Integrand size = 24, antiderivative size = 187 \[ \int \frac {A+B x}{(a+b x)^{3/2} (d+e x)^{7/2}} \, dx=-\frac {2 (A b-a B)}{b (b d-a e) \sqrt {a+b x} (d+e x)^{5/2}}+\frac {2 (b B d-6 A b e+5 a B e) \sqrt {a+b x}}{5 b (b d-a e)^2 (d+e x)^{5/2}}+\frac {8 (b B d-6 A b e+5 a B e) \sqrt {a+b x}}{15 (b d-a e)^3 (d+e x)^{3/2}}+\frac {16 b (b B d-6 A b e+5 a B e) \sqrt {a+b x}}{15 (b d-a e)^4 \sqrt {d+e x}} \]
-2*(A*b-B*a)/b/(-a*e+b*d)/(e*x+d)^(5/2)/(b*x+a)^(1/2)+2/5*(-6*A*b*e+5*B*a* e+B*b*d)*(b*x+a)^(1/2)/b/(-a*e+b*d)^2/(e*x+d)^(5/2)+8/15*(-6*A*b*e+5*B*a*e +B*b*d)*(b*x+a)^(1/2)/(-a*e+b*d)^3/(e*x+d)^(3/2)+16/15*b*(-6*A*b*e+5*B*a*e +B*b*d)*(b*x+a)^(1/2)/(-a*e+b*d)^4/(e*x+d)^(1/2)
Time = 0.26 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x}{(a+b x)^{3/2} (d+e x)^{7/2}} \, dx=-\frac {2 \left (-3 B d e^2 (a+b x)^3+3 A e^3 (a+b x)^3+10 b B d e (a+b x)^2 (d+e x)-15 A b e^2 (a+b x)^2 (d+e x)+5 a B e^2 (a+b x)^2 (d+e x)-15 b^2 B d (a+b x) (d+e x)^2+45 A b^2 e (a+b x) (d+e x)^2-30 a b B e (a+b x) (d+e x)^2+15 A b^3 (d+e x)^3-15 a b^2 B (d+e x)^3\right )}{15 (b d-a e)^4 \sqrt {a+b x} (d+e x)^{5/2}} \]
(-2*(-3*B*d*e^2*(a + b*x)^3 + 3*A*e^3*(a + b*x)^3 + 10*b*B*d*e*(a + b*x)^2 *(d + e*x) - 15*A*b*e^2*(a + b*x)^2*(d + e*x) + 5*a*B*e^2*(a + b*x)^2*(d + e*x) - 15*b^2*B*d*(a + b*x)*(d + e*x)^2 + 45*A*b^2*e*(a + b*x)*(d + e*x)^ 2 - 30*a*b*B*e*(a + b*x)*(d + e*x)^2 + 15*A*b^3*(d + e*x)^3 - 15*a*b^2*B*( d + e*x)^3))/(15*(b*d - a*e)^4*Sqrt[a + b*x]*(d + e*x)^(5/2))
Time = 0.25 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 55, 55, 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.
