3.7.52 \(\int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^3} \, dx\) [652]

Optimal. Leaf size=212 \[ \frac {b (11 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{4 c}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{4 c x}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}-\frac {\sqrt {a} \left (15 b^2 c^2+10 a b c d-a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 c^{3/2}}+\frac {b^{3/2} (b c+5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {d}} \]

[Out]

-1/4*(-a^2*d^2+10*a*b*c*d+15*b^2*c^2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))*a^(1/2)/c^(3/2)+b^(
3/2)*(5*a*d+b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(1/2)-1/4*(a*d+5*b*c)*(b*x+a)^(3/2)*(d
*x+c)^(1/2)/c/x-1/2*(b*x+a)^(5/2)*(d*x+c)^(1/2)/x^2+1/4*b*(a*d+11*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c

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Rubi [A]
time = 0.13, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {99, 154, 159, 163, 65, 223, 212, 95, 214} \begin {gather*} -\frac {\sqrt {a} \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 c^{3/2}}+\frac {b^{3/2} (5 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {d}}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{4 c x}+\frac {b \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(11*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*c) - ((5*b*c + a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(4*c*x) -
((a + b*x)^(5/2)*Sqrt[c + d*x])/(2*x^2) - (Sqrt[a]*(15*b^2*c^2 + 10*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[
a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*c^(3/2)) + (b^(3/2)*(b*c + 5*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt
[b]*Sqrt[c + d*x])])/Sqrt[d]

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 99

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

Rule 154

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] && ILtQ[m, -1] && GtQ[n, 0]

Rule 159

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 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} \sqrt {c+d x}}{x^3} \, dx &=-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}+\frac {1}{2} \int \frac {(a+b x)^{3/2} \left (\frac {1}{2} (5 b c+a d)+3 b d x\right )}{x^2 \sqrt {c+d x}} \, dx\\ &=-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{4 c x}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}+\frac {\int \frac {\sqrt {a+b x} \left (\frac {1}{4} \left (15 b^2 c^2+10 a b c d-a^2 d^2\right )+\frac {1}{2} b d (11 b c+a d) x\right )}{x \sqrt {c+d x}} \, dx}{2 c}\\ &=\frac {b (11 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{4 c}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{4 c x}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}+\frac {\int \frac {\frac {1}{4} a d \left (15 b^2 c^2+10 a b c d-a^2 d^2\right )+b^2 c d (b c+5 a d) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 c d}\\ &=\frac {b (11 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{4 c}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{4 c x}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}+\frac {1}{2} \left (b^2 (b c+5 a d)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx+\frac {\left (a \left (15 b^2 c^2+10 a b c d-a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 c}\\ &=\frac {b (11 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{4 c}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{4 c x}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}+(b (b c+5 a d)) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )+\frac {\left (a \left (15 b^2 c^2+10 a b c d-a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{4 c}\\ &=\frac {b (11 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{4 c}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{4 c x}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}-\frac {\sqrt {a} \left (15 b^2 c^2+10 a b c d-a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 c^{3/2}}+(b (b c+5 a d)) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )\\ &=\frac {b (11 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{4 c}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{4 c x}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{2 x^2}-\frac {\sqrt {a} \left (15 b^2 c^2+10 a b c d-a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 c^{3/2}}+\frac {b^{3/2} (b c+5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {d}}\\ \end {align*}

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Mathematica [A]
time = 0.79, size = 176, normalized size = 0.83 \begin {gather*} \frac {\sqrt {a} \left (-15 b^2 c^2-10 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )+\sqrt {c} \left (-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (9 a b c x-4 b^2 c x^2+a^2 (2 c+d x)\right )}{x^2}+\frac {4 b^{3/2} c (b c+5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{\sqrt {d}}\right )}{4 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*(-15*b^2*c^2 - 10*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])] + Sqrt[
c]*(-((Sqrt[a + b*x]*Sqrt[c + d*x]*(9*a*b*c*x - 4*b^2*c*x^2 + a^2*(2*c + d*x)))/x^2) + (4*b^(3/2)*c*(b*c + 5*a
*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/Sqrt[d]))/(4*c^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(440\) vs. \(2(168)=336\).
time = 0.07, size = 441, normalized size = 2.08

