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

Optimal. Leaf size=171 \[ \frac{\left (-15 a^2 d^2+6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{3/2} c^{7/2}}-\frac{d \sqrt{a+b x} (b c-15 a d)}{4 a c^3 \sqrt{c+d x}}-\frac{\sqrt{a+b x} (b c-5 a d)}{4 a c^2 x \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{2 c x^2 \sqrt{c+d x}} \]

[Out]

-(d*(b*c - 15*a*d)*Sqrt[a + b*x])/(4*a*c^3*Sqrt[c + d*x]) - Sqrt[a + b*x]/(2*c*x^2*Sqrt[c + d*x]) - ((b*c - 5*
a*d)*Sqrt[a + b*x])/(4*a*c^2*x*Sqrt[c + d*x]) + ((b^2*c^2 + 6*a*b*c*d - 15*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a +
b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(3/2)*c^(7/2))

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Rubi [A]  time = 0.132743, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {99, 151, 152, 12, 93, 208} \[ \frac{\left (-15 a^2 d^2+6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{3/2} c^{7/2}}-\frac{d \sqrt{a+b x} (b c-15 a d)}{4 a c^3 \sqrt{c+d x}}-\frac{\sqrt{a+b x} (b c-5 a d)}{4 a c^2 x \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{2 c x^2 \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(b*c - 15*a*d)*Sqrt[a + b*x])/(4*a*c^3*Sqrt[c + d*x]) - Sqrt[a + b*x]/(2*c*x^2*Sqrt[c + d*x]) - ((b*c - 5*
a*d)*Sqrt[a + b*x])/(4*a*c^2*x*Sqrt[c + d*x]) + ((b^2*c^2 + 6*a*b*c*d - 15*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a +
b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(3/2)*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 151

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] && IntegerQ[m]

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^3 (c+d x)^{3/2}} \, dx &=-\frac{\sqrt{a+b x}}{2 c x^2 \sqrt{c+d x}}+\frac{\int \frac{\frac{1}{2} (b c-5 a d)-2 b d x}{x^2 \sqrt{a+b x} (c+d x)^{3/2}} \, dx}{2 c}\\ &=-\frac{\sqrt{a+b x}}{2 c x^2 \sqrt{c+d x}}-\frac{(b c-5 a d) \sqrt{a+b x}}{4 a c^2 x \sqrt{c+d x}}-\frac{\int \frac{\frac{1}{4} \left (b^2 c^2+6 a b c d-15 a^2 d^2\right )+\frac{1}{2} b d (b c-5 a d) x}{x \sqrt{a+b x} (c+d x)^{3/2}} \, dx}{2 a c^2}\\ &=-\frac{d (b c-15 a d) \sqrt{a+b x}}{4 a c^3 \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{2 c x^2 \sqrt{c+d x}}-\frac{(b c-5 a d) \sqrt{a+b x}}{4 a c^2 x \sqrt{c+d x}}+\frac{\int -\frac{(b c-a d) \left (b^2 c^2+6 a b c d-15 a^2 d^2\right )}{8 x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{a c^3 (b c-a d)}\\ &=-\frac{d (b c-15 a d) \sqrt{a+b x}}{4 a c^3 \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{2 c x^2 \sqrt{c+d x}}-\frac{(b c-5 a d) \sqrt{a+b x}}{4 a c^2 x \sqrt{c+d x}}-\frac{\left (b^2 c^2+6 a b c d-15 a^2 d^2\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 a c^3}\\ &=-\frac{d (b c-15 a d) \sqrt{a+b x}}{4 a c^3 \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{2 c x^2 \sqrt{c+d x}}-\frac{(b c-5 a d) \sqrt{a+b x}}{4 a c^2 x \sqrt{c+d x}}-\frac{\left (b^2 c^2+6 a b c d-15 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{4 a c^3}\\ &=-\frac{d (b c-15 a d) \sqrt{a+b x}}{4 a c^3 \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{2 c x^2 \sqrt{c+d x}}-\frac{(b c-5 a d) \sqrt{a+b x}}{4 a c^2 x \sqrt{c+d x}}+\frac{\left (b^2 c^2+6 a b c d-15 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{3/2} c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0973808, size = 130, normalized size = 0.76 \[ \frac{\left (-15 a^2 d^2+6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{3/2} c^{7/2}}+\frac{\sqrt{a+b x} \left (a \left (-2 c^2+5 c d x+15 d^2 x^2\right )-b c x (c+d x)\right )}{4 a c^3 x^2 \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(-(b*c*x*(c + d*x)) + a*(-2*c^2 + 5*c*d*x + 15*d^2*x^2)))/(4*a*c^3*x^2*Sqrt[c + d*x]) + ((b^2*c
^2 + 6*a*b*c*d - 15*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(3/2)*c^(7/2))

