3.7.1 \(\int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x} \, dx\) [601]

Optimal. Leaf size=164 \[ \frac {(b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d}+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}-2 a^{3/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {\left (b^2 c^2-6 a b c d-3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.09, 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 = {103, 159, 163, 65, 223, 212, 95, 214} \begin {gather*} -2 a^{3/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {\left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{3/2}}+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}+\frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+b c)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c + 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*d) + ((a + b*x)^(3/2)*Sqrt[c + d*x])/2 - 2*a^(3/2)*Sqrt[c]*ArcT
anh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] - ((b^2*c^2 - 6*a*b*c*d - 3*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqr
t[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*Sqrt[b]*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 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)^{3/2} \sqrt {c+d x}}{x} \, dx &=\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}-\frac {1}{2} \int \frac {\sqrt {a+b x} \left (-2 a c+\frac {1}{2} (-b c-3 a d) x\right )}{x \sqrt {c+d x}} \, dx\\ &=\frac {(b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d}+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}-\frac {\int \frac {-2 a^2 c d+\frac {1}{4} \left (b^2 c^2-6 a b c d-3 a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 d}\\ &=\frac {(b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d}+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}+\left (a^2 c\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx-\frac {\left (b^2 c^2-6 a b c d-3 a^2 d^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 d}\\ &=\frac {(b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d}+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}+\left (2 a^2 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 {1}{4} \left (-6 a c+\frac {b c^2}{d}-\frac {3 a^2 d}{b}\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 )\\ &=\frac {(b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d}+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}-2 a^{3/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {1}{4} \left (-6 a c+\frac {b c^2}{d}-\frac {3 a^2 d}{b}\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 )\\ &=\frac {(b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d}+\frac {1}{2} (a+b x)^{3/2} \sqrt {c+d x}-2 a^{3/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {\left (b^2 c^2-6 a b c d-3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.49, size = 146, normalized size = 0.89 \begin {gather*} \frac {1}{4} \left (\frac {\sqrt {a+b x} \sqrt {c+d x} (5 a d+b (c+2 d x))}{d}-8 a^{3/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )+\frac {\left (-b^2 c^2+6 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{\sqrt {b} d^{3/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a + b*x]*Sqrt[c + d*x]*(5*a*d + b*(c + 2*d*x)))/d - 8*a^(3/2)*Sqrt[c]*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(
Sqrt[c]*Sqrt[a + b*x])] + ((-(b^2*c^2) + 6*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[
a + b*x])])/(Sqrt[b]*d^(3/2)))/4

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(332\) vs. \(2(126)=252\).
time = 0.07, size = 333, normalized size = 2.03

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{2} d^{2}+6 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a b c d -\ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, b^{2} c^{2}+4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b d x -8 \sqrt {b d}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{2} c d +10 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a d +2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b c \right )}{8 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, d \sqrt {b d}\, \sqrt {a c}}\)	\(333\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x+a)^(1/2)*(d*x+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))
*(a*c)^(1/2)*a^2*d^2+6*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*c*d-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^2*c^2+4*(b*
d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*b*d*x-8*(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*c*d+10*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*d+2*(b*d)^(1/2)*(a*c)^(1/2)*((
d*x+c)*(b*x+a))^(1/2)*b*c)/((d*x+c)*(b*x+a))^(1/2)/d/(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)^(3/2)*(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 = 2.17, size = 979, normalized size = 5.97 \begin {gather*} \left [\frac {8 \, \sqrt {a c} a b d^{2} \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 ) - {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\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 (2 \, b^{2} d^{2} x + b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b d^{2}}, \frac {4 \, \sqrt {a c} a b d^{2} \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 ) + {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\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 (2 \, b^{2} d^{2} x + b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b d^{2}}, \frac {16 \, \sqrt {-a c} a b d^{2} \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 ) - {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\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 (2 \, b^{2} d^{2} x + b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b d^{2}}, \frac {8 \, \sqrt {-a c} a b d^{2} \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 ) + {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\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 (2 \, b^{2} d^{2} x + b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(8*sqrt(a*c)*a*b*d^2*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)*sq
rt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - (b^2*c^2 - 6*a*b*c*d - 3*a^2*d^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*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*(2*b^2*d^2*x + b^2*c*d + 5*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^2
), 1/8*(4*sqrt(a*c)*a*b*d^2*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)*s
qrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + (b^2*c^2 - 6*a*b*c*d - 3*a^2*d^2)*sqrt(
-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*(2*b^2*d^2*x + b^2*c*d + 5*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^2), 1/16*(16*sqrt(-
a*c)*a*b*d^2*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)) - (b^2*c^2 - 6*a*b*c*d - 3*a^2*d^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*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*(2*
b^2*d^2*x + b^2*c*d + 5*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^2), 1/8*(8*sqrt(-a*c)*a*b*d^2*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))
+ (b^2*c^2 - 6*a*b*c*d - 3*a^2*d^2)*sqrt(-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*(2*b^2*d^2*x + b^2*c*d + 5*a*b*d^2)*sqrt(b*x + a
)*sqrt(d*x + c))/(b*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 {3}{2}} \sqrt {c + d x}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)**(3/2)*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)^(3/2)*(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 = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{3/2}\,\sqrt {c+d\,x}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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