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (20 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c d \,x^{2} \sqrt {a c}+4 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} x^{2} \sqrt {a c}+\ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{3} d^{2} x^{2} \sqrt {b d}-10 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{2} b c d \,x^{2} \sqrt {b d}-15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a \,b^{2} c^{2} x^{2} \sqrt {b d}+8 b^{2} c \,x^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}\, \sqrt {a c}-2 a^{2} d x \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}\, \sqrt {a c}-18 a b c x \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}\, \sqrt {a c}-4 a^{2} c \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}\, \sqrt {a c}\right )}{8 c \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, x^{2} \sqrt {b d}\, \sqrt {a c}}\)	\(441\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(d*x+c)^(1/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c*(20*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*b^2*c*d*x^2*(a*c)^(1/2)+4*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*x^2*(a*c)^(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^3*d^2*x^2*(b*d)^(1/2)-
10*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*x^2*(b*d)^(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*b^2*c^2*x^2*(b*d)^(1/2)+8*b^2*c*x^2*((d*x+c)*(b*x+a))^(1/2
)*(b*d)^(1/2)*(a*c)^(1/2)-2*a^2*d*x*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-18*a*b*c*x*((d*x+c)*(b*x+a
))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-4*a^2*c*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/((d*x+c)*(b*x+a))^(1
/2)/x^2/(b*d)^(1/2)/(a*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)^(5/2)*(d*x+c)^(1/2)/x^3,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 = 4.02, size = 1098, normalized size = 5.18 \begin {gather*} \left [\frac {4 \, {\left (b^{2} c^{2} + 5 \, a b c d\right )} x^{2} \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 (15 \, b^{2} c^{2} + 10 \, a b c d - a^{2} d^{2}\right )} x^{2} \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 (4 \, b^{2} c x^{2} - 2 \, a^{2} c - {\left (9 \, a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, c x^{2}}, -\frac {8 \, {\left (b^{2} c^{2} + 5 \, a b c d\right )} x^{2} \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 (15 \, b^{2} c^{2} + 10 \, a b c d - a^{2} d^{2}\right )} x^{2} \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 (4 \, b^{2} c x^{2} - 2 \, a^{2} c - {\left (9 \, a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, c x^{2}}, \frac {{\left (15 \, b^{2} c^{2} + 10 \, a b c d - a^{2} d^{2}\right )} x^{2} \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} + 5 \, a b c d\right )} x^{2} \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 (4 \, b^{2} c x^{2} - 2 \, a^{2} c - {\left (9 \, a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, c x^{2}}, \frac {{\left (15 \, b^{2} c^{2} + 10 \, a b c d - a^{2} d^{2}\right )} x^{2} \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 ) - 4 \, {\left (b^{2} c^{2} + 5 \, a b c d\right )} x^{2} \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 (4 \, b^{2} c x^{2} - 2 \, a^{2} c - {\left (9 \, a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, c x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(1/2)/x^3,x, algorithm="fricas")

[Out]

[1/16*(4*(b^2*c^2 + 5*a*b*c*d)*x^2*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) - (15*b^2*c^2 + 10*a*b*c*d -
a^2*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^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*(4*b^2*c*x^2 - 2*a^2*c - (9*a*b*c +
a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c*x^2), -1/16*(8*(b^2*c^2 + 5*a*b*c*d)*x^2*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)) + (15*b^2*c^
2 + 10*a*b*c*d - a^2*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^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*(4*b^2*c*x^2 - 2*a^
2*c - (9*a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c*x^2), 1/8*((15*b^2*c^2 + 10*a*b*c*d - a^2*d^2)*x^2*
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 + 5*a*b*c*d)*x^2*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*(4*b^2*c*x^2
- 2*a^2*c - (9*a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c*x^2), 1/8*((15*b^2*c^2 + 10*a*b*c*d - a^2*d^2
)*x^2*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)) - 4*(b^2*c^2 + 5*a*b*c*d)*x^2*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*(4*b^2*c*x^2 - 2*a^2*c - (9*a*b*c + a^2*
d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c*x^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{\frac {5}{2}} \sqrt {c + d x}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(d*x+c)**(1/2)/x**3,x)

