Integrand size = 24, antiderivative size = 95 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{9/2}} \, dx=-\frac {2 (B d-A e) (a+b x)^{5/2}}{7 e (b d-a e) (d+e x)^{7/2}}+\frac {2 (5 b B d+2 A b e-7 a B e) (a+b x)^{5/2}}{35 e (b d-a e)^2 (d+e x)^{5/2}} \] Output:
-2/7*(-A*e+B*d)*(b*x+a)^(5/2)/e/(-a*e+b*d)/(e*x+d)^(7/2)+2/35*(2*A*b*e-7*B *a*e+5*B*b*d)*(b*x+a)^(5/2)/e/(-a*e+b*d)^2/(e*x+d)^(5/2)
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{9/2}} \, dx=\frac {2 (a+b x)^{5/2} (B (-2 a d+5 b d x-7 a e x)+A (7 b d-5 a e+2 b e x))}{35 (b d-a e)^2 (d+e x)^{7/2}} \] Input:
Integrate[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(9/2),x]
Output:
(2*(a + b*x)^(5/2)*(B*(-2*a*d + 5*b*d*x - 7*a*e*x) + A*(7*b*d - 5*a*e + 2* b*e*x)))/(35*(b*d - a*e)^2*(d + e*x)^(7/2))
Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {87, 48}
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.
\(\displaystyle \int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{9/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(-7 a B e+2 A b e+5 b B d) \int \frac {(a+b x)^{3/2}}{(d+e x)^{7/2}}dx}{7 e (b d-a e)}-\frac {2 (a+b x)^{5/2} (B d-A e)}{7 e (d+e x)^{7/2} (b d-a e)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {2 (a+b x)^{5/2} (-7 a B e+2 A b e+5 b B d)}{35 e (d+e x)^{5/2} (b d-a e)^2}-\frac {2 (a+b x)^{5/2} (B d-A e)}{7 e (d+e x)^{7/2} (b d-a e)}\) |
Input:
Int[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(9/2),x]
Output:
(-2*(B*d - A*e)*(a + b*x)^(5/2))/(7*e*(b*d - a*e)*(d + e*x)^(7/2)) + (2*(5 *b*B*d + 2*A*b*e - 7*a*B*e)*(a + b*x)^(5/2))/(35*e*(b*d - a*e)^2*(d + e*x) ^(5/2))
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.78
method | result | size |
gosper | \(-\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-2 A b e x +7 B a e x -5 B b d x +5 A a e -7 A b d +2 B a d \right )}{35 \left (e x +d \right )^{\frac {7}{2}} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}\) | \(74\) |
orering | \(-\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-2 A b e x +7 B a e x -5 B b d x +5 A a e -7 A b d +2 B a d \right )}{35 \left (e x +d \right )^{\frac {7}{2}} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}\) | \(74\) |
default | \(-\frac {2 \left (-2 A \,b^{2} e \,x^{2}+7 B a b e \,x^{2}-5 B \,b^{2} d \,x^{2}+3 A a b e x -7 A \,b^{2} d x +7 B \,a^{2} e x -3 B a b d x +5 a^{2} A e -7 A a b d +2 B \,a^{2} d \right ) \left (b x +a \right )^{\frac {3}{2}}}{35 \left (e x +d \right )^{\frac {7}{2}} \left (a e -d b \right )^{2}}\) | \(107\) |
Input:
int((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(9/2),x,method=_RETURNVERBOSE)
Output:
-2/35*(b*x+a)^(5/2)*(-2*A*b*e*x+7*B*a*e*x-5*B*b*d*x+5*A*a*e-7*A*b*d+2*B*a* d)/(e*x+d)^(7/2)/(a^2*e^2-2*a*b*d*e+b^2*d^2)
Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (83) = 166\).
