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

Optimal. Leaf size=148 \[ -\frac{d \sqrt{a+b x} (13 b c-15 a d)}{3 c^3 \sqrt{c+d x} (b c-a d)}-\frac{5 d \sqrt{a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac{(b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{7/2}}-\frac{\sqrt{a+b x}}{c x (c+d x)^{3/2}} \]

[Out]

(-5*d*Sqrt[a + b*x])/(3*c^2*(c + d*x)^(3/2)) - Sqrt[a + b*x]/(c*x*(c + d*x)^(3/2)) - (d*(13*b*c - 15*a*d)*Sqrt
[a + b*x])/(3*c^3*(b*c - a*d)*Sqrt[c + d*x]) - ((b*c - 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c
+ d*x])])/(Sqrt[a]*c^(7/2))

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Rubi [A]  time = 0.132718, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {99, 152, 12, 93, 208} \[ -\frac{d \sqrt{a+b x} (13 b c-15 a d)}{3 c^3 \sqrt{c+d x} (b c-a d)}-\frac{5 d \sqrt{a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac{(b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{7/2}}-\frac{\sqrt{a+b x}}{c x (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*d*Sqrt[a + b*x])/(3*c^2*(c + d*x)^(3/2)) - Sqrt[a + b*x]/(c*x*(c + d*x)^(3/2)) - (d*(13*b*c - 15*a*d)*Sqrt
[a + b*x])/(3*c^3*(b*c - a*d)*Sqrt[c + d*x]) - ((b*c - 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c
+ d*x])])/(Sqrt[a]*c^(7/2))

Rule 99

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[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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^2 (c+d x)^{5/2}} \, dx &=-\frac{\sqrt{a+b x}}{c x (c+d x)^{3/2}}+\frac{\int \frac{\frac{1}{2} (b c-5 a d)-2 b d x}{x \sqrt{a+b x} (c+d x)^{5/2}} \, dx}{c}\\ &=-\frac{5 d \sqrt{a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{c x (c+d x)^{3/2}}-\frac{2 \int \frac{-\frac{3}{4} (b c-5 a d) (b c-a d)+\frac{5}{2} b d (b c-a d) x}{x \sqrt{a+b x} (c+d x)^{3/2}} \, dx}{3 c^2 (b c-a d)}\\ &=-\frac{5 d \sqrt{a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{c x (c+d x)^{3/2}}-\frac{d (13 b c-15 a d) \sqrt{a+b x}}{3 c^3 (b c-a d) \sqrt{c+d x}}+\frac{4 \int \frac{3 (b c-5 a d) (b c-a d)^2}{8 x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{3 c^3 (b c-a d)^2}\\ &=-\frac{5 d \sqrt{a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{c x (c+d x)^{3/2}}-\frac{d (13 b c-15 a d) \sqrt{a+b x}}{3 c^3 (b c-a d) \sqrt{c+d x}}+\frac{(b c-5 a d) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 c^3}\\ &=-\frac{5 d \sqrt{a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{c x (c+d x)^{3/2}}-\frac{d (13 b c-15 a d) \sqrt{a+b x}}{3 c^3 (b c-a d) \sqrt{c+d x}}+\frac{(b c-5 a d) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{c^3}\\ &=-\frac{5 d \sqrt{a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{c x (c+d x)^{3/2}}-\frac{d (13 b c-15 a d) \sqrt{a+b x}}{3 c^3 (b c-a d) \sqrt{c+d x}}-\frac{(b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.225474, size = 150, normalized size = 1.01 \[ \frac{\frac{3 (c+d x) (b c-5 a d) \left (\sqrt{c} \sqrt{a+b x}-\sqrt{a} \sqrt{c+d x} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )}{c^{5/2}}-\frac{d (a+b x)^{3/2} (3 b c-5 a d)}{c (b c-a d)}-\frac{3 (a+b x)^{3/2}}{x}}{3 a c (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.025, size = 653, normalized size = 4.4 \begin{align*}{\frac{1}{6\,{c}^{3}x \left ( ad-bc \right ) } \left ( 15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}{a}^{2}{d}^{4}-18\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}abc{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}^{3}{b}^{2}{c}^{2}{d}^{2}+30\,\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}c{d}^{3}-36\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}ab{c}^{2}{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}^{2}{b}^{2}{c}^{3}d+15\,\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}^{2}{d}^{2}-18\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) xab{c}^{3}d+3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) x{b}^{2}{c}^{4}-30\,{x}^{2}a{d}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+26\,{x}^{2}bc{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}-40\,xac{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+36\,xb{c}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}-6\,a{c}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+6\,b{c}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac} \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^2/(d*x+c)^(5/2),x)

