3.7.85 \(\int \frac {(a+b x)^{5/2}}{x (c+d x)^{3/2}} \, dx\) [685]

Optimal. Leaf size=163 \[ -\frac {2 (b c-a d) (a+b x)^{3/2}}{c d \sqrt {c+d x}}+\frac {b (3 b c-2 a d) \sqrt {a+b x} \sqrt {c+d x}}{c d^2}-\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}}-\frac {b^{3/2} (3 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 163, 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, 159, 163, 65, 223, 212, 95, 214} \begin {gather*} -\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}}-\frac {b^{3/2} (3 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{5/2}}+\frac {b \sqrt {a+b x} \sqrt {c+d x} (3 b c-2 a d)}{c d^2}-\frac {2 (a+b x)^{3/2} (b c-a d)}{c d \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.41, size = 146, normalized size = 0.90 \begin {gather*} \frac {\sqrt {a+b x} \left (-4 a b c d+2 a^2 d^2+b^2 c (3 c+d x)\right )}{c d^2 \sqrt {c+d x}}-\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}}-\frac {b^{3/2} (3 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(491\) vs. \(2(131)=262\).
time = 0.07, size = 492, normalized size = 3.02

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \left (2 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{3} d^{3} x \sqrt {b d}-5 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c \,d^{2} x \sqrt {a c}+3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} d x \sqrt {a c}+2 \sqrt {b d}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{3} c \,d^{2}-5 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a \,b^{2} c^{2} d +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, b^{3} c^{3}-2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{2} c d x -4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{2} d^{2}+8 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a b c d -6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{2} c^{2}\right )}{2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}\, \sqrt {a c}\, \sqrt {d x +c}\, c \,d^{2}}\)	\(492\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*x+a)^(1/2)*(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^3*x*(b*d)^(1/2)-5*
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^2*x*(a*c)^(1/2)+3*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*(a*c)^(1/2)+2*(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^3*c*d^2-5*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)*a*b^2*c^2*d+3*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^3-2*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*b^2*c*d*
x-4*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a^2*d^2+8*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*
a*b*c*d-6*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*b^2*c^2)/((d*x+c)*(b*x+a))^(1/2)/(b*d)^(1/2)/(a*c)^(
1/2)/(d*x+c)^(1/2)/c/d^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)/x/(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. 268 vs. \(2 (131) = 262\).
time = 3.18, size = 1181, normalized size = 7.25 \begin {gather*} \left [-\frac {{\left (3 \, b^{2} c^{3} - 5 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 5 \, a b c d^{2}\right )} 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} d^{3} x + a^{2} c d^{2}\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 (b^{2} c d x + 3 \, b^{2} c^{2} - 4 \, a b c d + 2 \, a^{2} d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, {\left (c d^{3} x + c^{2} d^{2}\right )}}, \frac {{\left (3 \, b^{2} c^{3} - 5 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 5 \, a b c d^{2}\right )} 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 (a^{2} d^{3} x + a^{2} c d^{2}\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 ) + 2 \, {\left (b^{2} c d x + 3 \, b^{2} c^{2} - 4 \, a b c d + 2 \, a^{2} d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (c d^{3} x + c^{2} d^{2}\right )}}, \frac {4 \, {\left (a^{2} d^{3} x + a^{2} c d^{2}\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 (3 \, b^{2} c^{3} - 5 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 5 \, a b c d^{2}\right )} 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 ) + 4 \, {\left (b^{2} c d x + 3 \, b^{2} c^{2} - 4 \, a b c d + 2 \, a^{2} d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, {\left (c d^{3} x + c^{2} d^{2}\right )}}, \frac {2 \, {\left (a^{2} d^{3} x + a^{2} c d^{2}\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 (3 \, b^{2} c^{3} - 5 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 5 \, a b c d^{2}\right )} 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 (b^{2} c d x + 3 \, b^{2} c^{2} - 4 \, a b c d + 2 \, a^{2} d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (c d^{3} x + c^{2} d^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((3*b^2*c^3 - 5*a*b*c^2*d + (3*b^2*c^2*d - 5*a*b*c*d^2)*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*d^3*x + a^2*c*d^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*(b^2*c*d*x + 3*b^2*
c^2 - 4*a*b*c*d + 2*a^2*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(c*d^3*x + c^2*d^2), 1/2*((3*b^2*c^3 - 5*a*b*c^2*d +
(3*b^2*c^2*d - 5*a*b*c*d^2)*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)) + (a^2*d^3*x + a^2*c*d^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) + 2*(b^2*c*d*x + 3*b^2*c^2 - 4*a*b*c*d + 2*a^2*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(c*d^3*x +
c^2*d^2), 1/4*(4*(a^2*d^3*x + a^2*c*d^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)) - (3*b^2*c^3 - 5*a*b*c^2*d + (3*b^2*c^2*d - 5*a*b*c*
d^2)*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) + 4*(b^2*c*d*x + 3*b^2*c^2 - 4*a*b*c*d + 2*a^2*d^2)*sqr
t(b*x + a)*sqrt(d*x + c))/(c*d^3*x + c^2*d^2), 1/2*(2*(a^2*d^3*x + a^2*c*d^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)) + (3*b^2*c^3 - 5
*a*b*c^2*d + (3*b^2*c^2*d - 5*a*b*c*d^2)*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*(b^2*c*d*x + 3*b^2*c^2 - 4*a*b*c*d + 2*a^2*d^2)*
sqrt(b*x + a)*sqrt(d*x + c))/(c*d^3*x + c^2*d^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}}}{x \left (c + d x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.00, size = 256, normalized size = 1.57 \begin {gather*} -\frac {2 \, \sqrt {b d} a^{3} b \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} c {\left | b \right |}} + \frac {\sqrt {b x + a} {\left (\frac {{\left (b x + a\right )} b^{3}}{d {\left | b \right |}} + \frac {3 \, b^{6} c^{2} d - 5 \, a b^{5} c d^{2} + 2 \, a^{2} b^{4} d^{3}}{b^{2} c d^{3} {\left | b \right |}}\right )}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {{\left (3 \, \sqrt {b d} b^{3} c - 5 \, \sqrt {b d} a b^{2} d\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 )}{2 \, d^{3} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(b*d)*a^3*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)*c*abs(b)) + sqrt(b*x + a)*((b*x + a)*b^3/(d*abs(b)) + (3*b^6*c^2*d -
5*a*b^5*c*d^2 + 2*a^2*b^4*d^3)/(b^2*c*d^3*abs(b)))/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) + 1/2*(3*sqrt(b*d)*b^3*
c - 5*sqrt(b*d)*a*b^2*d)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(d^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 )}^{5/2}}{x\,{\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*(c + d*x)^(3/2)),x)

[Out]

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

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