} \] Output:
4/3*c*(A*d+B*c)*(d*x+c)^(3/2)/d^2/(-d^2*x^2+c^2)^(3/2)-2*(A*d+3*B*c)*(d*x+ c)^(1/2)/d^2/(-d^2*x^2+c^2)^(1/2)-2*B*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(1/ 2)
Time = 1.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.65 \[ \int \frac {(A+B x) (c+d x)^{7/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {c+d x} \left (A d (c-3 d x)+B \left (10 c^2-15 c d x+3 d^2 x^2\right )\right )}{3 d^2 (-c+d x) \sqrt {c^2-d^2 x^2}} \] Input:
Integrate[((A + B*x)*(c + d*x)^(7/2))/(c^2 - d^2*x^2)^(5/2),x]
Output:
(2*Sqrt[c + d*x]*(A*d*(c - 3*d*x) + B*(10*c^2 - 15*c*d*x + 3*d^2*x^2)))/(3 *d^2*(-c + d*x)*Sqrt[c^2 - d^2*x^2])
Time = 0.39 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {669, 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.
\(\displaystyle \int \frac {(A+B x) (c+d x)^{7/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 669 |
\(\displaystyle \frac {(c+d x)^{7/2} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(A d+7 B c) \int \frac {(c+d x)^{5/2}}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{6 c d}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle \frac {(c+d x)^{7/2} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(A d+7 B c) \left (4 c \int \frac {(c+d x)^{3/2}}{\left (c^2-d^2 x^2\right )^{3/2}}dx-\frac {2 (c+d x)^{3/2}}{d \sqrt {c^2-d^2 x^2}}\right )}{6 c d}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {(c+d x)^{7/2} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\left (\frac {8 c \sqrt {c+d x}}{d \sqrt {c^2-d^2 x^2}}-\frac {2 (c+d x)^{3/2}}{d \sqrt {c^2-d^2 x^2}}\right ) (A d+7 B c)}{6 c d}\) |
Input:
Int[((A + B*x)*(c + d*x)^(7/2))/(c^2 - d^2*x^2)^(5/2),x]
Output:
((B*c + A*d)*(c + d*x)^(7/2))/(3*c*d^2*(c^2 - d^2*x^2)^(3/2)) - ((7*B*c + A*d)*((8*c*Sqrt[c + d*x])/(d*Sqrt[c^2 - d^2*x^2]) - (2*(c + d*x)^(3/2))/(d *Sqrt[c^2 - d^2*x^2])))/(6*c*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[(d*g + e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d* (p + 1))), x] - Simp[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))) Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0]
Time = 0.39 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.60
method | result | size |
gosper | \(-\frac {2 \left (-d x +c \right ) \left (3 B \,d^{2} x^{2}-3 A \,d^{2} x -15 B c d x +A c d +10 B \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\) | \(66\) |
orering | \(-\frac {2 \left (-d x +c \right ) \left (3 B \,d^{2} x^{2}-3 A \,d^{2} x -15 B c d x +A c d +10 B \,c^{2}\right ) \left (d x +c \right )^{\frac {5}{2}}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\) | \(66\) |
default | \(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (3 B \,d^{2} x^{2}-3 A \,d^{2} x -15 B c d x +A c d +10 B \,c^{2}\right )}{3 \sqrt {d x +c}\, \left (-d x +c \right )^{2} d^{2}}\) | \(68\) |
risch | \(-\frac {2 B \sqrt {-d x +c}\, \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}-\frac {2 \left (-3 A \,d^{2} x -9 B c d x +A c d +7 B \,c^{2}\right ) \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{3 d^{2} \left (-d x +c \right )^{\frac {3}{2}} \sqrt {-d^{2} x^{2}+c^{2}}}\) | \(139\) |
Input:
