\(\int \frac {(a+b x)^3 (A+B x)}{x^{7/2}} \, dx\) [342]

Optimal result

Integrand size = 18, antiderivative size = 81 \[ \int \frac {(a+b x)^3 (A+B x)}{x^{7/2}} \, dx=-\frac {2 a^3 A}{5 x^{5/2}}-\frac {2 a^2 (3 A b+a B)}{3 x^{3/2}}-\frac {6 a b (A b+a B)}{\sqrt {x}}+2 b^2 (A b+3 a B) \sqrt {x}+\frac {2}{3} b^3 B x^{3/2} \]

[Out]

-2/5*a^3*A/x^(5/2)-2/3*a^2*(3*A*b+B*a)/x^(3/2)+2/3*b^3*B*x^(3/2)-6*a*b*(A*b+B*a)/x^(1/2)+2*b^2*(A*b+3*B*a)*x^(
1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {77} \[ \int \frac {(a+b x)^3 (A+B x)}{x^{7/2}} \, dx=-\frac {2 a^3 A}{5 x^{5/2}}-\frac {2 a^2 (a B+3 A b)}{3 x^{3/2}}+2 b^2 \sqrt {x} (3 a B+A b)-\frac {6 a b (a B+A b)}{\sqrt {x}}+\frac {2}{3} b^3 B x^{3/2} \]

[In]

Int[((a + b*x)^3*(A + B*x))/x^(7/2),x]

[Out]

(-2*a^3*A)/(5*x^(5/2)) - (2*a^2*(3*A*b + a*B))/(3*x^(3/2)) - (6*a*b*(A*b + a*B))/Sqrt[x] + 2*b^2*(A*b + 3*a*B)
*Sqrt[x] + (2*b^3*B*x^(3/2))/3

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^3 A}{x^{7/2}}+\frac {a^2 (3 A b+a B)}{x^{5/2}}+\frac {3 a b (A b+a B)}{x^{3/2}}+\frac {b^2 (A b+3 a B)}{\sqrt {x}}+b^3 B \sqrt {x}\right ) \, dx \\ & = -\frac {2 a^3 A}{5 x^{5/2}}-\frac {2 a^2 (3 A b+a B)}{3 x^{3/2}}-\frac {6 a b (A b+a B)}{\sqrt {x}}+2 b^2 (A b+3 a B) \sqrt {x}+\frac {2}{3} b^3 B x^{3/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^3 (A+B x)}{x^{7/2}} \, dx=-\frac {2 \left (45 a b^2 x^2 (A-B x)-5 b^3 x^3 (3 A+B x)+15 a^2 b x (A+3 B x)+a^3 (3 A+5 B x)\right )}{15 x^{5/2}} \]

[In]

Integrate[((a + b*x)^3*(A + B*x))/x^(7/2),x]

[Out]

(-2*(45*a*b^2*x^2*(A - B*x) - 5*b^3*x^3*(3*A + B*x) + 15*a^2*b*x*(A + 3*B*x) + a^3*(3*A + 5*B*x)))/(15*x^(5/2)
)

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {2 b^{3} B \,x^{\frac {3}{2}}}{3}+2 A \,b^{3} \sqrt {x}+6 B a \,b^{2} \sqrt {x}-\frac {2 a^{3} A}{5 x^{\frac {5}{2}}}-\frac {6 a b \left (A b +B a \right )}{\sqrt {x}}-\frac {2 a^{2} \left (3 A b +B a \right )}{3 x^{\frac {3}{2}}}\)	\(69\)
default	\(\frac {2 b^{3} B \,x^{\frac {3}{2}}}{3}+2 A \,b^{3} \sqrt {x}+6 B a \,b^{2} \sqrt {x}-\frac {2 a^{3} A}{5 x^{\frac {5}{2}}}-\frac {6 a b \left (A b +B a \right )}{\sqrt {x}}-\frac {2 a^{2} \left (3 A b +B a \right )}{3 x^{\frac {3}{2}}}\)	\(69\)
gosper	\(-\frac {2 \left (-5 b^{3} B \,x^{4}-15 A \,b^{3} x^{3}-45 B a \,b^{2} x^{3}+45 a A \,b^{2} x^{2}+45 B \,a^{2} b \,x^{2}+15 a^{2} A b x +5 a^{3} B x +3 a^{3} A \right )}{15 x^{\frac {5}{2}}}\)	\(76\)
trager	\(-\frac {2 \left (-5 b^{3} B \,x^{4}-15 A \,b^{3} x^{3}-45 B a \,b^{2} x^{3}+45 a A \,b^{2} x^{2}+45 B \,a^{2} b \,x^{2}+15 a^{2} A b x +5 a^{3} B x +3 a^{3} A \right )}{15 x^{\frac {5}{2}}}\)	\(76\)
risch	\(-\frac {2 \left (-5 b^{3} B \,x^{4}-15 A \,b^{3} x^{3}-45 B a \,b^{2} x^{3}+45 a A \,b^{2} x^{2}+45 B \,a^{2} b \,x^{2}+15 a^{2} A b x +5 a^{3} B x +3 a^{3} A \right )}{15 x^{\frac {5}{2}}}\)	\(76\)

