Integrand size = 28, antiderivative size = 177 \[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{3/2} \, dx=\frac {a^3 c (e x)^{1+m} \sqrt {a-b x} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )}{e (1+m) \sqrt {1-\frac {b^2 x^2}{a^2}}}-\frac {a^2 b c (e x)^{2+m} \sqrt {a-b x} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {b^2 x^2}{a^2}\right )}{e^2 (2+m) \sqrt {1-\frac {b^2 x^2}{a^2}}} \] Output:
a^3*c*(e*x)^(1+m)*(-b*x+a)^(1/2)*(b*c*x+a*c)^(1/2)*hypergeom([-3/2, 1/2+1/ 2*m],[3/2+1/2*m],b^2*x^2/a^2)/e/(1+m)/(1-b^2*x^2/a^2)^(1/2)-a^2*b*c*(e*x)^ (2+m)*(-b*x+a)^(1/2)*(b*c*x+a*c)^(1/2)*hypergeom([-3/2, 1+1/2*m],[2+1/2*m] ,b^2*x^2/a^2)/e^2/(2+m)/(1-b^2*x^2/a^2)^(1/2)
Time = 10.05 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.72 \[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{3/2} \, dx=\frac {a^2 c x (e x)^m \sqrt {a-b x} \sqrt {c (a+b x)} \left (-b (1+m) x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1+\frac {m}{2},2+\frac {m}{2},\frac {b^2 x^2}{a^2}\right )+a (2+m) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )\right )}{(1+m) (2+m) \sqrt {1-\frac {b^2 x^2}{a^2}}} \] Input:
Integrate[(e*x)^m*(a - b*x)^(5/2)*(a*c + b*c*x)^(3/2),x]
Output:
(a^2*c*x*(e*x)^m*Sqrt[a - b*x]*Sqrt[c*(a + b*x)]*(-(b*(1 + m)*x*Hypergeome tric2F1[-3/2, 1 + m/2, 2 + m/2, (b^2*x^2)/a^2]) + a*(2 + m)*Hypergeometric 2F1[-3/2, (1 + m)/2, (3 + m)/2, (b^2*x^2)/a^2]))/((1 + m)*(2 + m)*Sqrt[1 - (b^2*x^2)/a^2])
Time = 0.27 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {92, 136, 279, 278}
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.
\(\displaystyle \int (a-b x)^{5/2} (e x)^m (a c+b c x)^{3/2} \, dx\) |
\(\Big \downarrow \) 92 |
\(\displaystyle a \int (e x)^m (a-b x)^{3/2} (a c+b x c)^{3/2}dx-\frac {b \int (e x)^{m+1} (a-b x)^{3/2} (a c+b x c)^{3/2}dx}{e}\) |
\(\Big \downarrow \) 136 |
\(\displaystyle \frac {a \sqrt {a-b x} \sqrt {a c+b c x} \int (e x)^m \left (a^2 c-b^2 c x^2\right )^{3/2}dx}{\sqrt {a^2 c-b^2 c x^2}}-\frac {b \sqrt {a-b x} \sqrt {a c+b c x} \int (e x)^{m+1} \left (a^2 c-b^2 c x^2\right )^{3/2}dx}{e \sqrt {a^2 c-b^2 c x^2}}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {a^3 c \sqrt {a-b x} \sqrt {a c+b c x} \int (e x)^m \left (1-\frac {b^2 x^2}{a^2}\right )^{3/2}dx}{\sqrt {1-\frac {b^2 x^2}{a^2}}}-\frac {a^2 b c \sqrt {a-b x} \sqrt {a c+b c x} \int (e x)^{m+1} \left (1-\frac {b^2 x^2}{a^2}\right )^{3/2}dx}{e \sqrt {1-\frac {b^2 x^2}{a^2}}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {a^3 c \sqrt {a-b x} (e x)^{m+1} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {b^2 x^2}{a^2}\right )}{e (m+1) \sqrt {1-\frac {b^2 x^2}{a^2}}}-\frac {a^2 b c \sqrt {a-b x} (e x)^{m+2} \sqrt {a c+b c x} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},\frac {b^2 x^2}{a^2}\right )}{e^2 (m+2) \sqrt {1-\frac {b^2 x^2}{a^2}}}\) |
Input:
Int[(e*x)^m*(a - b*x)^(5/2)*(a*c + b*c*x)^(3/2),x]
Output:
(a^3*c*(e*x)^(1 + m)*Sqrt[a - b*x]*Sqrt[a*c + b*c*x]*Hypergeometric2F1[-3/ 2, (1 + m)/2, (3 + m)/2, (b^2*x^2)/a^2])/(e*(1 + m)*Sqrt[1 - (b^2*x^2)/a^2 ]) - (a^2*b*c*(e*x)^(2 + m)*Sqrt[a - b*x]*Sqrt[a*c + b*c*x]*Hypergeometric 2F1[-3/2, (2 + m)/2, (4 + m)/2, (b^2*x^2)/a^2])/(e^2*(2 + m)*Sqrt[1 - (b^2 *x^2)/a^2])
Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_] :> Simp[a Int[(a + b*x)^n*(c + d*x)^n*(f*x)^p, x], x] + Simp[b/f In t[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] && !RationalQ[p] && !IGtQ[m, 0] && NeQ[m + n + p + 2, 0]
Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_] :> Simp[(a + b*x)^FracPart[m]*((c + d*x)^FracPart[m]/(a*c + b*d*x^2)^Fr acPart[m]) Int[(a*c + b*d*x^2)^m*(f*x)^p, x], x] /; FreeQ[{a, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
\[\int \left (e x \right )^{m} \left (-b x +a \right )^{\frac {5}{2}} \left (b c x +a c \right )^{\frac {3}{2}}d x\]
Input:
int((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(3/2),x)
Output:
int((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(3/2),x)
\[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{3/2} \, dx=\int { {\left (b c x + a c\right )}^{\frac {3}{2}} {\left (-b x + a\right )}^{\frac {5}{2}} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(3/2),x, algorithm="fricas")
Output:
integral((b^3*c*x^3 - a*b^2*c*x^2 - a^2*b*c*x + a^3*c)*sqrt(b*c*x + a*c)*s qrt(-b*x + a)*(e*x)^m, x)
Timed out. \[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{3/2} \, dx=\text {Timed out} \] Input:
integrate((e*x)**m*(-b*x+a)**(5/2)*(b*c*x+a*c)**(3/2),x)
Output:
Timed out
\[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{3/2} \, dx=\int { {\left (b c x + a c\right )}^{\frac {3}{2}} {\left (-b x + a\right )}^{\frac {5}{2}} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(3/2),x, algorithm="maxima")
Output:
integrate((b*c*x + a*c)^(3/2)*(-b*x + a)^(5/2)*(e*x)^m, x)
\[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{3/2} \, dx=\int { {\left (b c x + a c\right )}^{\frac {3}{2}} {\left (-b x + a\right )}^{\frac {5}{2}} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(3/2),x, algorithm="giac")
Output:
integrate((b*c*x + a*c)^(3/2)*(-b*x + a)^(5/2)*(e*x)^m, x)
Timed out. \[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{3/2} \, dx=\int {\left (a\,c+b\,c\,x\right )}^{3/2}\,{\left (e\,x\right )}^m\,{\left (a-b\,x\right )}^{5/2} \,d x \] Input:
int((a*c + b*c*x)^(3/2)*(e*x)^m*(a - b*x)^(5/2),x)
Output:
int((a*c + b*c*x)^(3/2)*(e*x)^m*(a - b*x)^(5/2), x)
\[ \int (e x)^m (a-b x)^{5/2} (a c+b c x)^{3/2} \, dx=e^{m} \sqrt {c}\, c \left (\left (\int x^{m} \sqrt {b x +a}\, \sqrt {-b x +a}\, x^{3}d x \right ) b^{3}-\left (\int x^{m} \sqrt {b x +a}\, \sqrt {-b x +a}\, x^{2}d x \right ) a \,b^{2}-\left (\int x^{m} \sqrt {b x +a}\, \sqrt {-b x +a}\, x d x \right ) a^{2} b +\left (\int x^{m} \sqrt {b x +a}\, \sqrt {-b x +a}d x \right ) a^{3}\right ) \] Input:
int((e*x)^m*(-b*x+a)^(5/2)*(b*c*x+a*c)^(3/2),x)
Output:
e**m*sqrt(c)*c*(int(x**m*sqrt(a + b*x)*sqrt(a - b*x)*x**3,x)*b**3 - int(x* *m*sqrt(a + b*x)*sqrt(a - b*x)*x**2,x)*a*b**2 - int(x**m*sqrt(a + b*x)*sqr t(a - b*x)*x,x)*a**2*b + int(x**m*sqrt(a + b*x)*sqrt(a - b*x),x)*a**3)