int((B*x+A)*(d*x+c)^(7/2)/(-d^2*x^2+c^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*(-d*x+c)*(3*B*d^2*x^2-3*A*d^2*x-15*B*c*d*x+A*c*d+10*B*c^2)*(d*x+c)^(5 /2)/d^2/(-d^2*x^2+c^2)^(5/2)
Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B x) (c+d x)^{7/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, B d^{2} x^{2} + 10 \, B c^{2} + A c d - 3 \, {\left (5 \, B c d + A d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{3 \, {\left (d^{5} x^{3} - c d^{4} x^{2} - c^{2} d^{3} x + c^{3} d^{2}\right )}} \] Input:
integrate((B*x+A)*(d*x+c)^(7/2)/(-d^2*x^2+c^2)^(5/2),x, algorithm="fricas" )
Output:
-2/3*(3*B*d^2*x^2 + 10*B*c^2 + A*c*d - 3*(5*B*c*d + A*d^2)*x)*sqrt(-d^2*x^ 2 + c^2)*sqrt(d*x + c)/(d^5*x^3 - c*d^4*x^2 - c^2*d^3*x + c^3*d^2)
\[ \int \frac {(A+B x) (c+d x)^{7/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \left (c + d x\right )^{\frac {7}{2}}}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((B*x+A)*(d*x+c)**(7/2)/(-d**2*x**2+c**2)**(5/2),x)
Output:
Integral((A + B*x)*(c + d*x)**(7/2)/(-(-c + d*x)*(c + d*x))**(5/2), x)
Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B x) (c+d x)^{7/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, d x - c\right )} A}{3 \, {\left (d^{2} x - c d\right )} \sqrt {-d x + c}} + \frac {2 \, {\left (3 \, d^{2} x^{2} - 15 \, c d x + 10 \, c^{2}\right )} B}{3 \, {\left (d^{3} x - c d^{2}\right )} \sqrt {-d x + c}} \] Input:
integrate((B*x+A)*(d*x+c)^(7/2)/(-d^2*x^2+c^2)^(5/2),x, algorithm="maxima" )
Output:
-2/3*(3*d*x - c)*A/((d^2*x - c*d)*sqrt(-d*x + c)) + 2/3*(3*d^2*x^2 - 15*c* d*x + 10*c^2)*B/((d^3*x - c*d^2)*sqrt(-d*x + c))
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.68 \[ \int \frac {(A+B x) (c+d x)^{7/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (\frac {3 \, \sqrt {-d x + c} B}{d} + \frac {9 \, {\left (d x - c\right )} B c + 2 \, B c^{2} + 3 \, {\left (d x - c\right )} A d + 2 \, A c d}{{\left (d x - c\right )} \sqrt {-d x + c} d}\right )}}{3 \, d} \] Input:
integrate((B*x+A)*(d*x+c)^(7/2)/(-d^2*x^2+c^2)^(5/2),x, algorithm="giac")
Output:
-2/3*(3*sqrt(-d*x + c)*B/d + (9*(d*x - c)*B*c + 2*B*c^2 + 3*(d*x - c)*A*d + 2*A*c*d)/((d*x - c)*sqrt(-d*x + c)*d))/d
Time = 9.66 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) (c+d x)^{7/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {\left (20\,B\,c^2+2\,A\,d\,c\right )\,\sqrt {c+d\,x}}{3\,d^5}-\frac {x\,\left (6\,A\,d^2+30\,B\,c\,d\right )\,\sqrt {c+d\,x}}{3\,d^5}+\frac {2\,B\,x^2\,\sqrt {c+d\,x}}{d^3}\right )}{x^3+\frac {c^3}{d^3}-\frac {c\,x^2}{d}-\frac {c^2\,x}{d^2}} \] Input:
int(((A + B*x)*(c + d*x)^(7/2))/(c^2 - d^2*x^2)^(5/2),x)
Output:
-((c^2 - d^2*x^2)^(1/2)*(((20*B*c^2 + 2*A*c*d)*(c + d*x)^(1/2))/(3*d^5) - (x*(6*A*d^2 + 30*B*c*d)*(c + d*x)^(1/2))/(3*d^5) + (2*B*x^2*(c + d*x)^(1/2 ))/d^3))/(x^3 + c^3/d^3 - (c*x^2)/d - (c^2*x)/d^2)
Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.51 \[ \int \frac {(A+B x) (c+d x)^{7/2}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {-2 b \,d^{2} x^{2}+2 a \,d^{2} x +10 b c d x -\frac {2}{3} a c d -\frac {20}{3} b \,c^{2}}{\sqrt {-d x +c}\, d^{2} \left (-d x +c \right )} \] Input:
int((B*x+A)*(d*x+c)^(7/2)/(-d^2*x^2+c^2)^(5/2),x)
Output:
(2*( - a*c*d + 3*a*d**2*x - 10*b*c**2 + 15*b*c*d*x - 3*b*d**2*x**2))/(3*sq rt(c - d*x)*d**2*(c - d*x))