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

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

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Rubi [A]  time = 0.08, antiderivative size = 124, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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 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 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*}

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Mathematica [A]  time = 0.12, size = 126, normalized size = 1.02 \begin {gather*} \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)} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.24, size = 146, normalized size = 1.18 \begin {gather*} \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 {\sqrt {a+b x} \left (3 a^2 d^2-\frac {2 a c d^2 (a+b x)}{c+d x}-2 a b c d+b^2 c^2\right )}{a c^2 \sqrt {c+d x} (a d-b c) \left (a-\frac {c (a+b x)}{c+d x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a + b*x]*(b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2 - (2*a*c*d^2*(a + b*x))/(c + d*x)))/(a*c^2*(-(b*c) + a*d)*Sq
rt[c + d*x]*(a - (c*(a + b*x))/(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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fricas [B]  time = 2.35, size = 492, normalized size = 3.97 \begin {gather*} \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 {gather*}

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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giac [B]  time = 3.99, size = 476, normalized size = 3.84 \begin {gather*} \frac {2 \, \sqrt {b x + a} b^{2} d^{2}}{{\left (b c^{3} {\left | b \right |} - a c^{2} d {\left | b \right |}\right )} \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {{\left (\sqrt {b d} b^{3} c + 3 \, \sqrt {b d} a b^{2} d\right )} \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} a b c^{2} {\left | b \right |}} - \frac {2 \, {\left (\sqrt {b d} b^{5} c^{2} - 2 \, \sqrt {b d} a b^{4} c d + \sqrt {b d} a^{2} b^{3} d^{2} - \sqrt {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} b^{3} c - \sqrt {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} a b^{2} d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} 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 )}^{4}\right )} a c^{2} {\left | b \right |}} \end {gather*}

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]

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

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maple [B]  time = 0.03, size = 441, normalized size = 3.56 \begin {gather*} \frac {\sqrt {b x +a}\, \left (3 a^{2} d^{3} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-2 a b c \,d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-b^{2} c^{2} d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+3 a^{2} c \,d^{2} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-2 a b \,c^{2} d x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-b^{2} c^{3} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-6 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,d^{2} x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b c d x -2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a c d +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b \,c^{2}\right )}{2 \left (a d -b c \right ) \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {d x +c}\, a \,c^{2} x} \end {gather*}

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{2}} x^{2}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \sqrt {a + b x} \left (c + d x\right )^{\frac {3}{2}}}\, dx \end {gather*}

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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