3.7.6 \(\int \frac {(a+b x)^{3/2}}{x^3 \sqrt {c+d x}} \, dx\)

Optimal. Leaf size=119 \[ -\frac {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a} c^{5/2}}-\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 c^2 x}-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 c x^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {94, 93, 208} \begin {gather*} -\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 c^2 x}-\frac {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a} c^{5/2}}-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/(x^3*Sqrt[c + d*x]),x]

[Out]

(-3*(b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*c^2*x) - ((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*c*x^2) - (3*(b*c -
a*d)^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*Sqrt[a]*c^(5/2))

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 208

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 98, normalized size = 0.82 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} (-2 a c+3 a d x-5 b c x)}{4 c^2 x^2}-\frac {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a} c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/(x^3*Sqrt[c + d*x]),x]

[Out]

(Sqrt[a + b*x]*Sqrt[c + d*x]*(-2*a*c - 5*b*c*x + 3*a*d*x))/(4*c^2*x^2) - (3*(b*c - a*d)^2*ArcTanh[(Sqrt[c]*Sqr
t[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*Sqrt[a]*c^(5/2))

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IntegrateAlgebraic [A]  time = 0.19, size = 134, normalized size = 1.13 \begin {gather*} -\frac {3 (a d-b c)^2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{4 \sqrt {a} c^{5/2}}-\frac {(b c-a d)^2 \left (\frac {5 c \sqrt {c+d x}}{\sqrt {a+b x}}-\frac {3 a (c+d x)^{3/2}}{(a+b x)^{3/2}}\right )}{4 c^2 \left (c-\frac {a (c+d x)}{a+b x}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(3/2)/(x^3*Sqrt[c + d*x]),x]

[Out]

-1/4*((b*c - a*d)^2*((5*c*Sqrt[c + d*x])/Sqrt[a + b*x] - (3*a*(c + d*x)^(3/2))/(a + b*x)^(3/2)))/(c^2*(c - (a*
(c + d*x))/(a + b*x))^2) - (3*(-(b*c) + a*d)^2*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(4*Sq
rt[a]*c^(5/2))

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fricas [A]  time = 1.88, size = 332, normalized size = 2.79 \begin {gather*} \left [\frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a c} x^{2} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} c^{2} + {\left (5 \, a b c^{2} - 3 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, a c^{3} x^{2}}, \frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-a c} x^{2} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (2 \, a^{2} c^{2} + {\left (5 \, a b c^{2} - 3 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, a c^{3} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^3/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(a*c)*x^2*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 -
4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a^2*c^2
+ (5*a*b*c^2 - 3*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^3*x^2), 1/8*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)
*sqrt(-a*c)*x^2*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c
^2 + (a*b*c^2 + a^2*c*d)*x)) - 2*(2*a^2*c^2 + (5*a*b*c^2 - 3*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^3*x
^2)]

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giac [B]  time = 3.71, size = 1037, normalized size = 8.71 \begin {gather*} -\frac {b {\left (\frac {3 \, {\left (\sqrt {b d} b^{3} c^{2} - 2 \, \sqrt {b d} a b^{2} c d + \sqrt {b d} a^{2} b d^{2}\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b c^{2}} + \frac {2 \, {\left (5 \, \sqrt {b d} b^{9} c^{5} - 23 \, \sqrt {b d} a b^{8} c^{4} d + 42 \, \sqrt {b d} a^{2} b^{7} c^{3} d^{2} - 38 \, \sqrt {b d} a^{3} b^{6} c^{2} d^{3} + 17 \, \sqrt {b d} a^{4} b^{5} c d^{4} - 3 \, \sqrt {b d} a^{5} b^{4} d^{5} - 15 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{7} c^{4} + 28 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{6} c^{3} d - 2 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{5} c^{2} d^{2} - 20 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{3} b^{4} c d^{3} + 9 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{4} b^{3} d^{4} + 15 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} b^{5} c^{3} + \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a b^{4} c^{2} d + 9 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a^{2} b^{3} c d^{2} - 9 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a^{3} b^{2} d^{3} - 5 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{6} b^{3} c^{2} - 6 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{6} a b^{2} c d + 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{6} a^{2} b d^{2}\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}\right )}^{2} c^{2}}\right )}}{4 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^3/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

-1/4*b*(3*(sqrt(b*d)*b^3*c^2 - 2*sqrt(b*d)*a*b^2*c*d + sqrt(b*d)*a^2*b*d^2)*arctan(-1/2*(b^2*c + a*b*d - (sqrt
(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b*c^2) + 2*(
5*sqrt(b*d)*b^9*c^5 - 23*sqrt(b*d)*a*b^8*c^4*d + 42*sqrt(b*d)*a^2*b^7*c^3*d^2 - 38*sqrt(b*d)*a^3*b^6*c^2*d^3 +
17*sqrt(b*d)*a^4*b^5*c*d^4 - 3*sqrt(b*d)*a^5*b^4*d^5 - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (
b*x + a)*b*d - a*b*d))^2*b^7*c^4 + 28*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)
)^2*a*b^6*c^3*d - 2*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^5*c^2*d^
2 - 20*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^4*c*d^3 + 9*sqrt(b*d)
*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^3*d^4 + 15*sqrt(b*d)*(sqrt(b*d)*sqrt(
b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^5*c^3 + sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^4*a*b^4*c^2*d + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*
b*d))^4*a^2*b^3*c*d^2 - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^2*
d^3 - 5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^3*c^2 - 6*sqrt(b*d)*(sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^2*c*d + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b*d^2)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)*sqrt
(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)
*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^2*c^2))/abs(b)

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maple [B]  time = 0.02, size = 255, normalized size = 2.14 \begin {gather*} -\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (3 a^{2} d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-6 a b c d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+3 b^{2} c^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a d x +10 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, b c x +4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a c \right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, c^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/x^3/(d*x+c)^(1/2),x)

[Out]

-1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c^2*(3*a^2*d^2*x^2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2
))/x)-6*a*b*c*d*x^2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)+3*b^2*c^2*x^2*ln((a*d*x+b*
c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)-6*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*d*x+10*((b*x+a)*(d
*x+c))^(1/2)*(a*c)^(1/2)*b*c*x+4*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*c)/((b*x+a)*(d*x+c))^(1/2)/x^2/(a*c)^(1
/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^3/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{3/2}}{x^3\,\sqrt {c+d\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(3/2)/(x^3*(c + d*x)^(1/2)),x)

[Out]

int((a + b*x)^(3/2)/(x^3*(c + d*x)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{\frac {3}{2}}}{x^{3} \sqrt {c + d x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)/x**3/(d*x+c)**(1/2),x)

[Out]

Integral((a + b*x)**(3/2)/(x**3*sqrt(c + d*x)), x)

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