[Out]

1/6*(15*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^3*a^2*d^4-18*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^3*a*b*c*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^3*b^2*c^2*d^2+30*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*c*d^3-36
*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^2*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^2*b^2*c^3*d+15*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^2*d^2-18*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^3*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)*x*b^2*c^4-30*x^2*a*d^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^
(1/2)+26*x^2*b*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-40*x*a*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+36*x
*b*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-6*a*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+6*b*c^3*((b*x+a)*(d
*x+c))^(1/2)*(a*c)^(1/2))/c^3*(b*x+a)^(1/2)/x/(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^2/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 9.27466, size = 1422, normalized size = 9.61 \begin{align*} \left [-\frac{3 \,{\left ({\left (b^{2} c^{2} d^{2} - 6 \, a b c d^{3} + 5 \, a^{2} d^{4}\right )} x^{3} + 2 \,{\left (b^{2} c^{3} d - 6 \, a b c^{2} d^{2} + 5 \, a^{2} c d^{3}\right )} x^{2} +{\left (b^{2} c^{4} - 6 \, a b c^{3} d + 5 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt{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 +{\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 (3 \, a b c^{4} - 3 \, a^{2} c^{3} d +{\left (13 \, a b c^{2} d^{2} - 15 \, a^{2} c d^{3}\right )} x^{2} + 2 \,{\left (9 \, a b c^{3} d - 10 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{12 \,{\left ({\left (a b c^{5} d^{2} - a^{2} c^{4} d^{3}\right )} x^{3} + 2 \,{\left (a b c^{6} d - a^{2} c^{5} d^{2}\right )} x^{2} +{\left (a b c^{7} - a^{2} c^{6} d\right )} x\right )}}, \frac{3 \,{\left ({\left (b^{2} c^{2} d^{2} - 6 \, a b c d^{3} + 5 \, a^{2} d^{4}\right )} x^{3} + 2 \,{\left (b^{2} c^{3} d - 6 \, a b c^{2} d^{2} + 5 \, a^{2} c d^{3}\right )} x^{2} +{\left (b^{2} c^{4} - 6 \, a b c^{3} d + 5 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt{-a c} \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 (3 \, a b c^{4} - 3 \, a^{2} c^{3} d +{\left (13 \, a b c^{2} d^{2} - 15 \, a^{2} c d^{3}\right )} x^{2} + 2 \,{\left (9 \, a b c^{3} d - 10 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{6 \,{\left ({\left (a b c^{5} d^{2} - a^{2} c^{4} d^{3}\right )} x^{3} + 2 \,{\left (a b c^{6} d - a^{2} c^{5} d^{2}\right )} x^{2} +{\left (a b c^{7} - a^{2} c^{6} d\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*((b^2*c^2*d^2 - 6*a*b*c*d^3 + 5*a^2*d^4)*x^3 + 2*(b^2*c^3*d - 6*a*b*c^2*d^2 + 5*a^2*c*d^3)*x^2 + (b^
2*c^4 - 6*a*b*c^3*d + 5*a^2*c^2*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 + (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*(3*a*b*c^4 - 3*
a^2*c^3*d + (13*a*b*c^2*d^2 - 15*a^2*c*d^3)*x^2 + 2*(9*a*b*c^3*d - 10*a^2*c^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x +
c))/((a*b*c^5*d^2 - a^2*c^4*d^3)*x^3 + 2*(a*b*c^6*d - a^2*c^5*d^2)*x^2 + (a*b*c^7 - a^2*c^6*d)*x), 1/6*(3*((b
^2*c^2*d^2 - 6*a*b*c*d^3 + 5*a^2*d^4)*x^3 + 2*(b^2*c^3*d - 6*a*b*c^2*d^2 + 5*a^2*c*d^3)*x^2 + (b^2*c^4 - 6*a*b
*c^3*d + 5*a^2*c^2*d^2)*x)*sqrt(-a*c)*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*(3*a*b*c^4 - 3*a^2*c^3*d + (13*a*b*c^2*d^2 - 15*a^2*c*d
^3)*x^2 + 2*(9*a*b*c^3*d - 10*a^2*c^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/((a*b*c^5*d^2 - a^2*c^4*d^3)*x^3 +
2*(a*b*c^6*d - a^2*c^5*d^2)*x^2 + (a*b*c^7 - a^2*c^6*d)*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**2/(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError