\(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx\) [789]

Optimal result

Integrand size = 31, antiderivative size = 118 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac {2 (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}-\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {784, 77} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{3 x^{3/2} (a+b x)}-\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)} \]

[In]

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

[Out]

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

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])

Rule 784

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.42 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (3 a A+5 A b x+5 a B x+15 b B x^2\right )}{15 x^{5/2} (a+b x)} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.29

method	result	size
default	\(-\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (15 B b \,x^{2}+5 A b x +5 a B x +3 a A \right )}{15 x^{\frac {5}{2}}}\)	\(34\)
gosper	\(-\frac {2 \left (15 B b \,x^{2}+5 A b x +5 a B x +3 a A \right ) \sqrt {\left (b x +a \right )^{2}}}{15 x^{\frac {5}{2}} \left (b x +a \right )}\)	\(44\)
risch	\(-\frac {2 \left (15 B b \,x^{2}+5 A b x +5 a B x +3 a A \right ) \sqrt {\left (b x +a \right )^{2}}}{15 x^{\frac {5}{2}} \left (b x +a \right )}\)	\(44\)

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.23 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 \, {\left (15 \, B b x^{2} + 3 \, A a + 5 \, {\left (B a + A b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.29 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 \, {\left (3 \, b x^{2} + a x\right )} B}{3 \, x^{\frac {5}{2}}} - \frac {2 \, {\left (5 \, b x^{2} + 3 \, a x\right )} A}{15 \, x^{\frac {7}{2}}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(7/2),x, algorithm="maxima")

[Out]

-2/3*(3*b*x^2 + a*x)*B/x^(5/2) - 2/15*(5*b*x^2 + 3*a*x)*A/x^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.43 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 \, {\left (15 \, B b x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a x \mathrm {sgn}\left (b x + a\right ) + 5 \, A b x \mathrm {sgn}\left (b x + a\right ) + 3 \, A a \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(7/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.46 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (2\,B\,x^2+\frac {2\,A\,a}{5\,b}+\frac {x\,\left (10\,A\,b+10\,B\,a\right )}{15\,b}\right )}{x^{7/2}+\frac {a\,x^{5/2}}{b}} \]

[In]

int((((a + b*x)^2)^(1/2)*(A + B*x))/x^(7/2),x)

[Out]

-(((a + b*x)^2)^(1/2)*(2*B*x^2 + (2*A*a)/(5*b) + (x*(10*A*b + 10*B*a))/(15*b)))/(x^(7/2) + (a*x^(5/2))/b)