| \(\displaystyle \int \frac {A+B x}{(a+b x)^{3/2} (d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(5 a B e-6 A b e+b B d) \int \frac {1}{\sqrt {a+b x} (d+e x)^{7/2}}dx}{b (b d-a e)}-\frac {2 (A b-a B)}{b \sqrt {a+b x} (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle \frac {(5 a B e-6 A b e+b B d) \left (\frac {4 b \int \frac {1}{\sqrt {a+b x} (d+e x)^{5/2}}dx}{5 (b d-a e)}+\frac {2 \sqrt {a+b x}}{5 (d+e x)^{5/2} (b d-a e)}\right )}{b (b d-a e)}-\frac {2 (A b-a B)}{b \sqrt {a+b x} (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle \frac {(5 a B e-6 A b e+b B d) \left (\frac {4 b \left (\frac {2 b \int \frac {1}{\sqrt {a+b x} (d+e x)^{3/2}}dx}{3 (b d-a e)}+\frac {2 \sqrt {a+b x}}{3 (d+e x)^{3/2} (b d-a e)}\right )}{5 (b d-a e)}+\frac {2 \sqrt {a+b x}}{5 (d+e x)^{5/2} (b d-a e)}\right )}{b (b d-a e)}-\frac {2 (A b-a B)}{b \sqrt {a+b x} (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle \frac {\left (\frac {4 b \left (\frac {4 b \sqrt {a+b x}}{3 \sqrt {d+e x} (b d-a e)^2}+\frac {2 \sqrt {a+b x}}{3 (d+e x)^{3/2} (b d-a e)}\right )}{5 (b d-a e)}+\frac {2 \sqrt {a+b x}}{5 (d+e x)^{5/2} (b d-a e)}\right ) (5 a B e-6 A b e+b B d)}{b (b d-a e)}-\frac {2 (A b-a B)}{b \sqrt {a+b x} (d+e x)^{5/2} (b d-a e)}\) |
(-2*(A*b - a*B))/(b*(b*d - a*e)*Sqrt[a + b*x]*(d + e*x)^(5/2)) + ((b*B*d - 6*A*b*e + 5*a*B*e)*((2*Sqrt[a + b*x])/(5*(b*d - a*e)*(d + e*x)^(5/2)) + ( 4*b*((2*Sqrt[a + b*x])/(3*(b*d - a*e)*(d + e*x)^(3/2)) + (4*b*Sqrt[a + b*x ])/(3*(b*d - a*e)^2*Sqrt[d + e*x])))/(5*(b*d - a*e))))/(b*(b*d - a*e))
3.23.51.3.1 Defintions of rubi rules used
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_))^(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] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 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 = 1.10 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.50
| method | result | size |
| default | \(-\frac {2 \left (48 A \,b^{3} e^{3} x^{3}-40 B a \,b^{2} e^{3} x^{3}-8 B \,b^{3} d \,e^{2} x^{3}+24 A a \,b^{2} e^{3} x^{2}+120 A \,b^{3} d \,e^{2} x^{2}-20 B \,a^{2} b \,e^{3} x^{2}-104 B a \,b^{2} d \,e^{2} x^{2}-20 B \,b^{3} d^{2} e \,x^{2}-6 A \,a^{2} b \,e^{3} x +60 A a \,b^{2} d \,e^{2} x +90 A \,b^{3} d^{2} e x +5 B \,a^{3} e^{3} x -49 B \,a^{2} b d \,e^{2} x -85 B a \,b^{2} d^{2} e x -15 b^{3} B \,d^{3} x +3 a^{3} A \,e^{3}-15 A \,a^{2} b d \,e^{2}+45 A a \,b^{2} d^{2} e +15 A \,b^{3} d^{3}+2 B \,a^{3} d \,e^{2}-20 B \,a^{2} b \,d^{2} e -30 B a \,b^{2} d^{3}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} \sqrt {b x +a}\, \left (a e -b d \right )^{4}}\) | \(281\) |
| gosper | \(-\frac {2 \left (48 A \,b^{3} e^{3} x^{3}-40 B a \,b^{2} e^{3} x^{3}-8 B \,b^{3} d \,e^{2} x^{3}+24 A a \,b^{2} e^{3} x^{2}+120 A \,b^{3} d \,e^{2} x^{2}-20 B \,a^{2} b \,e^{3} x^{2}-104 B a \,b^{2} d \,e^{2} x^{2}-20 B \,b^{3} d^{2} e \,x^{2}-6 A \,a^{2} b \,e^{3} x +60 A a \,b^{2} d \,e^{2} x +90 A \,b^{3} d^{2} e x +5 B \,a^{3} e^{3} x -49 B \,a^{2} b d \,e^{2} x -85 B a \,b^{2} d^{2} e x -15 b^{3} B \,d^{3} x +3 a^{3} A \,e^{3}-15 A \,a^{2} b d \,e^{2}+45 A a \,b^{2} d^{2} e +15 A \,b^{3} d^{3}+2 B \,a^{3} d \,e^{2}-20 B \,a^{2} b \,d^{2} e -30 B a \,b^{2} d^{3}\right )}{15 \sqrt {b x +a}\, \left (e x +d \right )^{\frac {5}{2}} \left (e^{4} a^{4}-4 b \,e^{3} d \,a^{3}+6 b^{2} e^{2} d^{2} a^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}\) | \(322\) |
-2/15*(48*A*b^3*e^3*x^3-40*B*a*b^2*e^3*x^3-8*B*b^3*d*e^2*x^3+24*A*a*b^2*e^ 3*x^2+120*A*b^3*d*e^2*x^2-20*B*a^2*b*e^3*x^2-104*B*a*b^2*d*e^2*x^2-20*B*b^ 3*d^2*e*x^2-6*A*a^2*b*e^3*x+60*A*a*b^2*d*e^2*x+90*A*b^3*d^2*e*x+5*B*a^3*e^ 3*x-49*B*a^2*b*d*e^2*x-85*B*a*b^2*d^2*e*x-15*B*b^3*d^3*x+3*A*a^3*e^3-15*A* a^2*b*d*e^2+45*A*a*b^2*d^2*e+15*A*b^3*d^3+2*B*a^3*d*e^2-20*B*a^2*b*d^2*e-3 0*B*a*b^2*d^3)/(e*x+d)^(5/2)/(b*x+a)^(1/2)/(a*e-b*d)^4
Leaf count of result is larger than twice the leaf count of optimal. 583 vs. \(2 (165) = 330\).
