3.5.17 \(\int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx\) [417]

Optimal. Leaf size=133 \[ \frac {5}{4} b (3 A b+4 a B) \sqrt {a+b x}+\frac {5 b (3 A b+4 a B) (a+b x)^{3/2}}{12 a}-\frac {(3 A b+4 a B) (a+b x)^{5/2}}{4 a x}-\frac {A (a+b x)^{7/2}}{2 a x^2}-\frac {5}{4} \sqrt {a} b (3 A b+4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

5/12*b*(3*A*b+4*B*a)*(b*x+a)^(3/2)/a-1/4*(3*A*b+4*B*a)*(b*x+a)^(5/2)/a/x-1/2*A*(b*x+a)^(7/2)/a/x^2-5/4*b*(3*A*
b+4*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))*a^(1/2)+5/4*b*(3*A*b+4*B*a)*(b*x+a)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {79, 43, 52, 65, 214} \begin {gather*} -\frac {(a+b x)^{5/2} (4 a B+3 A b)}{4 a x}+\frac {5 b (a+b x)^{3/2} (4 a B+3 A b)}{12 a}+\frac {5}{4} b \sqrt {a+b x} (4 a B+3 A b)-\frac {5}{4} \sqrt {a} b (4 a B+3 A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )-\frac {A (a+b x)^{7/2}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*(3*A*b + 4*a*B)*Sqrt[a + b*x])/4 + (5*b*(3*A*b + 4*a*B)*(a + b*x)^(3/2))/(12*a) - ((3*A*b + 4*a*B)*(a + b
*x)^(5/2))/(4*a*x) - (A*(a + b*x)^(7/2))/(2*a*x^2) - (5*Sqrt[a]*b*(3*A*b + 4*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a
]])/4

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(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] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx &=-\frac {A (a+b x)^{7/2}}{2 a x^2}+\frac {\left (\frac {3 A b}{2}+2 a B\right ) \int \frac {(a+b x)^{5/2}}{x^2} \, dx}{2 a}\\ &=-\frac {(3 A b+4 a B) (a+b x)^{5/2}}{4 a x}-\frac {A (a+b x)^{7/2}}{2 a x^2}+\frac {(5 b (3 A b+4 a B)) \int \frac {(a+b x)^{3/2}}{x} \, dx}{8 a}\\ &=\frac {5 b (3 A b+4 a B) (a+b x)^{3/2}}{12 a}-\frac {(3 A b+4 a B) (a+b x)^{5/2}}{4 a x}-\frac {A (a+b x)^{7/2}}{2 a x^2}+\frac {1}{8} (5 b (3 A b+4 a B)) \int \frac {\sqrt {a+b x}}{x} \, dx\\ &=\frac {5}{4} b (3 A b+4 a B) \sqrt {a+b x}+\frac {5 b (3 A b+4 a B) (a+b x)^{3/2}}{12 a}-\frac {(3 A b+4 a B) (a+b x)^{5/2}}{4 a x}-\frac {A (a+b x)^{7/2}}{2 a x^2}+\frac {1}{8} (5 a b (3 A b+4 a B)) \int \frac {1}{x \sqrt {a+b x}} \, dx\\ &=\frac {5}{4} b (3 A b+4 a B) \sqrt {a+b x}+\frac {5 b (3 A b+4 a B) (a+b x)^{3/2}}{12 a}-\frac {(3 A b+4 a B) (a+b x)^{5/2}}{4 a x}-\frac {A (a+b x)^{7/2}}{2 a x^2}+\frac {1}{4} (5 a (3 A b+4 a B)) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )\\ &=\frac {5}{4} b (3 A b+4 a B) \sqrt {a+b x}+\frac {5 b (3 A b+4 a B) (a+b x)^{3/2}}{12 a}-\frac {(3 A b+4 a B) (a+b x)^{5/2}}{4 a x}-\frac {A (a+b x)^{7/2}}{2 a x^2}-\frac {5}{4} \sqrt {a} b (3 A b+4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 91, normalized size = 0.68 \begin {gather*} \frac {\sqrt {a+b x} \left (8 b^2 x^2 (3 A+B x)-6 a^2 (A+2 B x)+a b x (-27 A+56 B x)\right )}{12 x^2}-\frac {5}{4} \sqrt {a} b (3 A b+4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(8*b^2*x^2*(3*A + B*x) - 6*a^2*(A + 2*B*x) + a*b*x*(-27*A + 56*B*x)))/(12*x^2) - (5*Sqrt[a]*b*(
3*A*b + 4*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/4

