3.7.86 \(\int \frac {(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx\) [686]

Optimal. Leaf size=164 \[ -\frac {(2 b c-3 a d) (b c-a d) \sqrt {a+b x}}{c^2 d \sqrt {c+d x}}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}}-\frac {a^{3/2} (5 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {100, 155, 163, 65, 223, 212, 95, 214} \begin {gather*} -\frac {a^{3/2} (5 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}}-\frac {\sqrt {a+b x} (2 b c-3 a d) (b c-a d)}{c^2 d \sqrt {c+d x}}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/(x^2*(c + d*x)^(3/2)),x]

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 155

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*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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]

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [A]
time = 0.40, size = 150, normalized size = 0.91 \begin {gather*} -\frac {\sqrt {a+b x} \left (2 b^2 c^2 x-4 a b c d x+a^2 d (c+3 d x)\right )}{c^2 d x \sqrt {c+d x}}+\frac {a^{3/2} (-5 b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/(x^2*(c + d*x)^(3/2)),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(501\) vs. \(2(132)=264\).
time = 0.07, size = 502, normalized size = 3.06

method	result	size
default	\(\frac {\sqrt {b x +a}\, \left (3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{3} d^{3} x^{2} \sqrt {b d}-5 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{2} b c \,d^{2} x^{2} \sqrt {b d}+2 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} d \,x^{2} \sqrt {a c}+3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) \sqrt {b d}\, a^{3} c \,d^{2} x -5 \sqrt {b d}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{2} b \,c^{2} d x +2 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, b^{3} c^{3} x -6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{2} d^{2} x +8 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a b c d x -4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{2} c^{2} x -2 a^{2} c d \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}\right )}{2 c^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}\, x \sqrt {a c}\, \sqrt {d x +c}\, d}\)	\(502\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2*(b*x+a)^(1/2)*(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)*a^3*d^3*x^2*(b*d)^(1/2)-5
*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*b*c*d^2*x^2*(b*d)^(1/2)+2*ln(1/2*(2*b*d*x
+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^3*c^2*d*x^2*(a*c)^(1/2)+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*d)^(1/2)*a^3*c*d^2*x-5*(b*d)^(1/2)*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*b*c^2*d*x+2*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b
*c)/(b*d)^(1/2))*(a*c)^(1/2)*b^3*c^3*x-6*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a^2*d^2*x+8*(b*d)^(1/
2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*b*c*d*x-4*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*b^2*c^2*x-2
*a^2*c*d*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2))/c^2/((d*x+c)*(b*x+a))^(1/2)/(b*d)^(1/2)/x/(a*c)^(1/2
)/(d*x+c)^(1/2)/d

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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)^(5/2)/x^2/(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (132) = 264\).
time = 2.40, size = 1233, normalized size = 7.52 \begin {gather*} \left [\frac {2 \, {\left (b^{2} c^{2} d x^{2} + b^{2} c^{3} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - {\left ({\left (5 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{2} + {\left (5 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x\right )} \sqrt {\frac {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^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (a^{2} c d + {\left (2 \, b^{2} c^{2} - 4 \, a b c d + 3 \, a^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, {\left (c^{2} d^{2} x^{2} + c^{3} d x\right )}}, -\frac {4 \, {\left (b^{2} c^{2} d x^{2} + b^{2} c^{3} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + {\left ({\left (5 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{2} + {\left (5 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x\right )} \sqrt {\frac {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^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (a^{2} c d + {\left (2 \, b^{2} c^{2} - 4 \, a b c d + 3 \, a^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, {\left (c^{2} d^{2} x^{2} + c^{3} d x\right )}}, \frac {{\left ({\left (5 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{2} + {\left (5 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x\right )} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) + {\left (b^{2} c^{2} d x^{2} + b^{2} c^{3} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 2 \, {\left (a^{2} c d + {\left (2 \, b^{2} c^{2} - 4 \, a b c d + 3 \, a^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (c^{2} d^{2} x^{2} + c^{3} d x\right )}}, \frac {{\left ({\left (5 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{2} + {\left (5 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x\right )} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) - 2 \, {\left (b^{2} c^{2} d x^{2} + b^{2} c^{3} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) - 2 \, {\left (a^{2} c d + {\left (2 \, b^{2} c^{2} - 4 \, a b c d + 3 \, a^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (c^{2} d^{2} x^{2} + c^{3} d x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

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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)**(5/2)/x**2/(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. 560 vs. \(2 (132) = 264\).
time = 2.31, size = 560, normalized size = 3.41 \begin {gather*} -\frac {\sqrt {b d} b^{3} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{d^{2} {\left | b \right |}} - \frac {{\left (5 \, \sqrt {b d} a^{2} b^{3} c - 3 \, \sqrt {b d} a^{3} 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} b c^{2} {\left | b \right |}} - \frac {2 \, {\left (\sqrt {b d} a^{2} b^{5} c^{2} - 2 \, \sqrt {b d} a^{3} b^{4} c d + \sqrt {b d} a^{4} 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} a^{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^{3} 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 )} c^{2} {\left | b \right |}} - \frac {2 \, {\left (b^{4} c^{2} {\left | b \right |} - 2 \, a b^{3} c d {\left | b \right |} + a^{2} b^{2} d^{2} {\left | b \right |}\right )} \sqrt {b x + a}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} b^{2} c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

-sqrt(b*d)*b^3*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(d^2*abs(b)) - (5*sqrt(b
*d)*a^2*b^3*c - 3*sqrt(b*d)*a^3*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)*b*c^2*abs(b)) - 2*(sqrt(b*d)*a^2*b^5*c^2 - 2*sqrt
(b*d)*a^3*b^4*c*d + sqrt(b*d)*a^4*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*a^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^3*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) - sq
rt(b^2*c + (b*x + a)*b*d - a*b*d))^4)*c^2*abs(b)) - 2*(b^4*c^2*abs(b) - 2*a*b^3*c*d*abs(b) + a^2*b^2*d^2*abs(b
))*sqrt(b*x + a)/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*b^2*c^2*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{5/2}}{x^2\,{\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)^(5/2)/(x^2*(c + d*x)^(3/2)),x)

[Out]

int((a + b*x)^(5/2)/(x^2*(c + d*x)^(3/2)), x)

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