Time = 3.84 (sec) , antiderivative size = 583, normalized size of antiderivative = 3.12 \[ \int \frac {A+B x}{(a+b x)^{3/2} (d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (3 \, A a^{3} e^{3} - 15 \, {\left (2 \, B a b^{2} - A b^{3}\right )} d^{3} - 5 \, {\left (4 \, B a^{2} b - 9 \, A a b^{2}\right )} d^{2} e + {\left (2 \, B a^{3} - 15 \, A a^{2} b\right )} d e^{2} - 8 \, {\left (B b^{3} d e^{2} + {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} e^{3}\right )} x^{3} - 4 \, {\left (5 \, B b^{3} d^{2} e + 2 \, {\left (13 \, B a b^{2} - 15 \, A b^{3}\right )} d e^{2} + {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} e^{3}\right )} x^{2} - {\left (15 \, B b^{3} d^{3} + 5 \, {\left (17 \, B a b^{2} - 18 \, A b^{3}\right )} d^{2} e + {\left (49 \, B a^{2} b - 60 \, A a b^{2}\right )} d e^{2} - {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{15 \, {\left (a b^{4} d^{7} - 4 \, a^{2} b^{3} d^{6} e + 6 \, a^{3} b^{2} d^{5} e^{2} - 4 \, a^{4} b d^{4} e^{3} + a^{5} d^{3} e^{4} + {\left (b^{5} d^{4} e^{3} - 4 \, a b^{4} d^{3} e^{4} + 6 \, a^{2} b^{3} d^{2} e^{5} - 4 \, a^{3} b^{2} d e^{6} + a^{4} b e^{7}\right )} x^{4} + {\left (3 \, b^{5} d^{5} e^{2} - 11 \, a b^{4} d^{4} e^{3} + 14 \, a^{2} b^{3} d^{3} e^{4} - 6 \, a^{3} b^{2} d^{2} e^{5} - a^{4} b d e^{6} + a^{5} e^{7}\right )} x^{3} + 3 \, {\left (b^{5} d^{6} e - 3 \, a b^{4} d^{5} e^{2} + 2 \, a^{2} b^{3} d^{4} e^{3} + 2 \, a^{3} b^{2} d^{3} e^{4} - 3 \, a^{4} b d^{2} e^{5} + a^{5} d e^{6}\right )} x^{2} + {\left (b^{5} d^{7} - a b^{4} d^{6} e - 6 \, a^{2} b^{3} d^{5} e^{2} + 14 \, a^{3} b^{2} d^{4} e^{3} - 11 \, a^{4} b d^{3} e^{4} + 3 \, a^{5} d^{2} e^{5}\right )} x\right )}} \]
-2/15*(3*A*a^3*e^3 - 15*(2*B*a*b^2 - A*b^3)*d^3 - 5*(4*B*a^2*b - 9*A*a*b^2 )*d^2*e + (2*B*a^3 - 15*A*a^2*b)*d*e^2 - 8*(B*b^3*d*e^2 + (5*B*a*b^2 - 6*A *b^3)*e^3)*x^3 - 4*(5*B*b^3*d^2*e + 2*(13*B*a*b^2 - 15*A*b^3)*d*e^2 + (5*B *a^2*b - 6*A*a*b^2)*e^3)*x^2 - (15*B*b^3*d^3 + 5*(17*B*a*b^2 - 18*A*b^3)*d ^2*e + (49*B*a^2*b - 60*A*a*b^2)*d*e^2 - (5*B*a^3 - 6*A*a^2*b)*e^3)*x)*sqr t(b*x + a)*sqrt(e*x + d)/(a*b^4*d^7 - 4*a^2*b^3*d^6*e + 6*a^3*b^2*d^5*e^2 - 4*a^4*b*d^4*e^3 + a^5*d^3*e^4 + (b^5*d^4*e^3 - 4*a*b^4*d^3*e^4 + 6*a^2*b ^3*d^2*e^5 - 4*a^3*b^2*d*e^6 + a^4*b*e^7)*x^4 + (3*b^5*d^5*e^2 - 11*a*b^4* d^4*e^3 + 14*a^2*b^3*d^3*e^4 - 6*a^3*b^2*d^2*e^5 - a^4*b*d*e^6 + a^5*e^7)* x^3 + 3*(b^5*d^6*e - 3*a*b^4*d^5*e^2 + 2*a^2*b^3*d^4*e^3 + 2*a^3*b^2*d^3*e ^4 - 3*a^4*b*d^2*e^5 + a^5*d*e^6)*x^2 + (b^5*d^7 - a*b^4*d^6*e - 6*a^2*b^3 *d^5*e^2 + 14*a^3*b^2*d^4*e^3 - 11*a^4*b*d^3*e^4 + 3*a^5*d^2*e^5)*x)
\[ \int \frac {A+B x}{(a+b x)^{3/2} (d+e x)^{7/2}} \, dx=\int \frac {A + B x}{\left (a + b x\right )^{\frac {3}{2}} \left (d + e x\right )^{\frac {7}{2}}}\, dx \]
Exception generated. \[ \int \frac {A+B x}{(a+b x)^{3/2} (d+e x)^{7/2}} \, dx=\text {Exception raised: ValueError} \]
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. 1353 vs. \(2 (165) = 330\).
Time = 0.54 (sec) , antiderivative size = 1353, normalized size of antiderivative = 7.24 \[ \int \frac {A+B x}{(a+b x)^{3/2} (d+e x)^{7/2}} \, dx=\text {Too large to display} \]
4*(B^2*a^2*b^7*e - 2*A*B*a*b^8*e + A^2*b^9*e)/((sqrt(b*e)*B*a*b^5*d - sqrt (b*e)*A*b^6*d - sqrt(b*e)*B*a^2*b^4*e + sqrt(b*e)*A*a*b^5*e - sqrt(b*e)*(s qrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*a*b^3 + sqrt(b*e)*(sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^ 2*A*b^4)*(b^3*d^3*abs(b) - 3*a*b^2*d^2*e*abs(b) + 3*a^2*b*d*e^2*abs(b) - a ^3*e^3*abs(b))) + 2/15*((b*x + a)*((8*B*b^13*d^6*e^4 - 15*B*a*b^12*d^5*e^5 - 33*A*b^13*d^5*e^5 - 45*B*a^2*b^11*d^4*e^6 + 165*A*a*b^12*d^4*e^6 + 170* B*a^3*b^10*d^3*e^7 - 330*A*a^2*b^11*d^3*e^7 - 210*B*a^4*b^9*d^2*e^8 + 330* A*a^3*b^10*d^2*e^8 + 117*B*a^5*b^8*d*e^9 - 165*A*a^4*b^9*d*e^9 - 25*B*a^6* b^7*e^10 + 33*A*a^5*b^8*e^10)*(b*x + a)/(b^11*d^9*e^2*abs(b) - 9*a*b^10*d^ 8*e^3*abs(b) + 36*a^2*b^9*d^7*e^4*abs(b) - 84*a^3*b^8*d^6*e^5*abs(b) + 126 *a^4*b^7*d^5*e^6*abs(b) - 126*a^5*b^6*d^4*e^7*abs(b) + 84*a^6*b^5*d^3*e^8* abs(b) - 36*a^7*b^4*d^2*e^9*abs(b) + 9*a^8*b^3*d*e^10*abs(b) - a^9*b^2*e^1 1*abs(b)) + 5*(4*B*b^14*d^7*e^3 - 13*B*a*b^13*d^6*e^4 - 15*A*b^14*d^6*e^4 - 6*B*a^2*b^12*d^5*e^5 + 90*A*a*b^13*d^5*e^5 + 85*B*a^3*b^11*d^4*e^6 - 225 *A*a^2*b^12*d^4*e^6 - 160*B*a^4*b^10*d^3*e^7 + 300*A*a^3*b^11*d^3*e^7 + 14 1*B*a^5*b^9*d^2*e^8 - 225*A*a^4*b^10*d^2*e^8 - 62*B*a^6*b^8*d*e^9 + 90*A*a ^5*b^9*d*e^9 + 11*B*a^7*b^7*e^10 - 15*A*a^6*b^8*e^10)/(b^11*d^9*e^2*abs(b) - 9*a*b^10*d^8*e^3*abs(b) + 36*a^2*b^9*d^7*e^4*abs(b) - 84*a^3*b^8*d^6*e^ 5*abs(b) + 126*a^4*b^7*d^5*e^6*abs(b) - 126*a^5*b^6*d^4*e^7*abs(b) + 84...
Time = 2.88 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.54 \[ \int \frac {A+B x}{(a+b x)^{3/2} (d+e x)^{7/2}} \, dx=\frac {\sqrt {d+e\,x}\,\left (\frac {2\,x\,\left (-a^2\,e^2+10\,a\,b\,d\,e+15\,b^2\,d^2\right )\,\left (5\,B\,a\,e-6\,A\,b\,e+B\,b\,d\right )}{15\,e^3\,{\left (a\,e-b\,d\right )}^4}-\frac {4\,B\,a^3\,d\,e^2+6\,A\,a^3\,e^3-40\,B\,a^2\,b\,d^2\,e-30\,A\,a^2\,b\,d\,e^2-60\,B\,a\,b^2\,d^3+90\,A\,a\,b^2\,d^2\,e+30\,A\,b^3\,d^3}{15\,e^3\,{\left (a\,e-b\,d\right )}^4}+\frac {16\,b^2\,x^3\,\left (5\,B\,a\,e-6\,A\,b\,e+B\,b\,d\right )}{15\,e\,{\left (a\,e-b\,d\right )}^4}+\frac {8\,b\,x^2\,\left (a\,e+5\,b\,d\right )\,\left (5\,B\,a\,e-6\,A\,b\,e+B\,b\,d\right )}{15\,e^2\,{\left (a\,e-b\,d\right )}^4}\right )}{x^3\,\sqrt {a+b\,x}+\frac {d^3\,\sqrt {a+b\,x}}{e^3}+\frac {3\,d\,x^2\,\sqrt {a+b\,x}}{e}+\frac {3\,d^2\,x\,\sqrt {a+b\,x}}{e^2}} \]
((d + e*x)^(1/2)*((2*x*(15*b^2*d^2 - a^2*e^2 + 10*a*b*d*e)*(5*B*a*e - 6*A* b*e + B*b*d))/(15*e^3*(a*e - b*d)^4) - (6*A*a^3*e^3 + 30*A*b^3*d^3 - 60*B* a*b^2*d^3 + 4*B*a^3*d*e^2 + 90*A*a*b^2*d^2*e - 30*A*a^2*b*d*e^2 - 40*B*a^2 *b*d^2*e)/(15*e^3*(a*e - b*d)^4) + (16*b^2*x^3*(5*B*a*e - 6*A*b*e + B*b*d) )/(15*e*(a*e - b*d)^4) + (8*b*x^2*(a*e + 5*b*d)*(5*B*a*e - 6*A*b*e + B*b*d ))/(15*e^2*(a*e - b*d)^4)))/(x^3*(a + b*x)^(1/2) + (d^3*(a + b*x)^(1/2))/e ^3 + (3*d*x^2*(a + b*x)^(1/2))/e + (3*d^2*x*(a + b*x)^(1/2))/e^2)