3.7.12 \(\int \frac {(a+b x)^{3/2}}{x (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=119 \[ -\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}}+\frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}}-\frac {2 \sqrt {a+b x} (b c-a d)}{c d \sqrt {c+d x}} \]

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Rubi [A]  time = 0.07, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {98, 157, 63, 217, 206, 93, 208} \begin {gather*} -\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}}+\frac {2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}}-\frac {2 \sqrt {a+b x} (b c-a d)}{c d \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

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 93

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 98

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 157

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 206

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

Rule 208

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

Rule 217

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}}{x (c+d x)^{3/2}} \, dx &=-\frac {2 (b c-a d) \sqrt {a+b x}}{c d \sqrt {c+d x}}+\frac {2 \int \frac {\frac {a^2 d}{2}+\frac {1}{2} b^2 c x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{c d}\\ &=-\frac {2 (b c-a d) \sqrt {a+b x}}{c d \sqrt {c+d x}}+\frac {a^2 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{c}+\frac {b^2 \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{d}\\ &=-\frac {2 (b c-a d) \sqrt {a+b x}}{c d \sqrt {c+d x}}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{c}+\frac {(2 b) \operatorname {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 {2 (b c-a d) \sqrt {a+b x}}{c d \sqrt {c+d x}}-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}}+\frac {(2 b) \operatorname {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-a d) \sqrt {a+b x}}{c d \sqrt {c+d x}}-\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}}+\frac {2 b^{3/2} \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 = 0.83, size = 162, normalized size = 1.36 \begin {gather*} \frac {2 \left (\frac {\sqrt {d} \left (\sqrt {c} \sqrt {a+b x} (a d-b c)-a^{3/2} d \sqrt {c+d x} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )}{c^{3/2}}+b \sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )\right )}{d^{3/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*Sqrt[b*c - a*d]*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]] + (Sqrt
[d]*(Sqrt[c]*(-(b*c) + a*d)*Sqrt[a + b*x] - a^(3/2)*d*Sqrt[c + d*x]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*S
qrt[c + d*x])]))/c^(3/2)))/(d^(3/2)*Sqrt[c + d*x])

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IntegrateAlgebraic [B]  time = 0.80, size = 252, normalized size = 2.12 \begin {gather*} -\frac {2 a^{3/2} \sqrt {d} \sqrt {\frac {b}{d}} \tanh ^{-1}\left (-\frac {\sqrt {b} (c+d x)}{\sqrt {a} \sqrt {c} \sqrt {d}}+\frac {\sqrt {d} \sqrt {\frac {b}{d}} \sqrt {c+d x} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}{\sqrt {a} \sqrt {b} \sqrt {c}}+\frac {\sqrt {b} \sqrt {c}}{\sqrt {a} \sqrt {d}}\right )}{\sqrt {b} c^{3/2}}+\frac {2 (a d-b c) \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}{c d \sqrt {c+d x}}-\frac {2 b \sqrt {\frac {b}{d}} \log \left (\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}-\sqrt {\frac {b}{d}} \sqrt {c+d x}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-(b*c) + a*d)*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])/(c*d*Sqrt[c + d*x]) - (2*a^(3/2)*Sqrt[b/d]*Sqrt[d]*ArcT
anh[(Sqrt[b]*Sqrt[c])/(Sqrt[a]*Sqrt[d]) - (Sqrt[b]*(c + d*x))/(Sqrt[a]*Sqrt[c]*Sqrt[d]) + (Sqrt[b/d]*Sqrt[d]*S
qrt[c + d*x]*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])/(Sqrt[a]*Sqrt[b]*Sqrt[c])])/(Sqrt[b]*c^(3/2)) - (2*b*Sqrt[b/
d]*Log[-(Sqrt[b/d]*Sqrt[c + d*x]) + Sqrt[a - (b*c)/d + (b*(c + d*x))/d]])/d

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fricas [B]  time = 3.53, size = 956, normalized size = 8.03 \begin {gather*} \left [\frac {{\left (b c d x + b c^{2}\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 (a d^{2} x + a c d\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 c - a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (c d^{2} x + c^{2} d\right )}}, -\frac {2 \, {\left (b c d x + b c^{2}\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 d^{2} x + a c d\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 c - a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (c d^{2} x + c^{2} d\right )}}, \frac {2 \, {\left (a d^{2} x + a c d\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 c d x + b c^{2}\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 c - a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (c d^{2} x + c^{2} d\right )}}, \frac {{\left (a d^{2} x + a c d\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 c d x + b c^{2}\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 c - a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{c d^{2} x + c^{2} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((b*c*d*x + b*c^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) + (a*d^2*x + a*c*d)*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*c - a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(c*d^2*x + c^2*d), -1/2*(2*
(b*c*d*x + b*c^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)) - (a*d^2*x + a*c*d)*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*c - a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(c*d^2*x + c^2*d), 1/2*(2*(a*d^2*x + a*c*d)*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*c
*d*x + b*c^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) - 4*(b*c - a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(c*d
^2*x + c^2*d), ((a*d^2*x + a*c*d)*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sq
rt(-a/c)/(a*b*d*x^2 + a^2*c + (a*b*c + a^2*d)*x)) - (b*c*d*x + b*c^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*(b*c - a*d)*sqrt(b*x +
a)*sqrt(d*x + c))/(c*d^2*x + c^2*d)]

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giac [B]  time = 1.64, size = 205, normalized size = 1.72 \begin {gather*} -\frac {2 \, \sqrt {b d} a^{2} 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 d} b^{2} \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 {2 \, {\left (b^{3} c {\left | b \right |} - a b^{2} d {\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 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(b*d)*a^2*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*d)*b^2*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)) - 2*(b^3*c*abs(b) - a*b^2*d*abs(b))*sqrt(b*x + a)/(sqrt(b^2*c + (b*x
+ a)*b*d - a*b*d)*b^2*c*d)

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maple [B]  time = 0.02, size = 306, normalized size = 2.57 \begin {gather*} \frac {\left (-\sqrt {b d}\, a^{2} d^{2} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+\sqrt {a c}\, b^{2} c d x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-\sqrt {b d}\, a^{2} c d \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+\sqrt {a c}\, b^{2} c^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a d -2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b c \right ) \sqrt {b x +a}}{\sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, \sqrt {d x +c}\, c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x*a^2*d^2*(b*d)^(1/2)+ln(1/2*(2*b*d*x+a*d+b*
c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x*b^2*c*d*(a*c)^(1/2)-(b*d)^(1/2)*ln((a*d*x+b*c*x+2*a*c+
2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*a^2*c*d+(a*c)^(1/2)*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2
)*(b*d)^(1/2))/(b*d)^(1/2))*b^2*c^2+2*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*d-2*(b*d)^(1/2)*(a*c)^
(1/2)*((b*x+a)*(d*x+c))^(1/2)*b*c)*(b*x+a)^(1/2)/c/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(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)^(3/2)/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 details)Is a*d-b*c zero or nonzero?

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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}}{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)^(3/2)/(x*(c + d*x)^(3/2)),x)

[Out]

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

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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}}}{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)**(3/2)/x/(d*x+c)**(3/2),x)

[Out]

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

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