[Out]

Integral((a + b*x)**(5/2)*sqrt(c + d*x)/x**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1197 vs. \(2 (168) = 336\).
time = 2.54, size = 1197, normalized size = 5.65 \begin {gather*} \frac {4 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} b {\left | b \right |} - \frac {2 \, {\left (\sqrt {b d} b^{2} c {\left | b \right |} + 5 \, \sqrt {b d} a b d {\left | b \right |}\right )} \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} - \frac {{\left (15 \, \sqrt {b d} a b^{3} c^{2} {\left | b \right |} + 10 \, \sqrt {b d} a^{2} b^{2} c d {\left | b \right |} - \sqrt {b d} a^{3} b d^{2} {\left | b \right |}\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} - \frac {2 \, {\left (9 \, \sqrt {b d} a b^{9} c^{5} {\left | b \right |} - 35 \, \sqrt {b d} a^{2} b^{8} c^{4} d {\left | b \right |} + 50 \, \sqrt {b d} a^{3} b^{7} c^{3} d^{2} {\left | b \right |} - 30 \, \sqrt {b d} a^{4} b^{6} c^{2} d^{3} {\left | b \right |} + 5 \, \sqrt {b d} a^{5} b^{5} c d^{4} {\left | b \right |} + \sqrt {b d} a^{6} b^{4} d^{5} {\left | b \right |} - 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 )}^{2} a b^{7} c^{4} {\left | b \right |} + 28 \, \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^{3} d {\left | b \right |} + 22 \, \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^{2} d^{2} {\left | b \right |} - 20 \, \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} c d^{3} {\left | b \right |} - 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 )}^{2} a^{5} b^{3} d^{4} {\left | b \right |} + 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^{3} {\left | b \right |} + 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^{2} b^{4} c^{2} d {\left | b \right |} + 29 \, \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} c d^{2} {\left | b \right |} + 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 )}^{4} a^{4} b^{2} d^{3} {\left | b \right |} - 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 )}^{6} a b^{3} c^{2} {\left | b \right |} - 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 )}^{6} a^{2} b^{2} c d {\left | b \right |} - \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^{3} b d^{2} {\left | b \right |}\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 )}^{2} c}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(1/2)/x^3,x, algorithm="giac")

[Out]

1/4*(4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*b*abs(b) - 2*(sqrt(b*d)*b^2*c*abs(b) + 5*sqrt(b*d)*a*
b*d*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/d - (15*sqrt(b*d)*a*b^3*c^2
*abs(b) + 10*sqrt(b*d)*a^2*b^2*c*d*abs(b) - sqrt(b*d)*a^3*b*d^2*abs(b))*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*(9*sqrt
(b*d)*a*b^9*c^5*abs(b) - 35*sqrt(b*d)*a^2*b^8*c^4*d*abs(b) + 50*sqrt(b*d)*a^3*b^7*c^3*d^2*abs(b) - 30*sqrt(b*d
)*a^4*b^6*c^2*d^3*abs(b) + 5*sqrt(b*d)*a^5*b^5*c*d^4*abs(b) + sqrt(b*d)*a^6*b^4*d^5*abs(b) - 27*sqrt(b*d)*(sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^7*c^4*abs(b) + 28*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^3*d*abs(b) + 22*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^2*d^2*abs(b) - 20*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr
t(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^4*c*d^3*abs(b) - 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^2*a^5*b^3*d^4*abs(b) + 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^3*abs(b) + 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^2*b^4*c^2*d*abs(b) + 29*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*c*d^2*abs(b) + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^4*b^2*d^3*
abs(b) - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^3*c^2*abs(b) - 14*s
qrt(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*c*d*abs(b) - sqrt(b*d)*(sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^3*b*d^2*abs(b))/((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))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^(5/2)*(c + d*x)^(1/2))/x^3,x)

[Out]

int(((a + b*x)^(5/2)*(c + d*x)^(1/2))/x^3, x)

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