3.5.4 \(\int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx\) [404]

Optimal. Leaf size=95 \[ (3 A b+2 a B) \sqrt {a+b x}+\frac {(3 A b+2 a B) (a+b x)^{3/2}}{3 a}-\frac {A (a+b x)^{5/2}}{a x}-\sqrt {a} (3 A b+2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {79, 52, 65, 214} \begin {gather*} \frac {(a+b x)^{3/2} (2 a B+3 A b)}{3 a}+\sqrt {a+b x} (2 a B+3 A b)-\sqrt {a} (2 a B+3 A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )-\frac {A (a+b x)^{5/2}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.11, size = 71, normalized size = 0.75 \begin {gather*} \frac {\sqrt {a+b x} (2 b x (3 A+B x)+a (-3 A+8 B x))}{3 x}-\sqrt {a} (3 A b+2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(2*b*x*(3*A + B*x) + a*(-3*A + 8*B*x)))/(3*x) - Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x]/S
qrt[a]]

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Maple [A]
time = 0.10, size = 77, normalized size = 0.81

method	result	size
risch	\(-\frac {a A \sqrt {b x +a}}{x}+\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-\left (3 A b +2 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {a}\)	\(74\)
derivativedivides	\(\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-2 a \left (\frac {A \sqrt {b x +a}}{2 x}+\frac {\left (3 A b +2 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\)	\(77\)
default	\(\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-2 a \left (\frac {A \sqrt {b x +a}}{2 x}+\frac {\left (3 A b +2 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 97, normalized size = 1.02 \begin {gather*} \frac {1}{6} \, {\left (\frac {3 \, {\left (2 \, 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 \, \sqrt {b x + a} A a}{b x} + \frac {4 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} B + 3 \, {\left (B a + A b\right )} \sqrt {b x + a}\right )}}{b}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(2*B*a + 3*A*b)*sqrt(a)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/b - 6*sqrt(b*x + a)*A*
a/(b*x) + 4*((b*x + a)^(3/2)*B + 3*(B*a + A*b)*sqrt(b*x + a))/b)*b

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Fricas [A]
time = 1.20, size = 151, normalized size = 1.59 \begin {gather*} \left [\frac {3 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {a} x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, B b x^{2} - 3 \, A a + 2 \, {\left (4 \, B a + 3 \, A b\right )} x\right )} \sqrt {b x + a}}{6 \, x}, \frac {3 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {-a} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (2 \, B b x^{2} - 3 \, A a + 2 \, {\left (4 \, B a + 3 \, A b\right )} x\right )} \sqrt {b x + a}}{3 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(2*B*a + 3*A*b)*sqrt(a)*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(2*B*b*x^2 - 3*A*a + 2*(4*B
*a + 3*A*b)*x)*sqrt(b*x + a))/x, 1/3*(3*(2*B*a + 3*A*b)*sqrt(-a)*x*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (2*B*b*x
^2 - 3*A*a + 2*(4*B*a + 3*A*b)*x)*sqrt(b*x + a))/x]

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Sympy [A]
time = 12.14, size = 202, normalized size = 2.13 \begin {gather*} - \frac {A a^{2} b \sqrt {\frac {1}{a^{3}}} \log {\left (- a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {A a^{2} b \sqrt {\frac {1}{a^{3}}} \log {\left (a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {4 A a b \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} - \frac {A a \sqrt {a + b x}}{x} + 2 A b \sqrt {a + b x} + \frac {2 B a^{2} \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 B a \sqrt {a + b x} + B b \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)**(3/2)*(B*x+A)/x**2,x)

[Out]

-A*a**2*b*sqrt(a**(-3))*log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + A*a**2*b*sqrt(a**(-3))*log(a**2*sqrt(a**(
-3)) + sqrt(a + b*x))/2 + 4*A*a*b*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - A*a*sqrt(a + b*x)/x + 2*A*b*sqrt(a +
b*x) + 2*B*a**2*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2*B*a*sqrt(a + b*x) + B*b*Piecewise((sqrt(a)*x, Eq(b,
0)), (2*(a + b*x)**(3/2)/(3*b), True))

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Giac [A]
time = 1.27, size = 93, normalized size = 0.98 \begin {gather*} \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} B b + 6 \, \sqrt {b x + a} B a b + 6 \, \sqrt {b x + a} A b^{2} - \frac {3 \, \sqrt {b x + a} A a b}{x} + \frac {3 \, {\left (2 \, B a^{2} b + 3 \, A a b^{2}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.40, size = 96, normalized size = 1.01 \begin {gather*} \left (2\,A\,b+2\,B\,a\right )\,\sqrt {a+b\,x}+\frac {2\,B\,{\left (a+b\,x\right )}^{3/2}}{3}+2\,\mathrm {atan}\left (\frac {2\,\left (3\,A\,b+2\,B\,a\right )\,\sqrt {-\frac {a}{4}}\,\sqrt {a+b\,x}}{2\,B\,a^2+3\,A\,b\,a}\right )\,\left (3\,A\,b+2\,B\,a\right )\,\sqrt {-\frac {a}{4}}-\frac {A\,a\,\sqrt {a+b\,x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*b + 2*B*a)*(a + b*x)^(1/2) + (2*B*(a + b*x)^(3/2))/3 + 2*atan((2*(3*A*b + 2*B*a)*(-a/4)^(1/2)*(a + b*x)^(
1/2))/(2*B*a^2 + 3*A*a*b))*(3*A*b + 2*B*a)*(-a/4)^(1/2) - (A*a*(a + b*x)^(1/2))/x

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