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Maple [A]
time = 0.09, size = 111, normalized size = 0.83

method	result	size
risch	\(-\frac {a \sqrt {b x +a}\, \left (9 A b x +4 B a x +2 A a \right )}{4 x^{2}}+\frac {b \left (\frac {16 B \left (b x +a \right )^{\frac {3}{2}}}{3}+16 A b \sqrt {b x +a}+32 B a \sqrt {b x +a}-10 \sqrt {a}\, \left (3 A b +4 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right )}{8}\)	\(92\)
derivativedivides	\(2 b \left (\frac {B \left (b x +a \right )^{\frac {3}{2}}}{3}+A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-a \left (\frac {\left (\frac {9 A b}{8}+\frac {B a}{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a b A -\frac {1}{2} a^{2} B \right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {5 \left (3 A b +4 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\)	\(111\)
default	\(2 b \left (\frac {B \left (b x +a \right )^{\frac {3}{2}}}{3}+A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-a \left (\frac {\left (\frac {9 A b}{8}+\frac {B a}{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a b A -\frac {1}{2} a^{2} B \right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {5 \left (3 A b +4 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^3,x,method=_RETURNVERBOSE)

[Out]

2*b*(1/3*B*(b*x+a)^(3/2)+A*b*(b*x+a)^(1/2)+2*B*a*(b*x+a)^(1/2)-a*(((9/8*A*b+1/2*B*a)*(b*x+a)^(3/2)+(-7/8*a*b*A
-1/2*a^2*B)*(b*x+a)^(1/2))/b^2/x^2+5/8*(3*A*b+4*B*a)/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2))))

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Maxima [A]
time = 0.49, size = 155, normalized size = 1.17 \begin {gather*} \frac {1}{24} \, {\left (\frac {15 \, {\left (4 \, B a + 3 \, A b\right )} \sqrt {a} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{b} - \frac {6 \, {\left ({\left (4 \, B a^{2} + 9 \, A a b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - {\left (4 \, B a^{3} + 7 \, A a^{2} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{2} b - 2 \, {\left (b x + a\right )} a b + a^{2} b} + \frac {16 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} B + 3 \, {\left (2 \, B a + A b\right )} \sqrt {b x + a}\right )}}{b}\right )} b^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^3,x, algorithm="maxima")

[Out]

1/24*(15*(4*B*a + 3*A*b)*sqrt(a)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/b - 6*((4*B*a^2 + 9*
A*a*b)*(b*x + a)^(3/2) - (4*B*a^3 + 7*A*a^2*b)*sqrt(b*x + a))/((b*x + a)^2*b - 2*(b*x + a)*a*b + a^2*b) + 16*(
(b*x + a)^(3/2)*B + 3*(2*B*a + A*b)*sqrt(b*x + a))/b)*b^2

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Fricas [A]
time = 1.21, size = 209, normalized size = 1.57 \begin {gather*} \left [\frac {15 \, {\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt {a} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, B b^{2} x^{3} - 6 \, A a^{2} + 8 \, {\left (7 \, B a b + 3 \, A b^{2}\right )} x^{2} - 3 \, {\left (4 \, B a^{2} + 9 \, A a b\right )} x\right )} \sqrt {b x + a}}{24 \, x^{2}}, \frac {15 \, {\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (8 \, B b^{2} x^{3} - 6 \, A a^{2} + 8 \, {\left (7 \, B a b + 3 \, A b^{2}\right )} x^{2} - 3 \, {\left (4 \, B a^{2} + 9 \, A a b\right )} x\right )} \sqrt {b x + a}}{12 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^3,x, algorithm="fricas")

[Out]

[1/24*(15*(4*B*a*b + 3*A*b^2)*sqrt(a)*x^2*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*B*b^2*x^3 - 6*A*
a^2 + 8*(7*B*a*b + 3*A*b^2)*x^2 - 3*(4*B*a^2 + 9*A*a*b)*x)*sqrt(b*x + a))/x^2, 1/12*(15*(4*B*a*b + 3*A*b^2)*sq
rt(-a)*x^2*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (8*B*b^2*x^3 - 6*A*a^2 + 8*(7*B*a*b + 3*A*b^2)*x^2 - 3*(4*B*a^2
+ 9*A*a*b)*x)*sqrt(b*x + a))/x^2]

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Sympy [A]
time = 34.78, size = 488, normalized size = 3.67 \begin {gather*} - \frac {10 A a^{4} b^{2} \sqrt {a + b x}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac {6 A a^{3} b^{2} \left (a + b x\right )^{\frac {3}{2}}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac {3 A a^{3} b^{2} \sqrt {\frac {1}{a^{5}}} \log {\left (- a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {3 A a^{3} b^{2} \sqrt {\frac {1}{a^{5}}} \log {\left (a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {3 A a^{2} b^{2} \sqrt {\frac {1}{a^{3}}} \log {\left (- a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {3 A a^{2} b^{2} \sqrt {\frac {1}{a^{3}}} \log {\left (a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {6 A a b^{2} \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} - \frac {3 A a b \sqrt {a + b x}}{x} + 2 A b^{2} \sqrt {a + b x} - \frac {B a^{3} b \sqrt {\frac {1}{a^{3}}} \log {\left (- a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {B a^{3} b \sqrt {\frac {1}{a^{3}}} \log {\left (a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {6 B a^{2} b \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} - \frac {B a^{2} \sqrt {a + b x}}{x} + 4 B a b \sqrt {a + b x} + B b^{2} \left (\begin {cases} \sqrt {a} x & \text {for}\: b = 0 \\\frac {2 \left (a + b x\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(B*x+A)/x**3,x)

[Out]

-10*A*a**4*b**2*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 6*A*a**3*b**2*(a + b*x)**(3/2)/(
-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 3*A*a**3*b**2*sqrt(a**(-5))*log(-a**3*sqrt(a**(-5)) + sqrt(a +
b*x))/8 - 3*A*a**3*b**2*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3*A*a**2*b**2*sqrt(a**(-3))*
log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + 3*A*a**2*b**2*sqrt(a**(-3))*log(a**2*sqrt(a**(-3)) + sqrt(a + b*x
))/2 + 6*A*a*b**2*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - 3*A*a*b*sqrt(a + b*x)/x + 2*A*b**2*sqrt(a + b*x) - B
*a**3*b*sqrt(a**(-3))*log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + B*a**3*b*sqrt(a**(-3))*log(a**2*sqrt(a**(-3
)) + sqrt(a + b*x))/2 + 6*B*a**2*b*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - B*a**2*sqrt(a + b*x)/x + 4*B*a*b*sq
rt(a + b*x) + B*b**2*Piecewise((sqrt(a)*x, Eq(b, 0)), (2*(a + b*x)**(3/2)/(3*b), True))

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Giac [A]
time = 2.16, size = 155, normalized size = 1.17 \begin {gather*} \frac {8 \, {\left (b x + a\right )}^{\frac {3}{2}} B b^{2} + 48 \, \sqrt {b x + a} B a b^{2} + 24 \, \sqrt {b x + a} A b^{3} + \frac {15 \, {\left (4 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3 \, {\left (4 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{2} b^{2} - 4 \, \sqrt {b x + a} B a^{3} b^{2} + 9 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b^{3} - 7 \, \sqrt {b x + a} A a^{2} b^{3}\right )}}{b^{2} x^{2}}}{12 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^3,x, algorithm="giac")

[Out]

1/12*(8*(b*x + a)^(3/2)*B*b^2 + 48*sqrt(b*x + a)*B*a*b^2 + 24*sqrt(b*x + a)*A*b^3 + 15*(4*B*a^2*b^2 + 3*A*a*b^
3)*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - 3*(4*(b*x + a)^(3/2)*B*a^2*b^2 - 4*sqrt(b*x + a)*B*a^3*b^2 + 9*(b
*x + a)^(3/2)*A*a*b^3 - 7*sqrt(b*x + a)*A*a^2*b^3)/(b^2*x^2))/b

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Mupad [B]
time = 0.42, size = 162, normalized size = 1.22 \begin {gather*} \left (2\,A\,b^2+4\,B\,a\,b\right )\,\sqrt {a+b\,x}-\frac {\left (B\,a^2\,b+\frac {9\,A\,a\,b^2}{4}\right )\,{\left (a+b\,x\right )}^{3/2}-\left (B\,a^3\,b+\frac {7\,A\,a^2\,b^2}{4}\right )\,\sqrt {a+b\,x}}{{\left (a+b\,x\right )}^2-2\,a\,\left (a+b\,x\right )+a^2}+\frac {2\,B\,b\,{\left (a+b\,x\right )}^{3/2}}{3}+2\,b\,\mathrm {atan}\left (\frac {2\,b\,\left (3\,A\,b+4\,B\,a\right )\,\sqrt {-\frac {25\,a}{64}}\,\sqrt {a+b\,x}}{5\,B\,a^2\,b+\frac {15\,A\,a\,b^2}{4}}\right )\,\left (3\,A\,b+4\,B\,a\right )\,\sqrt {-\frac {25\,a}{64}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(5/2))/x^3,x)

[Out]

(2*A*b^2 + 4*B*a*b)*(a + b*x)^(1/2) - (((9*A*a*b^2)/4 + B*a^2*b)*(a + b*x)^(3/2) - ((7*A*a^2*b^2)/4 + B*a^3*b)
*(a + b*x)^(1/2))/((a + b*x)^2 - 2*a*(a + b*x) + a^2) + (2*B*b*(a + b*x)^(3/2))/3 + 2*b*atan((2*b*(3*A*b + 4*B
*a)*(-(25*a)/64)^(1/2)*(a + b*x)^(1/2))/((15*A*a*b^2)/4 + 5*B*a^2*b))*(3*A*b + 4*B*a)*(-(25*a)/64)^(1/2)

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