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

Optimal. Leaf size=194 \[ \frac {d \sqrt {a+b x} (3 b c-5 a d) (3 a d+b c)}{4 a^2 c^3 \sqrt {c+d x} (b c-a d)}+\frac {\sqrt {a+b x} (5 a d+3 b c)}{4 a^2 c^2 x \sqrt {c+d x}}-\frac {3 \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{7/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 \sqrt {c+d x}} \]

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Rubi [A]  time = 0.16, antiderivative size = 194, normalized size of antiderivative = 1.00, 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 = {103, 151, 152, 12, 93, 208} \begin {gather*} -\frac {3 \left (5 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 )}{4 a^{5/2} c^{7/2}}+\frac {d \sqrt {a+b x} (3 b c-5 a d) (3 a d+b c)}{4 a^2 c^3 \sqrt {c+d x} (b c-a d)}+\frac {\sqrt {a+b x} (5 a d+3 b c)}{4 a^2 c^2 x \sqrt {c+d x}}-\frac {\sqrt {a+b x}}{2 a c x^2 \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.19, size = 186, normalized size = 0.96 \begin {gather*} \frac {\frac {\sqrt {a} \sqrt {c} \sqrt {a+b x} \left (a^2 d \left (2 c^2-5 c d x-15 d^2 x^2\right )+2 a b c \left (-c^2+c d x+2 d^2 x^2\right )+3 b^2 c^2 x (c+d x)\right )}{x^2 \sqrt {c+d x}}-3 \left (-5 a^3 d^3+3 a^2 b c d^2+a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{7/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.44, size = 276, normalized size = 1.42 \begin {gather*} \frac {\sqrt {a+b x} \left (15 a^4 d^3-\frac {25 a^3 c d^3 (a+b x)}{c+d x}-9 a^3 b c d^2-3 a^2 b^2 c^2 d+\frac {8 a^2 c^2 d^3 (a+b x)^2}{(c+d x)^2}+\frac {15 a^2 b c^2 d^2 (a+b x)}{c+d x}-\frac {3 b^3 c^4 (a+b x)}{c+d x}+5 a b^3 c^3-\frac {3 a b^2 c^3 d (a+b x)}{c+d x}\right )}{4 a^2 c^3 \sqrt {c+d x} (a d-b c) \left (a-\frac {c (a+b x)}{c+d x}\right )^2}-\frac {3 \left (5 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 )}{4 a^{5/2} c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(5*a*b^3*c^3 - 3*a^2*b^2*c^2*d - 9*a^3*b*c*d^2 + 15*a^4*d^3 + (8*a^2*c^2*d^3*(a + b*x)^2)/(c +
d*x)^2 - (3*b^3*c^4*(a + b*x))/(c + d*x) - (3*a*b^2*c^3*d*(a + b*x))/(c + d*x) + (15*a^2*b*c^2*d^2*(a + b*x))/
(c + d*x) - (25*a^3*c*d^3*(a + b*x))/(c + d*x)))/(4*a^2*c^3*(-(b*c) + a*d)*Sqrt[c + d*x]*(a - (c*(a + b*x))/(c
+ d*x))^2) - (3*(b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(
4*a^(5/2)*c^(7/2))

