3.7.50 \(\int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x} \, dx\) [650]

Optimal. Leaf size=218 \[ -\frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 d^2}+\frac {(b c+5 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d}+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}-2 a^{5/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{5/2}} \]

[Out]

1/8*(5*a^3*d^3+15*a^2*b*c*d^2-5*a*b^2*c^2*d+b^3*c^3)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(5
/2)/b^(1/2)-2*a^(5/2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))*c^(1/2)+1/12*(5*a*d+b*c)*(b*x+a)^(3
/2)*(d*x+c)^(1/2)/d+1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)-1/8*(-5*a*d+b*c)*(a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^2

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Rubi [A]
time = 0.16, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {103, 159, 163, 65, 223, 212, 95, 214} \begin {gather*} -2 a^{5/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{5/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{8 d^2}+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((b*c - 5*a*d)*(b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/d^2 + ((b*c + 5*a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x
])/(12*d) + ((a + b*x)^(5/2)*Sqrt[c + d*x])/3 - 2*a^(5/2)*Sqrt[c]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqr
t[c + d*x])] + ((b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b
]*Sqrt[c + d*x])])/(8*Sqrt[b]*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 103

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b
*x)^m*(c + d*x)^n*((e + f*x)^(p + 1)/(f*(m + n + p + 1))), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[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} \sqrt {c+d x}}{x} \, dx &=\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}-\frac {1}{3} \int \frac {(a+b x)^{3/2} \left (-3 a c+\frac {1}{2} (-b c-5 a d) x\right )}{x \sqrt {c+d x}} \, dx\\ &=\frac {(b c+5 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d}+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}-\frac {\int \frac {\sqrt {a+b x} \left (-6 a^2 c d+\frac {3}{4} (b c-5 a d) (b c+a d) x\right )}{x \sqrt {c+d x}} \, dx}{6 d}\\ &=-\frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 d^2}+\frac {(b c+5 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d}+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}-\frac {\int \frac {-6 a^3 c d^2-\frac {3}{8} \left (16 a^2 b c d^2+(b c-5 a d) (b c-a d) (b c+a d)\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{6 d^2}\\ &=-\frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 d^2}+\frac {(b c+5 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d}+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}+\left (a^3 c\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx+\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 d^2}\\ &=-\frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 d^2}+\frac {(b c+5 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d}+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}+\left (2 a^3 c\right ) \text {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )+\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\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 )}{8 b d^2}\\ &=-\frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 d^2}+\frac {(b c+5 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d}+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}-2 a^{5/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\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 )}{8 b d^2}\\ &=-\frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 d^2}+\frac {(b c+5 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d}+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}-2 a^{5/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.54, size = 193, normalized size = 0.89 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} \left (33 a^2 d^2+2 a b d (7 c+13 d x)+b^2 \left (-3 c^2+2 c d x+8 d^2 x^2\right )\right )}{24 d^2}-2 a^{5/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )+\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 \sqrt {b} d^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*Sqrt[c + d*x]*(33*a^2*d^2 + 2*a*b*d*(7*c + 13*d*x) + b^2*(-3*c^2 + 2*c*d*x + 8*d^2*x^2)))/(24*d
^2) - 2*a^(5/2)*Sqrt[c]*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])] + ((b^3*c^3 - 5*a*b^2*c^2*d +
15*a^2*b*c*d^2 + 5*a^3*d^3)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(8*Sqrt[b]*d^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(498\) vs. \(2(174)=348\).
time = 0.07, size = 499, normalized size = 2.29

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (16 b^{2} d^{2} x^{2} \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{3} d^{3}+45 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{2} b c \,d^{2}-15 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a \,b^{2} c^{2} d +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, b^{3} c^{3}-48 \sqrt {b d}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{3} c \,d^{2}+52 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a b \,d^{2} x +4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{2} c d x +66 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{2} d^{2}+28 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a b c d -6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{2} c^{2}\right )}{48 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, d^{2} \sqrt {b d}\, \sqrt {a c}}\)	\(499\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(16*b^2*d^2*x^2*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+15*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^3*d^3+45*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^2*b*c*d^2-15*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^3-48*(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+52*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*b*d^2*x+4*(b*d)^(1
/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*b^2*c*d*x+66*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a^2*d^2+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)/d^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,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 = 5.25, size = 1197, normalized size = 5.49 \begin {gather*} \left [\frac {48 \, \sqrt {a c} a^{2} b d^{3} \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 ) + 3 \, {\left (b^{3} c^{3} - 5 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {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 x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d + 14 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} + 13 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b d^{3}}, \frac {24 \, \sqrt {a c} a^{2} b d^{3} \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 ) - 3 \, {\left (b^{3} c^{3} - 5 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d + 14 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} + 13 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b d^{3}}, \frac {96 \, \sqrt {-a c} a^{2} b d^{3} \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 ) + 3 \, {\left (b^{3} c^{3} - 5 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {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 x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d + 14 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} + 13 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b d^{3}}, \frac {48 \, \sqrt {-a c} a^{2} b d^{3} \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 ) - 3 \, {\left (b^{3} c^{3} - 5 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d + 14 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} + 13 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b d^{3}}\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,x, algorithm="fricas")

[Out]

[1/96*(48*sqrt(a*c)*a^2*b*d^3*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) + 3*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b
*c*d^2 + 5*a^3*d^3)*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*x + b*c + a*d)*sqrt
(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(8*b^3*d^3*x^2 - 3*b^3*c^2*d + 14*a*b^2*c*d^2
+ 33*a^2*b*d^3 + 2*(b^3*c*d^2 + 13*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^3), 1/48*(24*sqrt(a*c)*a^2
*b*d^3*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) - 3*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*sqr
t(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*
c*d + a*b*d^2)*x)) + 2*(8*b^3*d^3*x^2 - 3*b^3*c^2*d + 14*a*b^2*c*d^2 + 33*a^2*b*d^3 + 2*(b^3*c*d^2 + 13*a*b^2*
d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^3), 1/96*(96*sqrt(-a*c)*a^2*b*d^3*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)) + 3*(b^3*c^3 - 5*a*b^
2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*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*x
+ b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(8*b^3*d^3*x^2 - 3*b^3*c^2*
d + 14*a*b^2*c*d^2 + 33*a^2*b*d^3 + 2*(b^3*c*d^2 + 13*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^3), 1/48
*(48*sqrt(-a*c)*a^2*b*d^3*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)) - 3*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(-b*d)*a
rctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b
*d^2)*x)) + 2*(8*b^3*d^3*x^2 - 3*b^3*c^2*d + 14*a*b^2*c*d^2 + 33*a^2*b*d^3 + 2*(b^3*c*d^2 + 13*a*b^2*d^3)*x)*s
qrt(b*x + a)*sqrt(d*x + c))/(b*d^3)]

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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}\, 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,x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

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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} \,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,x)

[Out]

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

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