3.753 \(\int \frac{1}{x \sqrt{a+b x} (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=124 \[ -\frac{2 d \sqrt{a+b x} (5 b c-3 a d)}{3 c^2 \sqrt{c+d x} (b c-a d)^2}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{5/2}}-\frac{2 d \sqrt{a+b x}}{3 c (c+d x)^{3/2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0862209, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {104, 152, 12, 93, 208} \[ -\frac{2 d \sqrt{a+b x} (5 b c-3 a d)}{3 c^2 \sqrt{c+d x} (b c-a d)^2}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{5/2}}-\frac{2 d \sqrt{a+b x}}{3 c (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[a + b*x]*(c + d*x)^(5/2)),x]

[Out]

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

Rule 104

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[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*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegersQ[2*m, 2*n, 2*p]

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

Mathematica [A]  time = 0.166861, size = 125, normalized size = 1.01 \[ \frac{2 \left (\frac{3 (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{3/2}}+\frac{d \sqrt{a+b x} (b c (6 c+5 d x)-a d (4 c+3 d x))}{c (c+d x)^{3/2} (b c-a d)}\right )}{3 c (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[a + b*x]*(c + d*x)^(5/2)),x]

[Out]

(2*((d*Sqrt[a + b*x]*(-(a*d*(4*c + 3*d*x)) + b*c*(6*c + 5*d*x)))/(c*(b*c - a*d)*(c + d*x)^(3/2)) + (3*(b*c - a
*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(Sqrt[a]*c^(3/2))))/(3*c*(-(b*c) + a*d))

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Maple [B]  time = 0.028, size = 586, normalized size = 4.7 \begin{align*} -{\frac{1}{3\, \left ( ad-bc \right ) ^{2}{c}^{2}}\sqrt{bx+a} \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}^{4}-6\,\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}^{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}{b}^{2}{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{a}^{2}c{d}^{3}-12\,\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}^{2}+6\,\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}^{3}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}^{2}-6\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) ab{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 ){b}^{2}{c}^{4}-6\,xa{d}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+10\,xbc{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}-8\,ac{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+12\,b{c}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac} \right ){\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(1/x/(d*x+c)^(5/2)/(b*x+a)^(1/2),x)

[Out]

-1/3*(b*x+a)^(1/2)/c^2*(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^4-6*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*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^2*b^2*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*a^2*c*d^3-12*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^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*b^2*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)*a^2*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)*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)*b^2*c^4-6*x*a*d^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(
1/2)+10*x*b*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-8*a*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+12*b*c^2*d
*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2)/(a*d-b*c)^2/((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x + a}{\left (d x + c\right )}^{\frac{5}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 6.27799, size = 1432, normalized size = 11.55 \begin{align*} \left [\frac{3 \,{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} +{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \,{\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\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 (6 \, a b c^{3} d - 4 \, a^{2} c^{2} d^{2} +{\left (5 \, a b c^{2} d^{2} - 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{6 \,{\left (a b^{2} c^{7} - 2 \, a^{2} b c^{6} d + a^{3} c^{5} d^{2} +{\left (a b^{2} c^{5} d^{2} - 2 \, a^{2} b c^{4} d^{3} + a^{3} c^{3} d^{4}\right )} x^{2} + 2 \,{\left (a b^{2} c^{6} d - 2 \, a^{2} b c^{5} d^{2} + a^{3} c^{4} d^{3}\right )} x\right )}}, \frac{3 \,{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} +{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \,{\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\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 (6 \, a b c^{3} d - 4 \, a^{2} c^{2} d^{2} +{\left (5 \, a b c^{2} d^{2} - 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{3 \,{\left (a b^{2} c^{7} - 2 \, a^{2} b c^{6} d + a^{3} c^{5} d^{2} +{\left (a b^{2} c^{5} d^{2} - 2 \, a^{2} b c^{4} d^{3} + a^{3} c^{3} d^{4}\right )} x^{2} + 2 \,{\left (a b^{2} c^{6} d - 2 \, a^{2} b c^{5} d^{2} + a^{3} c^{4} d^{3}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^2 + 2*(b^2*c^3*d - 2*a*
b*c^2*d^2 + a^2*c*d^3)*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*(6*a*b*c^3*d - 4*a^2*c^2*d^2
+ (5*a*b*c^2*d^2 - 3*a^2*c*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^2*c^7 - 2*a^2*b*c^6*d + a^3*c^5*d^2 + (a*
b^2*c^5*d^2 - 2*a^2*b*c^4*d^3 + a^3*c^3*d^4)*x^2 + 2*(a*b^2*c^6*d - 2*a^2*b*c^5*d^2 + a^3*c^4*d^3)*x), 1/3*(3*
(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^2 + 2*(b^2*c^3*d - 2*a*b*c^2*d^
2 + a^2*c*d^3)*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*(6*a*b*c^3*d - 4*a^2*c^2*d^2 + (5*a*b*c^2*d^2 - 3*a^2*c*d^3)*x)*
sqrt(b*x + a)*sqrt(d*x + c))/(a*b^2*c^7 - 2*a^2*b*c^6*d + a^3*c^5*d^2 + (a*b^2*c^5*d^2 - 2*a^2*b*c^4*d^3 + a^3
*c^3*d^4)*x^2 + 2*(a*b^2*c^6*d - 2*a^2*b*c^5*d^2 + a^3*c^4*d^3)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \sqrt{a + b x} \left (c + d x\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.33405, size = 358, normalized size = 2.89 \begin{align*} \frac{\sqrt{b x + a}{\left (\frac{{\left (5 \, b^{4} c^{3} d^{3}{\left | b \right |} - 3 \, a b^{3} c^{2} d^{4}{\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 (2 \, b^{5} c^{4} d^{2}{\left | b \right |} - 3 \, a b^{4} c^{3} d^{3}{\left | b \right |} + a^{2} b^{3} c^{2} d^{4}{\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} 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(1/x/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/12*sqrt(b*x + a)*((5*b^4*c^3*d^3*abs(b) - 3*a*b^3*c^2*d^4*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*(2*b^5*c^4*d^2*abs(b) - 3*a*b^4*c^3*d^3*abs(b) + a^2*b^3*c^2*d^4*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)*b*arctan(-1/2*(b^2*c + a*b*d - (sq
rt(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)
)