3.7.39 \(\int \frac {(a+b x)^{3/2}}{x^3 (c+d x)^{3/2}} \, dx\) [639]

Optimal. Leaf size=175 \[ \frac {3 (b c-5 a d) (b c-a d) \sqrt {a+b x}}{4 a c^3 \sqrt {c+d x}}-\frac {(b c-5 a d) (a+b x)^{3/2}}{4 a c^2 x \sqrt {c+d x}}-\frac {(a+b x)^{5/2}}{2 a c x^2 \sqrt {c+d x}}-\frac {3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a} c^{7/2}} \]

[Out]

-3/4*(-5*a*d+b*c)*(-a*d+b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/c^(7/2)/a^(1/2)-1/4*(-5*a*d+
b*c)*(b*x+a)^(3/2)/a/c^2/x/(d*x+c)^(1/2)-1/2*(b*x+a)^(5/2)/a/c/x^2/(d*x+c)^(1/2)+3/4*(-5*a*d+b*c)*(-a*d+b*c)*(
b*x+a)^(1/2)/a/c^3/(d*x+c)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {98, 96, 95, 214} \begin {gather*} -\frac {3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a} c^{7/2}}+\frac {3 \sqrt {a+b x} (b c-5 a d) (b c-a d)}{4 a c^3 \sqrt {c+d x}}-\frac {(a+b x)^{3/2} (b c-5 a d)}{4 a c^2 x \sqrt {c+d x}}-\frac {(a+b x)^{5/2}}{2 a c x^2 \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 95

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 96

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[n*((d*e - c*f)/((m + 1)*(b*e - a*f
))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 98

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps

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

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Mathematica [A]
time = 1.55, size = 174, normalized size = 0.99 \begin {gather*} \frac {\sqrt {a+b x} \left (-b c x (5 c+13 d x)+a \left (-2 c^2+5 c d x+15 d^2 x^2\right )\right )}{4 c^3 x^2 \sqrt {c+d x}}-\frac {3 \sqrt {\frac {b}{d}} \sqrt {d} \left (b^2 c^2-6 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \left (-b x+\sqrt {\frac {b}{d}} \sqrt {a+b x} \sqrt {c+d x}\right )}{\sqrt {a} \sqrt {b} \sqrt {c}}\right )}{4 \sqrt {a} \sqrt {b} c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(-(b*c*x*(5*c + 13*d*x)) + a*(-2*c^2 + 5*c*d*x + 15*d^2*x^2)))/(4*c^3*x^2*Sqrt[c + d*x]) - (3*S
qrt[b/d]*Sqrt[d]*(b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[d]*(-(b*x) + Sqrt[b/d]*Sqrt[a + b*x]*Sqrt[c +
d*x]))/(Sqrt[a]*Sqrt[b]*Sqrt[c])])/(4*Sqrt[a]*Sqrt[b]*c^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(463\) vs. \(2(143)=286\).
time = 0.07, size = 464, normalized size = 2.65

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \left (15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{2} d^{3} x^{3}-18 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a b c \,d^{2} x^{3}+3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) b^{2} c^{2} d \,x^{3}+15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{2} c \,d^{2} x^{2}-18 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a b \,c^{2} d \,x^{2}+3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) b^{2} c^{3} x^{2}-30 a \,d^{2} x^{2} \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+26 b c d \,x^{2} \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}-10 a c d x \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+10 b \,c^{2} x \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+4 a \,c^{2} \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\right )}{8 c^{3} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, x^{2} \sqrt {a c}\, \sqrt {d x +c}}\)	\(464\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/x^3/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/8*(b*x+a)^(1/2)*(15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*a^2*d^3*x^3-18*ln((a*d*
x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*a*b*c*d^2*x^3+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)
*(b*x+a))^(1/2)+2*a*c)/x)*b^2*c^2*d*x^3+15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*a^2
*c*d^2*x^2-18*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*a*b*c^2*d*x^2+3*ln((a*d*x+b*c*x+
2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*b^2*c^3*x^2-30*a*d^2*x^2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2
6*b*c*d*x^2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)-10*a*c*d*x*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+10*b*c^2*x*(a*c
)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+4*a*c^2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2))/c^3/((d*x+c)*(b*x+a))^(1/2)/x^2/(
a*c)^(1/2)/(d*x+c)^(1/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

