3.7.19 \(\int \frac {(a+b x)^{3/2}}{x (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=92 \[ -\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {2 a \sqrt {a+b x}}{c^2 \sqrt {c+d x}}+\frac {2 (a+b x)^{3/2}}{3 c (c+d x)^{3/2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 92, 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 {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {2 a \sqrt {a+b x}}{c^2 \sqrt {c+d x}}+\frac {2 (a+b x)^{3/2}}{3 c (c+d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 82, normalized size = 0.89 \begin {gather*} \frac {2 \sqrt {a+b x} (4 a c+3 a d x+b c x)}{3 c^2 (c+d x)^{3/2}}-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.14, size = 85, normalized size = 0.92 \begin {gather*} \frac {2 (a+b x)^{3/2} \left (\frac {3 a (c+d x)}{a+b x}+c\right )}{3 c^2 (c+d x)^{3/2}}-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(3/2)/(x*(c + d*x)^(5/2)),x]

[Out]

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

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fricas [B]  time = 2.25, size = 337, normalized size = 3.66 \begin {gather*} \left [\frac {3 \, {\left (a d^{2} x^{2} + 2 \, a c d x + a c^{2}\right )} \sqrt {\frac {a}{c}} \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^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (4 \, a c + {\left (b c + 3 \, a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (c^{2} d^{2} x^{2} + 2 \, c^{3} d x + c^{4}\right )}}, \frac {3 \, {\left (a d^{2} x^{2} + 2 \, a c d x + a c^{2}\right )} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) + 2 \, {\left (4 \, a c + {\left (b c + 3 \, a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (c^{2} d^{2} x^{2} + 2 \, c^{3} d x + c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 2.01, size = 261, normalized size = 2.84 \begin {gather*} -\frac {2 \, \sqrt {b d} a^{2} b \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} c^{2} {\left | b \right |}} + \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (b^{5} c^{4} d {\left | b \right |} + 2 \, a b^{4} c^{3} d^{2} {\left | b \right |} - 3 \, a^{2} b^{3} c^{2} d^{3} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{3} c^{5} d - a b^{2} c^{4} d^{2}} + \frac {3 \, {\left (a b^{5} c^{4} d {\left | b \right |} - 2 \, a^{2} b^{4} c^{3} d^{2} {\left | b \right |} + a^{3} b^{3} c^{2} d^{3} {\left | b \right |}\right )}}{b^{3} c^{5} d - a b^{2} c^{4} d^{2}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(b*d)*a^2*b*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)*c^2*abs(b)) + 2/3*sqrt(b*x + a)*((b^5*c^4*d*abs(b) + 2*a*b^4*c^3*d^2*
abs(b) - 3*a^2*b^3*c^2*d^3*abs(b))*(b*x + a)/(b^3*c^5*d - a*b^2*c^4*d^2) + 3*(a*b^5*c^4*d*abs(b) - 2*a^2*b^4*c
^3*d^2*abs(b) + a^3*b^3*c^2*d^3*abs(b))/(b^3*c^5*d - a*b^2*c^4*d^2))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2)

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maple [B]  time = 0.03, size = 248, normalized size = 2.70 \begin {gather*} -\frac {\sqrt {b x +a}\, \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^{2} c d x \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 a^{2} c^{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 -2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, b c x -8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a c \right )}{3 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \left (d x +c \right )^{\frac {3}{2}} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/x/(d*x+c)^(5/2),x)

[Out]

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

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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/(d*x+c)^(5/2),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\,{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(3/2)/(x*(c + d*x)^(5/2)),x)

[Out]

int((a + b*x)^(3/2)/(x*(c + d*x)^(5/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)/x/(d*x+c)**(5/2),x)

[Out]

Timed out

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