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fricas [A]  time = 3.38, size = 664, normalized size = 3.42 \begin {gather*} \left [\frac {3 \, {\left ({\left (b^{3} c^{3} d + a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - 5 \, a^{3} d^{4}\right )} x^{3} + {\left (b^{3} c^{4} + a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - 5 \, a^{3} c d^{3}\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} b c^{4} - 2 \, a^{3} c^{3} d - {\left (3 \, a b^{2} c^{3} d + 4 \, a^{2} b c^{2} d^{2} - 15 \, a^{3} c d^{3}\right )} x^{2} - {\left (3 \, a b^{2} c^{4} + 2 \, a^{2} b c^{3} d - 5 \, a^{3} c^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left ({\left (a^{3} b c^{5} d - a^{4} c^{4} d^{2}\right )} x^{3} + {\left (a^{3} b c^{6} - a^{4} c^{5} d\right )} x^{2}\right )}}, \frac {3 \, {\left ({\left (b^{3} c^{3} d + a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - 5 \, a^{3} d^{4}\right )} x^{3} + {\left (b^{3} c^{4} + a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - 5 \, a^{3} c d^{3}\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} b c^{4} - 2 \, a^{3} c^{3} d - {\left (3 \, a b^{2} c^{3} d + 4 \, a^{2} b c^{2} d^{2} - 15 \, a^{3} c d^{3}\right )} x^{2} - {\left (3 \, a b^{2} c^{4} + 2 \, a^{2} b c^{3} d - 5 \, a^{3} c^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left ({\left (a^{3} b c^{5} d - a^{4} c^{4} d^{2}\right )} x^{3} + {\left (a^{3} b c^{6} - a^{4} c^{5} d\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 8.90, size = 1105, normalized size = 5.70 \begin {gather*} -\frac {2 \, \sqrt {b x + a} b^{2} d^{3}}{{\left (b c^{4} {\left | b \right |} - a c^{3} d {\left | b \right |}\right )} \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {3 \, {\left (\sqrt {b d} b^{4} c^{2} + 2 \, \sqrt {b d} a b^{3} c d + 5 \, \sqrt {b d} a^{2} b^{2} d^{2}\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 )}{4 \, \sqrt {-a b c d} a^{2} b c^{3} {\left | b \right |}} + \frac {3 \, \sqrt {b d} b^{10} c^{5} - 5 \, \sqrt {b d} a b^{9} c^{4} d - 10 \, \sqrt {b d} a^{2} b^{8} c^{3} d^{2} + 30 \, \sqrt {b d} a^{3} b^{7} c^{2} d^{3} - 25 \, \sqrt {b d} a^{4} b^{6} c d^{4} + 7 \, \sqrt {b d} a^{5} b^{5} d^{5} - 9 \, \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^{8} c^{4} - 16 \, \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^{7} c^{3} d + 38 \, \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^{2} b^{6} c^{2} d^{2} + 8 \, \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^{3} b^{5} c d^{3} - 21 \, \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^{4} b^{4} d^{4} + 9 \, \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 )}^{4} b^{6} c^{3} + 27 \, \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 )}^{4} a b^{5} c^{2} d + 23 \, \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 )}^{4} a^{2} b^{4} c d^{2} + 21 \, \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 )}^{4} a^{3} b^{3} d^{3} - 3 \, \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 )}^{6} b^{4} c^{2} - 6 \, \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 )}^{6} a b^{3} c d - 7 \, \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 )}^{6} a^{2} b^{2} d^{2}}{2 \, {\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 )}^{2} a^{2} c^{3} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(b*x + a)*b^2*d^3/((b*c^4*abs(b) - a*c^3*d*abs(b))*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) - 3/4*(sqrt(b*d
)*b^4*c^2 + 2*sqrt(b*d)*a*b^3*c*d + 5*sqrt(b*d)*a^2*b^2*d^2)*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^2*b*c^3*abs(b)) + 1/2*(3*
sqrt(b*d)*b^10*c^5 - 5*sqrt(b*d)*a*b^9*c^4*d - 10*sqrt(b*d)*a^2*b^8*c^3*d^2 + 30*sqrt(b*d)*a^3*b^7*c^2*d^3 - 2
5*sqrt(b*d)*a^4*b^6*c*d^4 + 7*sqrt(b*d)*a^5*b^5*d^5 - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x
+ a)*b*d - a*b*d))^2*b^8*c^4 - 16*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^7*c^3*d + 38*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^6*c^2*d^2
+ 8*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^5*c*d^3 - 21*sqrt(b*d)*(
sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^4*d^4 + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x
+ a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^6*c^3 + 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^4*a*b^5*c^2*d + 23*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a
*b*d))^4*a^2*b^4*c*d^2 + 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^
3*d^3 - 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^4*c^2 - 6*sqrt(b*d)*(s
qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^3*c*d - 7*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x +
a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b^2*d^2)/((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)^2*a^2*c^3*abs(
b))

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maple [B]  time = 0.03, size = 683, normalized size = 3.52 \begin {gather*} -\frac {\sqrt {b x +a}\, \left (15 a^{3} d^{4} x^{3} \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 )-9 a^{2} b c \,d^{3} x^{3} \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 \,b^{2} c^{2} d^{2} x^{3} \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 b^{3} c^{3} d \,x^{3} \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 )+15 a^{3} c \,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 )-9 a^{2} b \,c^{2} 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 )-3 a \,b^{2} c^{3} 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 b^{3} c^{4} 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 )-30 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{3} x^{2}+8 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c \,d^{2} x^{2}+6 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2} d \,x^{2}-10 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c \,d^{2} x +4 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,c^{2} d x +6 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{3} x +4 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c^{2} d -4 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,c^{3}\right )}{8 \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^{2} c^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b*x+a)^(1/2)/a^2/c^3*(15*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^3*a^3*d^4-9*l
n((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a^2*b*c*d^3-3*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^3*a*b^2*c^2*d^2-3*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^3*b^3*c^3*d+15*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*a^3*c*d^3-9*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*a^2*b*c^2*d^2-3*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*a*b^2*c^3*d-3*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^3*c^4-30*x^2*a^2*d^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+8*x^2*a*b*c*d^2*(a*c)^(1/2)*((b*x+a)*(d*x
+c))^(1/2)+6*x^2*b^2*c^2*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-10*x*a^2*c*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1
/2)+4*x*a*b*c^2*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+6*x*b^2*c^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*a^2*c^
2*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-4*a*b*c^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/(a*d-b*c)/(a*c)^(1/2)/x
^2/((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^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(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^3), x)

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

[Out]

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

[Out]

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

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