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

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

[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)

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Rubi [A]  time = 0.0450964, antiderivative size = 102, normalized size of antiderivative = 1., 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} \[ \frac{2 \sqrt{a+b x}}{c^2 \sqrt{c+d x}}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{5/2}}-\frac{2 d (a+b x)^{3/2}}{3 c (c+d x)^{3/2} (b c-a d)} \]

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 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 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 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 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*}

Mathematica [A]  time = 0.196622, size = 102, normalized size = 1. \[ \frac{2 \sqrt{a+b x} (b c (3 c+2 d x)-a d (4 c+3 d x))}{3 c^2 (c+d x)^{3/2} (b c-a d)}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{5/2}} \]

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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Maple [B]  time = 0.021, size = 430, normalized size = 4.2 \begin{align*} -{\frac{1}{3\, \left ( ad-bc \right ){c}^{2}} \left ( 3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{a}^{2}{d}^{3}-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}abc{d}^{2}+6\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) x{a}^{2}c{d}^{2}-6\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) xab{c}^{2}d+3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){a}^{2}{c}^{2}d-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) ab{c}^{3}-6\,xa{d}^{2}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+4\,xbcd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }-8\,acd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+6\,b{c}^{2}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) } \right ) \sqrt{bx+a}{\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}} \left ( dx+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

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)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a^2*d^3-3*ln((a*d*x+b*c*x+2*(a*c)^
(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a*b*c*d^2+6*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+
2*a*c)/x)*x*a^2*c*d^2-6*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a*b*c^2*d+3*ln((a*d*
x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a^2*c^2*d-3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*
x+c))^(1/2)+2*a*c)/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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 5.75018, size = 1019, normalized size = 9.99 \begin{align*} \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{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [B]  time = 2.21754, size = 355, normalized size = 3.48 \begin{align*} -\frac{\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^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}} + \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^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}}\right )}}{12 \,{\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{align*}

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]

-1/12*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^8*c^2*d^4 - 2*a*b^7*c*d^5 +
a^2*b^6*d^6) + 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^8*c^2*d^4 - 2*a*b^7*c
*d^5 + a^2*b^6*d^6))/(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 - (sqr
t(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))