Time = 5.94 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.22 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{9/2}} \, dx=-\frac {2 \, {\left (5 \, A a^{3} e - {\left (5 \, B b^{3} d - {\left (7 \, B a b^{2} - 2 \, A b^{3}\right )} e\right )} x^{3} - {\left ({\left (8 \, B a b^{2} + 7 \, A b^{3}\right )} d - {\left (14 \, B a^{2} b + A a b^{2}\right )} e\right )} x^{2} + {\left (2 \, B a^{3} - 7 \, A a^{2} b\right )} d - {\left ({\left (B a^{2} b + 14 \, A a b^{2}\right )} d - {\left (7 \, B a^{3} + 8 \, A a^{2} b\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{35 \, {\left (b^{2} d^{6} - 2 \, a b d^{5} e + a^{2} d^{4} e^{2} + {\left (b^{2} d^{2} e^{4} - 2 \, a b d e^{5} + a^{2} e^{6}\right )} x^{4} + 4 \, {\left (b^{2} d^{3} e^{3} - 2 \, a b d^{2} e^{4} + a^{2} d e^{5}\right )} x^{3} + 6 \, {\left (b^{2} d^{4} e^{2} - 2 \, a b d^{3} e^{3} + a^{2} d^{2} e^{4}\right )} x^{2} + 4 \, {\left (b^{2} d^{5} e - 2 \, a b d^{4} e^{2} + a^{2} d^{3} e^{3}\right )} x\right )}} \] Input:
integrate((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(9/2),x, algorithm="fricas")
Output:
-2/35*(5*A*a^3*e - (5*B*b^3*d - (7*B*a*b^2 - 2*A*b^3)*e)*x^3 - ((8*B*a*b^2 + 7*A*b^3)*d - (14*B*a^2*b + A*a*b^2)*e)*x^2 + (2*B*a^3 - 7*A*a^2*b)*d - ((B*a^2*b + 14*A*a*b^2)*d - (7*B*a^3 + 8*A*a^2*b)*e)*x)*sqrt(b*x + a)*sqrt (e*x + d)/(b^2*d^6 - 2*a*b*d^5*e + a^2*d^4*e^2 + (b^2*d^2*e^4 - 2*a*b*d*e^ 5 + a^2*e^6)*x^4 + 4*(b^2*d^3*e^3 - 2*a*b*d^2*e^4 + a^2*d*e^5)*x^3 + 6*(b^ 2*d^4*e^2 - 2*a*b*d^3*e^3 + a^2*d^2*e^4)*x^2 + 4*(b^2*d^5*e - 2*a*b*d^4*e^ 2 + a^2*d^3*e^3)*x)
\[ \int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{9/2}} \, dx=\int \frac {\left (A + B x\right ) \left (a + b x\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {9}{2}}}\, dx \] Input:
integrate((b*x+a)**(3/2)*(B*x+A)/(e*x+d)**(9/2),x)
Output:
Integral((A + B*x)*(a + b*x)**(3/2)/(d + e*x)**(9/2), x)
Exception generated. \[ \int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{9/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(9/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e*(a*e-b*d)>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (83) = 166\).
Time = 0.31 (sec) , antiderivative size = 285, normalized size of antiderivative = 3.00 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{9/2}} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}} {\left (\frac {{\left (5 \, B b^{9} d^{2} e^{3} {\left | b \right |} - 12 \, B a b^{8} d e^{4} {\left | b \right |} + 2 \, A b^{9} d e^{4} {\left | b \right |} + 7 \, B a^{2} b^{7} e^{5} {\left | b \right |} - 2 \, A a b^{8} e^{5} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{5} d^{3} e^{3} - 3 \, a b^{4} d^{2} e^{4} + 3 \, a^{2} b^{3} d e^{5} - a^{3} b^{2} e^{6}} - \frac {7 \, {\left (B a b^{9} d^{2} e^{3} {\left | b \right |} - A b^{10} d^{2} e^{3} {\left | b \right |} - 2 \, B a^{2} b^{8} d e^{4} {\left | b \right |} + 2 \, A a b^{9} d e^{4} {\left | b \right |} + B a^{3} b^{7} e^{5} {\left | b \right |} - A a^{2} b^{8} e^{5} {\left | b \right |}\right )}}{b^{5} d^{3} e^{3} - 3 \, a b^{4} d^{2} e^{4} + 3 \, a^{2} b^{3} d e^{5} - a^{3} b^{2} e^{6}}\right )}}{35 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {7}{2}}} \] Input:
integrate((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(9/2),x, algorithm="giac")
Output:
2/35*(b*x + a)^(5/2)*((5*B*b^9*d^2*e^3*abs(b) - 12*B*a*b^8*d*e^4*abs(b) + 2*A*b^9*d*e^4*abs(b) + 7*B*a^2*b^7*e^5*abs(b) - 2*A*a*b^8*e^5*abs(b))*(b*x + a)/(b^5*d^3*e^3 - 3*a*b^4*d^2*e^4 + 3*a^2*b^3*d*e^5 - a^3*b^2*e^6) - 7* (B*a*b^9*d^2*e^3*abs(b) - A*b^10*d^2*e^3*abs(b) - 2*B*a^2*b^8*d*e^4*abs(b) + 2*A*a*b^9*d*e^4*abs(b) + B*a^3*b^7*e^5*abs(b) - A*a^2*b^8*e^5*abs(b))/( b^5*d^3*e^3 - 3*a*b^4*d^2*e^4 + 3*a^2*b^3*d*e^5 - a^3*b^2*e^6))/(b^2*d + ( b*x + a)*b*e - a*b*e)^(7/2)
Time = 1.46 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.71 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{9/2}} \, dx=-\frac {\sqrt {d+e\,x}\,\left (\frac {\sqrt {a+b\,x}\,\left (10\,A\,a^3\,e+4\,B\,a^3\,d-14\,A\,a^2\,b\,d\right )}{35\,e^4\,{\left (a\,e-b\,d\right )}^2}-\frac {x^3\,\sqrt {a+b\,x}\,\left (4\,A\,b^3\,e+10\,B\,b^3\,d-14\,B\,a\,b^2\,e\right )}{35\,e^4\,{\left (a\,e-b\,d\right )}^2}+\frac {x\,\sqrt {a+b\,x}\,\left (14\,B\,a^3\,e-28\,A\,a\,b^2\,d+16\,A\,a^2\,b\,e-2\,B\,a^2\,b\,d\right )}{35\,e^4\,{\left (a\,e-b\,d\right )}^2}-\frac {x^2\,\sqrt {a+b\,x}\,\left (14\,A\,b^3\,d-2\,A\,a\,b^2\,e+16\,B\,a\,b^2\,d-28\,B\,a^2\,b\,e\right )}{35\,e^4\,{\left (a\,e-b\,d\right )}^2}\right )}{x^4+\frac {d^4}{e^4}+\frac {4\,d\,x^3}{e}+\frac {4\,d^3\,x}{e^3}+\frac {6\,d^2\,x^2}{e^2}} \] Input:
int(((A + B*x)*(a + b*x)^(3/2))/(d + e*x)^(9/2),x)
Output:
-((d + e*x)^(1/2)*(((a + b*x)^(1/2)*(10*A*a^3*e + 4*B*a^3*d - 14*A*a^2*b*d ))/(35*e^4*(a*e - b*d)^2) - (x^3*(a + b*x)^(1/2)*(4*A*b^3*e + 10*B*b^3*d - 14*B*a*b^2*e))/(35*e^4*(a*e - b*d)^2) + (x*(a + b*x)^(1/2)*(14*B*a^3*e - 28*A*a*b^2*d + 16*A*a^2*b*e - 2*B*a^2*b*d))/(35*e^4*(a*e - b*d)^2) - (x^2* (a + b*x)^(1/2)*(14*A*b^3*d - 2*A*a*b^2*e + 16*B*a*b^2*d - 28*B*a^2*b*e))/ (35*e^4*(a*e - b*d)^2)))/(x^4 + d^4/e^4 + (4*d*x^3)/e + (4*d^3*x)/e^3 + (6 *d^2*x^2)/e^2)
Time = 0.44 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.81 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{(d+e x)^{9/2}} \, dx=\frac {-\frac {2 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{3} e^{4}}{7}-\frac {6 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{2} b \,e^{4} x}{7}-\frac {6 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{2} e^{4} x^{2}}{7}-\frac {2 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{3} e^{4} x^{3}}{7}-\frac {2 \sqrt {e}\, \sqrt {b}\, b^{3} d^{4}}{7}-\frac {8 \sqrt {e}\, \sqrt {b}\, b^{3} d^{3} e x}{7}-\frac {12 \sqrt {e}\, \sqrt {b}\, b^{3} d^{2} e^{2} x^{2}}{7}-\frac {8 \sqrt {e}\, \sqrt {b}\, b^{3} d \,e^{3} x^{3}}{7}-\frac {2 \sqrt {e}\, \sqrt {b}\, b^{3} e^{4} x^{4}}{7}}{e^{4} \left (a \,e^{5} x^{4}-b d \,e^{4} x^{4}+4 a d \,e^{4} x^{3}-4 b \,d^{2} e^{3} x^{3}+6 a \,d^{2} e^{3} x^{2}-6 b \,d^{3} e^{2} x^{2}+4 a \,d^{3} e^{2} x -4 b \,d^{4} e x +a \,d^{4} e -b \,d^{5}\right )} \] Input:
int((b*x+a)^(3/2)*(B*x+A)/(e*x+d)^(9/2),x)
Output:
(2*( - sqrt(d + e*x)*sqrt(a + b*x)*a**3*e**4 - 3*sqrt(d + e*x)*sqrt(a + b* x)*a**2*b*e**4*x - 3*sqrt(d + e*x)*sqrt(a + b*x)*a*b**2*e**4*x**2 - sqrt(d + e*x)*sqrt(a + b*x)*b**3*e**4*x**3 - sqrt(e)*sqrt(b)*b**3*d**4 - 4*sqrt( e)*sqrt(b)*b**3*d**3*e*x - 6*sqrt(e)*sqrt(b)*b**3*d**2*e**2*x**2 - 4*sqrt( e)*sqrt(b)*b**3*d*e**3*x**3 - sqrt(e)*sqrt(b)*b**3*e**4*x**4))/(7*e**4*(a* d**4*e + 4*a*d**3*e**2*x + 6*a*d**2*e**3*x**2 + 4*a*d*e**4*x**3 + a*e**5*x **4 - b*d**5 - 4*b*d**4*e*x - 6*b*d**3*e**2*x**2 - 4*b*d**2*e**3*x**3 - b* d*e**4*x**4))