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

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

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Rubi [A]  time = 0.15, 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 = {98, 154, 157, 63, 217, 206, 93, 208} \begin {gather*} -\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}+\frac {d^{3/2} (5 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}}-\frac {d \sqrt {a+b x} \sqrt {c+d x} (2 b c-3 a d)}{a b^2}+\frac {2 (c+d x)^{3/2} (b c-a d)}{a b \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 154

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

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Mathematica [C]  time = 1.47, size = 274, normalized size = 1.68 \begin {gather*} -\frac {2 \left (b c \left (\sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}} \left (b c^{3/2} \sqrt {a+b x} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\sqrt {a} \sqrt {c+d x} (a d-b c)\right )-a^{3/2} d^{3/2} \sqrt {a+b x} \sqrt {c+d x} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )\right )+a^{3/2} d \sqrt {c+d x} (b c-a d)^{3/2} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};\frac {d (a+b x)}{a d-b c}\right )\right )}{a^{3/2} b^2 \sqrt {a+b x} \sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*c*(-(a^(3/2)*d^(3/2)*Sqrt[a + b*x]*Sqrt[c + d*x]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]]) + Sq
rt[b*c - a*d]*Sqrt[(b*(c + d*x))/(b*c - a*d)]*(Sqrt[a]*(-(b*c) + a*d)*Sqrt[c + d*x] + b*c^(3/2)*Sqrt[a + b*x]*
ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])) + a^(3/2)*d*(b*c - a*d)^(3/2)*Sqrt[c + d*x]*Hyperge
ometric2F1[-3/2, -1/2, 1/2, (d*(a + b*x))/(-(b*c) + a*d)]))/(a^(3/2)*b^2*Sqrt[b*c - a*d]*Sqrt[a + b*x]*Sqrt[(b
*(c + d*x))/(b*c - a*d)])

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IntegrateAlgebraic [A]  time = 0.37, size = 229, normalized size = 1.40 \begin {gather*} -\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{a^{3/2}}+\frac {\sqrt {c+d x} \left (\frac {2 a^2 b d^2 (c+d x)}{a+b x}-3 a^2 d^3+\frac {2 b^3 c^2 (c+d x)}{a+b x}-\frac {4 a b^2 c d (c+d x)}{a+b x}+5 a b c d^2-2 b^2 c^2 d\right )}{a b^2 \sqrt {a+b x} \left (\frac {b (c+d x)}{a+b x}-d\right )}+\frac {\left (5 b c d^{3/2}-3 a d^{5/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 6.67, size = 1183, normalized size = 7.26

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((5*a^2*b*c*d - 3*a^3*d^2 + (5*a*b^2*c*d - 3*a^2*b*d^2)*x)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b
*c*d + a^2*d^2 - 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x
) - 2*(b^3*c^2*x + a*b^2*c^2)*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a
*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(a*b*d^2*x + 2*b^2*
c^2 - 4*a*b*c*d + 3*a^2*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^3*x + a^2*b^2), -1/2*((5*a^2*b*c*d - 3*a^3*d^2
+ (5*a*b^2*c*d - 3*a^2*b*d^2)*x)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(
-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) - (b^3*c^2*x + a*b^2*c^2)*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 +
6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2
+ a^2*c*d)*x)/x^2) - 2*(a*b*d^2*x + 2*b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^3*x
+ a^2*b^2), 1/4*(4*(b^3*c^2*x + a*b^2*c^2)*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*
x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) - (5*a^2*b*c*d - 3*a^3*d^2 + (5*a*b^2*c*d - 3*a^2*b
*d^2)*x)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x
+ a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(a*b*d^2*x + 2*b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*sq
rt(b*x + a)*sqrt(d*x + c))/(a*b^3*x + a^2*b^2), 1/2*(2*(b^3*c^2*x + a*b^2*c^2)*sqrt(-c/a)*arctan(1/2*(2*a*c +
(b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) - (5*a^2*b*c*d
- 3*a^3*d^2 + (5*a*b^2*c*d - 3*a^2*b*d^2)*x)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*
x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) + 2*(a*b*d^2*x + 2*b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)
*sqrt(b*x + a)*sqrt(d*x + c))/(a*b^3*x + a^2*b^2)]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 0.49

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maple [B]  time = 0.02, size = 492, normalized size = 3.02 \begin {gather*} -\frac {\sqrt {d x +c}\, \left (3 \sqrt {a c}\, a^{2} b \,d^{3} 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 )-5 \sqrt {a c}\, a \,b^{2} c \,d^{2} 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 )+2 \sqrt {b d}\, b^{3} c^{3} 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 )+3 \sqrt {a c}\, a^{3} d^{3} \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 )-5 \sqrt {a c}\, a^{2} b c \,d^{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}\, a \,b^{2} c^{3} \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 )-2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,d^{2} x -6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2}+8 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d -4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2}\right )}{2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, \sqrt {b x +a}\, a \,b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d*x+c)^(1/2)*(3*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*a^2*b*d^3*
(a*c)^(1/2)-5*(a*c)^(1/2)*a*b^2*c*d^2*x*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))+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)*x*b^3*c^3*(b*d)^(1/2)+3*(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))*a^3*d^3-5*(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))*a^2*b*c*d^2+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*b^2*c^3-2*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a*b*d^
2-6*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2+8*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*
a*b*c*d-4*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c^2)/a/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(a*c)
^(1/2)/(b*x+a)^(1/2)/b^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((d*x+c)^(5/2)/x/(b*x+a)^(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 (c+d\,x\right )}^{5/2}}{x\,{\left (a+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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