3.746 \(\int \frac{1}{x^2 \sqrt{a+b x} (c+d x)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.07997, 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 = {103, 152, 12, 93, 208} \[ \frac{(3 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{3/2} c^{5/2}}-\frac{d \sqrt{a+b x} (b c-3 a d)}{a c^2 \sqrt{c+d x} (b c-a d)}-\frac{\sqrt{a+b x}}{a c x \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[a + b*x]*(c + d*x)^(3/2)),x]

[Out]

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

Rule 103

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] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[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^2 \sqrt{a+b x} (c+d x)^{3/2}} \, dx &=-\frac{\sqrt{a+b x}}{a c x \sqrt{c+d x}}-\frac{\int \frac{\frac{1}{2} (b c+3 a d)+b d x}{x \sqrt{a+b x} (c+d x)^{3/2}} \, dx}{a c}\\ &=-\frac{d (b c-3 a d) \sqrt{a+b x}}{a c^2 (b c-a d) \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{a c x \sqrt{c+d x}}+\frac{2 \int -\frac{(b c-a d) (b c+3 a d)}{4 x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{a c^2 (b c-a d)}\\ &=-\frac{d (b c-3 a d) \sqrt{a+b x}}{a c^2 (b c-a d) \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{a c x \sqrt{c+d x}}-\frac{(b c+3 a d) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 a c^2}\\ &=-\frac{d (b c-3 a d) \sqrt{a+b x}}{a c^2 (b c-a d) \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{a c x \sqrt{c+d x}}-\frac{(b c+3 a d) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{a c^2}\\ &=-\frac{d (b c-3 a d) \sqrt{a+b x}}{a c^2 (b c-a d) \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{a c x \sqrt{c+d x}}+\frac{(b c+3 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{3/2} c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.118086, size = 126, normalized size = 1.02 \[ \frac{\left (-3 a^2 d^2+2 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{\sqrt{a} \sqrt{c} \sqrt{a+b x} (a d (c+3 d x)-b c (c+d x))}{x \sqrt{c+d x}}}{a^{3/2} c^{5/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[a + b*x]*(c + d*x)^(3/2)),x]

[Out]

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

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Maple [B]  time = 0.027, size = 441, normalized size = 3.6 \begin{align*}{\frac{1}{2\,{c}^{2}ax \left ( ad-bc \right ) }\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}^{3}-2\,\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}-\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}^{2}d+3\,\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}-2\,\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-\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{b}^{2}{c}^{3}-6\,xa{d}^{2}\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,xbcd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }-2\,acd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,b{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(1/x^2/(d*x+c)^(3/2)/(b*x+a)^(1/2),x)

[Out]

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

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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{3}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 4.55144, size = 1050, normalized size = 8.47 \begin{align*} \left [\frac{{\left ({\left (b^{2} c^{2} d + 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{2} +{\left (b^{2} c^{3} + 2 \, a b c^{2} d - 3 \, a^{2} c 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 (a b c^{3} - a^{2} c^{2} d +{\left (a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{4 \,{\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{2} +{\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x\right )}}, -\frac{{\left ({\left (b^{2} c^{2} d + 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{2} +{\left (b^{2} c^{3} + 2 \, a b c^{2} d - 3 \, a^{2} c 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 (a b c^{3} - a^{2} c^{2} d +{\left (a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{2} +{\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \sqrt{a + b x} \left (c + d x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**2*sqrt(a + b*x)*(c + d*x)**(3/2)), x)

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

[Out]

Exception raised: TypeError