[In]

int((b*x+a)^3*(B*x+A)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/3*b^3*B*x^(3/2)+2*A*b^3*x^(1/2)+6*B*a*b^2*x^(1/2)-2/5*a^3*A/x^(5/2)-6*a*b*(A*b+B*a)/x^(1/2)-2/3*a^2*(3*A*b+B
*a)/x^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^3 (A+B x)}{x^{7/2}} \, dx=\frac {2 \, {\left (5 \, B b^{3} x^{4} - 3 \, A a^{3} + 15 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} - 45 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} - 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/x^(7/2),x, algorithm="fricas")

[Out]

2/15*(5*B*b^3*x^4 - 3*A*a^3 + 15*(3*B*a*b^2 + A*b^3)*x^3 - 45*(B*a^2*b + A*a*b^2)*x^2 - 5*(B*a^3 + 3*A*a^2*b)*
x)/x^(5/2)

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.30 \[ \int \frac {(a+b x)^3 (A+B x)}{x^{7/2}} \, dx=- \frac {2 A a^{3}}{5 x^{\frac {5}{2}}} - \frac {2 A a^{2} b}{x^{\frac {3}{2}}} - \frac {6 A a b^{2}}{\sqrt {x}} + 2 A b^{3} \sqrt {x} - \frac {2 B a^{3}}{3 x^{\frac {3}{2}}} - \frac {6 B a^{2} b}{\sqrt {x}} + 6 B a b^{2} \sqrt {x} + \frac {2 B b^{3} x^{\frac {3}{2}}}{3} \]

[In]

integrate((b*x+a)**3*(B*x+A)/x**(7/2),x)

[Out]

-2*A*a**3/(5*x**(5/2)) - 2*A*a**2*b/x**(3/2) - 6*A*a*b**2/sqrt(x) + 2*A*b**3*sqrt(x) - 2*B*a**3/(3*x**(3/2)) -
6*B*a**2*b/sqrt(x) + 6*B*a*b**2*sqrt(x) + 2*B*b**3*x**(3/2)/3

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b x)^3 (A+B x)}{x^{7/2}} \, dx=\frac {2}{3} \, B b^{3} x^{\frac {3}{2}} + 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \sqrt {x} - \frac {2 \, {\left (3 \, A a^{3} + 45 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/x^(7/2),x, algorithm="maxima")

[Out]

2/3*B*b^3*x^(3/2) + 2*(3*B*a*b^2 + A*b^3)*sqrt(x) - 2/15*(3*A*a^3 + 45*(B*a^2*b + A*a*b^2)*x^2 + 5*(B*a^3 + 3*
A*a^2*b)*x)/x^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94 \[ \int \frac {(a+b x)^3 (A+B x)}{x^{7/2}} \, dx=\frac {2}{3} \, B b^{3} x^{\frac {3}{2}} + 6 \, B a b^{2} \sqrt {x} + 2 \, A b^{3} \sqrt {x} - \frac {2 \, {\left (45 \, B a^{2} b x^{2} + 45 \, A a b^{2} x^{2} + 5 \, B a^{3} x + 15 \, A a^{2} b x + 3 \, A a^{3}\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((b*x+a)^3*(B*x+A)/x^(7/2),x, algorithm="giac")

[Out]

2/3*B*b^3*x^(3/2) + 6*B*a*b^2*sqrt(x) + 2*A*b^3*sqrt(x) - 2/15*(45*B*a^2*b*x^2 + 45*A*a*b^2*x^2 + 5*B*a^3*x +
15*A*a^2*b*x + 3*A*a^3)/x^(5/2)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x)^3 (A+B x)}{x^{7/2}} \, dx=-\frac {10\,B\,a^3\,x+6\,A\,a^3+90\,B\,a^2\,b\,x^2+30\,A\,a^2\,b\,x-90\,B\,a\,b^2\,x^3+90\,A\,a\,b^2\,x^2-10\,B\,b^3\,x^4-30\,A\,b^3\,x^3}{15\,x^{5/2}} \]

[In]

int(((A + B*x)*(a + b*x)^3)/x^(7/2),x)

[Out]

-(6*A*a^3 - 30*A*b^3*x^3 - 10*B*b^3*x^4 + 10*B*a^3*x + 30*A*a^2*b*x + 90*A*a*b^2*x^2 + 90*B*a^2*b*x^2 - 90*B*a
*b^2*x^3)/(15*x^(5/2))