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Maple [B]  time = 0.025, size = 467, normalized size = 2.7 \begin{align*} -{\frac{1}{8\,a{c}^{3}{x}^{2}}\sqrt{bx+a} \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}^{3}-6\,\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}^{2}-\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac \right ) } \right ){x}^{3}{b}^{2}{c}^{2}d+15\,\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}^{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}ab{c}^{2}d-\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac \right ) } \right ){x}^{2}{b}^{2}{c}^{3}-30\,{x}^{2}a{d}^{2}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,{x}^{2}bcd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }-10\,xacd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,xb{c}^{2}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+4\,a{c}^{2}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) } \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)/x^3/(d*x+c)^(3/2),x)

[Out]

-1/8*(b*x+a)^(1/2)/a/c^3*(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^3-6*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^3*a*b*c*d^2-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+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^2*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^2*a*b*c^2*d-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^3-30*x^2*a*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2
*x^2*b*c*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-10*x*a*c*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*x*b*c^2*(a*c)^
(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*a*c^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x^2/(a*c)^(1/2)/((b*x+a)*(d*x+c))^(
1/2)/(d*x+c)^(1/2)

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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^3/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 8.46069, size = 1034, normalized size = 6.05 \begin{align*} \left [-\frac{{\left ({\left (b^{2} c^{2} d + 6 \, a b c d^{2} - 15 \, a^{2} d^{3}\right )} x^{3} +{\left (b^{2} c^{3} + 6 \, a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2}\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 (2 \, a^{2} c^{3} +{\left (a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2} +{\left (a b c^{3} - 5 \, a^{2} c^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \,{\left (a^{2} c^{4} d x^{3} + a^{2} c^{5} x^{2}\right )}}, -\frac{{\left ({\left (b^{2} c^{2} d + 6 \, a b c d^{2} - 15 \, a^{2} d^{3}\right )} x^{3} +{\left (b^{2} c^{3} + 6 \, a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2}\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 (2 \, a^{2} c^{3} +{\left (a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2} +{\left (a b c^{3} - 5 \, a^{2} c^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \,{\left (a^{2} c^{4} d x^{3} + a^{2} c^{5} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)/x^3/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(((b^2*c^2*d + 6*a*b*c*d^2 - 15*a^2*d^3)*x^3 + (b^2*c^3 + 6*a*b*c^2*d - 15*a^2*c*d^2)*x^2)*sqrt(a*c)*lo
g((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*(2*a^2*c^3 + (a*b*c^2*d - 15*a^2*c*d^2)*x^2 + (a*b*c^3 - 5*a^2*c^2*
d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^4*d*x^3 + a^2*c^5*x^2), -1/8*(((b^2*c^2*d + 6*a*b*c*d^2 - 15*a^2*d^3
)*x^3 + (b^2*c^3 + 6*a*b*c^2*d - 15*a^2*c*d^2)*x^2)*sqrt(-a*c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*s
qrt(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*(2*a^2*c^3 + (a*b*c^2*d - 15*a
^2*c*d^2)*x^2 + (a*b*c^3 - 5*a^2*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^4*d*x^3 + a^2*c^5*x^2)]

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

[Out]

Exception raised: TypeError