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Fricas [A]
time = 1.86, size = 472, normalized size = 2.70 \begin {gather*} \left [\frac {3 \, {\left ({\left (b^{2} c^{2} d - 6 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} x^{3} + {\left (b^{2} c^{3} - 6 \, a b c^{2} d + 5 \, 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 (13 \, a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2} + 5 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (a c^{4} d x^{3} + a c^{5} x^{2}\right )}}, \frac {3 \, {\left ({\left (b^{2} c^{2} d - 6 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} x^{3} + {\left (b^{2} c^{3} - 6 \, a b c^{2} d + 5 \, 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 (13 \, a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2} + 5 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (a c^{4} d x^{3} + a c^{5} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*((b^2*c^2*d - 6*a*b*c*d^2 + 5*a^2*d^3)*x^3 + (b^2*c^3 - 6*a*b*c^2*d + 5*a^2*c*d^2)*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*c^3 + (13*a*b*c^2*d - 15*a^2*c*d^2)*x^2 + 5*(a*b*c^3 - a^2*c^
2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^4*d*x^3 + a*c^5*x^2), 1/8*(3*((b^2*c^2*d - 6*a*b*c*d^2 + 5*a^2*d^3)*
x^3 + (b^2*c^3 - 6*a*b*c^2*d + 5*a^2*c*d^2)*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*c^3 + (13*a*b*c^2*d - 15*a
^2*c*d^2)*x^2 + 5*(a*b*c^3 - a^2*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^4*d*x^3 + a*c^5*x^2)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1096 vs. \(2 (143) = 286\).
time = 4.44, size = 1096, normalized size = 6.26 \begin {gather*} -\frac {2 \, {\left (b^{3} c d - a b^{2} d^{2}\right )} \sqrt {b x + a}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} c^{3} {\left | b \right |}} - \frac {3 \, {\left (\sqrt {b d} b^{4} c^{2} - 6 \, \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} b c^{3} {\left | b \right |}} - \frac {5 \, \sqrt {b d} b^{10} c^{5} - 27 \, \sqrt {b d} a b^{9} c^{4} d + 58 \, \sqrt {b d} a^{2} b^{8} c^{3} d^{2} - 62 \, \sqrt {b d} a^{3} b^{7} c^{2} d^{3} + 33 \, \sqrt {b d} a^{4} b^{6} c d^{4} - 7 \, \sqrt {b d} a^{5} b^{5} d^{5} - 15 \, \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} + 40 \, \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 - 14 \, \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} - 32 \, \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} + 15 \, \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} - 11 \, \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 + \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} - 5 \, \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} - 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 )}^{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} c^{3} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b^3*c*d - a*b^2*d^2)*sqrt(b*x + a)/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*c^3*abs(b)) - 3/4*(sqrt(b*d)*b^4*c
^2 - 6*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)*b*c^3*abs(b)) - 1/2*(5*sqrt(b*d)*b
^10*c^5 - 27*sqrt(b*d)*a*b^9*c^4*d + 58*sqrt(b*d)*a^2*b^8*c^3*d^2 - 62*sqrt(b*d)*a^3*b^7*c^2*d^3 + 33*sqrt(b*d
)*a^4*b^6*c*d^4 - 7*sqrt(b*d)*a^5*b^5*d^5 - 15*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 + 40*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 - 14*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 - 32*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 + 15*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 - 11*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 + 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 - 5*
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 - 2*sqrt(b*d)*(sqrt(b*d)*s
qrt(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*c^3*abs(b))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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