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

Optimal. Leaf size=102 \[ -\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {2 \sqrt {a+b x}}{c^2 \sqrt {c+d x}}-\frac {2 d (a+b x)^{3/2}}{3 c (c+d x)^{3/2} (b c-a d)} \]

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Rubi [A]  time = 0.05, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {96, 94, 93, 208} \begin {gather*} \frac {2 \sqrt {a+b x}}{c^2 \sqrt {c+d x}}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}-\frac {2 d (a+b x)^{3/2}}{3 c (c+d x)^{3/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*(a + b*x)^(3/2))/(3*c*(b*c - a*d)*(c + d*x)^(3/2)) + (2*Sqrt[a + b*x])/(c^2*Sqrt[c + d*x]) - (2*Sqrt[a]*
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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[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 {\sqrt {a+b x}}{x (c+d x)^{5/2}} \, dx &=-\frac {2 d (a+b x)^{3/2}}{3 c (b c-a d) (c+d x)^{3/2}}+\frac {\int \frac {\sqrt {a+b x}}{x (c+d x)^{3/2}} \, dx}{c}\\ &=-\frac {2 d (a+b x)^{3/2}}{3 c (b c-a d) (c+d x)^{3/2}}+\frac {2 \sqrt {a+b x}}{c^2 \sqrt {c+d x}}+\frac {a \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{c^2}\\ &=-\frac {2 d (a+b x)^{3/2}}{3 c (b c-a d) (c+d x)^{3/2}}+\frac {2 \sqrt {a+b x}}{c^2 \sqrt {c+d x}}+\frac {(2 a) \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 d (a+b x)^{3/2}}{3 c (b c-a d) (c+d x)^{3/2}}+\frac {2 \sqrt {a+b x}}{c^2 \sqrt {c+d x}}-\frac {2 \sqrt {a} \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 = 0.26, size = 102, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {a+b x} (b c (3 c+2 d x)-a d (4 c+3 d x))}{3 c^2 (c+d x)^{3/2} (b c-a d)}-\frac {2 \sqrt {a} \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[Sqrt[a + b*x]/(x*(c + d*x)^(5/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.15, size = 103, normalized size = 1.01 \begin {gather*} \frac {2 \sqrt {a+b x} \left (-\frac {c d (a+b x)}{c+d x}-3 a d+3 b c\right )}{3 c^2 \sqrt {c+d x} (b c-a d)}-\frac {2 \sqrt {a} \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]

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

[Out]

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

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fricas [B]  time = 2.28, size = 479, normalized size = 4.70 \begin {gather*} \left [\frac {3 \, {\left (b c^{3} - a c^{2} d + {\left (b c d^{2} - a d^{3}\right )} x^{2} + 2 \, {\left (b c^{2} d - a c d^{2}\right )} x\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 (3 \, b c^{2} - 4 \, a c d + {\left (2 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (b c^{5} - a c^{4} d + {\left (b c^{3} d^{2} - a c^{2} d^{3}\right )} x^{2} + 2 \, {\left (b c^{4} d - a c^{3} d^{2}\right )} x\right )}}, \frac {3 \, {\left (b c^{3} - a c^{2} d + {\left (b c d^{2} - a d^{3}\right )} x^{2} + 2 \, {\left (b c^{2} d - a c d^{2}\right )} x\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 (3 \, b c^{2} - 4 \, a c d + {\left (2 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (b c^{5} - a c^{4} d + {\left (b c^{3} d^{2} - a c^{2} d^{3}\right )} x^{2} + 2 \, {\left (b c^{4} d - a c^{3} d^{2}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)/x/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.61, size = 243, normalized size = 2.38 \begin {gather*} \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (2 \, b^{4} c^{3} d^{2} {\left | b \right |} - 3 \, a 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 (b^{5} c^{4} d {\left | b \right |} - 2 \, a b^{4} c^{3} d^{2} {\left | b \right |} + a^{2} 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}}} - \frac {2 \, \sqrt {b d} a 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 |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)/x/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.03, size = 430, normalized size = 4.22 \begin {gather*} -\frac {\left (3 a^{2} d^{3} 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 a b c \,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^{2} 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 )-6 a b \,c^{2} 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} d \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 b \,c^{3} \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 {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,d^{2} x +4 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b c d x -8 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a c d +6 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b \,c^{2}\right ) \sqrt {b x +a}}{3 \left (a d -b c \right ) \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)^(1/2)/x/(d*x+c)^(5/2),x)

[Out]

-1/3*(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)*x^2*a^2*d^3-3*ln((a*d*x+b*c*x+2*a*c+2*
(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a*b*c*d^2+6*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^
(1/2))/x)*x*a^2*c*d^2-6*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x*a*b*c^2*d+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*d-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*b*c^3-6*x*a*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*x*b*c*d*(a*c)^(1/2)*((b*x+a)*(d*
x+c))^(1/2)-8*a*c*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+6*b*c^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/c^2*(b*x+
a)^(1/2)/(a*d-b*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)^(1/2)/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 {\sqrt {a+b\,x}}{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)^(1/2)/(x*(c + d*x)^(5/2)),x)

[Out]

int((a + b*x)^(1/2)/(x*(c + d*x)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x}}{x \left (c + d x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)/x/(d*x+c)**(5/2),x)

[Out]

Integral(sqrt(a + b*x)/(x*(c + d*x)**(5/2)), x)

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