3.8.90 \(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{9/2}} \, dx\) [790]

Optimal. Leaf size=120 \[ -\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^{7/2} (a+b x)}-\frac {2 (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)} \]

[Out]

-2/7*a*A*((b*x+a)^2)^(1/2)/x^(7/2)/(b*x+a)-2/5*(A*b+B*a)*((b*x+a)^2)^(1/2)/x^(5/2)/(b*x+a)-2/3*b*B*((b*x+a)^2)
^(1/2)/x^(3/2)/(b*x+a)

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Rubi [A]
time = 0.03, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {784, 77} \begin {gather*} -\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{5 x^{5/2} (a+b x)}-\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^{7/2} (a+b x)}-\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^(9/2),x]

[Out]

(-2*a*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*x^(7/2)*(a + b*x)) - (2*(A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
5*x^(5/2)*(a + b*x)) - (2*b*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^(3/2)*(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*} \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{9/2}} \, dx &=\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^{9/2}} \, 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^{9/2}}+\frac {b (A b+a B)}{x^{7/2}}+\frac {b^2 B}{x^{5/2}}\right ) \, dx}{a b+b^2 x}\\ &=-\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^{7/2} (a+b x)}-\frac {2 (A b+a B) \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}}{3 x^{3/2} (a+b x)}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 49, normalized size = 0.41 \begin {gather*} -\frac {2 \sqrt {(a+b x)^2} \left (15 a A+21 A b x+21 a B x+35 b B x^2\right )}{105 x^{7/2} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^(9/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(15*a*A + 21*A*b*x + 21*a*B*x + 35*b*B*x^2))/(105*x^(7/2)*(a + b*x))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 2.
time = 0.29, size = 34, normalized size = 0.28

method	result	size
default	\(-\frac {2 \,\mathrm {csgn}\left (b x +a \right ) \left (35 b B \,x^{2}+21 A b x +21 B a x +15 A a \right )}{105 x^{\frac {7}{2}}}\)	\(34\)
gosper	\(-\frac {2 \left (35 b B \,x^{2}+21 A b x +21 B a x +15 A a \right ) \sqrt {\left (b x +a \right )^{2}}}{105 x^{\frac {7}{2}} \left (b x +a \right )}\)	\(44\)
risch	\(-\frac {2 \left (35 b B \,x^{2}+21 A b x +21 B a x +15 A a \right ) \sqrt {\left (b x +a \right )^{2}}}{105 x^{\frac {7}{2}} \left (b x +a \right )}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/x^(9/2),x,method=_RETURNVERBOSE)

[Out]

-2/105*csgn(b*x+a)*(35*B*b*x^2+21*A*b*x+21*B*a*x+15*A*a)/x^(7/2)

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Maxima [A]
time = 0.30, size = 35, normalized size = 0.29 \begin {gather*} -\frac {2 \, {\left (5 \, b x^{2} + 3 \, a x\right )} B}{15 \, x^{\frac {7}{2}}} - \frac {2 \, {\left (7 \, b x^{2} + 5 \, a x\right )} A}{35 \, x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(9/2),x, algorithm="maxima")

[Out]

-2/15*(5*b*x^2 + 3*a*x)*B/x^(7/2) - 2/35*(7*b*x^2 + 5*a*x)*A/x^(9/2)

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Fricas [A]
time = 2.12, size = 27, normalized size = 0.22 \begin {gather*} -\frac {2 \, {\left (35 \, B b x^{2} + 15 \, A a + 21 \, {\left (B a + A b\right )} x\right )}}{105 \, x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(9/2),x, algorithm="fricas")

[Out]

-2/105*(35*B*b*x^2 + 15*A*a + 21*(B*a + A*b)*x)/x^(7/2)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**(9/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5988 deep

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Giac [A]
time = 0.78, size = 51, normalized size = 0.42 \begin {gather*} -\frac {2 \, {\left (35 \, B b x^{2} \mathrm {sgn}\left (b x + a\right ) + 21 \, B a x \mathrm {sgn}\left (b x + a\right ) + 21 \, A b x \mathrm {sgn}\left (b x + a\right ) + 15 \, A a \mathrm {sgn}\left (b x + a\right )\right )}}{105 \, x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(9/2),x, algorithm="giac")

[Out]

-2/105*(35*B*b*x^2*sgn(b*x + a) + 21*B*a*x*sgn(b*x + a) + 21*A*b*x*sgn(b*x + a) + 15*A*a*sgn(b*x + a))/x^(7/2)

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Mupad [B]
time = 1.35, size = 54, normalized size = 0.45 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,B\,x^2}{3}+\frac {2\,A\,a}{7\,b}+\frac {x\,\left (42\,A\,b+42\,B\,a\right )}{105\,b}\right )}{x^{9/2}+\frac {a\,x^{7/2}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((a + b*x)^2)^(1/2)*(A + B*x))/x^(9/2),x)

[Out]

-(((a + b*x)^2)^(1/2)*((2*B*x^2)/3 + (2*A*a)/(7*b) + (x*(42*A*b + 42*B*a))/(105*b)))/(x^(9/2) + (a*x^(7/2))/b)

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