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

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

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Rubi [A]  time = 0.14, 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 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 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 {(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 ) \operatorname {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)) \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^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)) \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^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 [C]  time = 1.66, size = 226, normalized size = 1.39 \begin {gather*} \frac {2 \left (5 a \left (-\frac {a^{3/2} \sqrt {c+d x} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{3/2}}+\frac {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 )}{d^{3/2}}+\frac {\sqrt {a+b x} (a d-b c)}{c d}\right )+\frac {(a+b x)^{5/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/2} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};\frac {d (a+b x)}{a d-b c}\right )}{c+d x}\right )}{5 \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*a*(((-(b*c) + a*d)*Sqrt[a + b*x])/(c*d) + (b*Sqrt[b*c - a*d]*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sq
rt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/d^(3/2) - (a^(3/2)*Sqrt[c + d*x]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[
a]*Sqrt[c + d*x])])/c^(3/2)) + ((a + b*x)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(3/2)*Hypergeometric2F1[3/2, 5/2,
7/2, (d*(a + b*x))/(-(b*c) + a*d)])/(c + d*x)))/(5*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.40, size = 229, normalized size = 1.40 \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 {\sqrt {a+b x} \left (\frac {2 a^2 d^3 (a+b x)}{c+d x}-2 a^2 b d^2+\frac {2 b^2 c^2 d (a+b x)}{c+d x}+5 a b^2 c d-\frac {4 a b c d^2 (a+b x)}{c+d x}-3 b^3 c^2\right )}{c d^2 \sqrt {c+d x} \left (\frac {d (a+b x)}{c+d x}-b\right )}+\frac {\left (5 a b^{3/2} d-3 b^{5/2} c\right ) \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]

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

[Out]

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

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fricas [B]  time = 5.11, size = 1181, normalized size = 7.25

result too large to display

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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giac [A]  time = 1.68, 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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maple [B]  time = 0.02, size = 492, normalized size = 3.02 \begin {gather*} \frac {\sqrt {b x +a}\, \left (-2 \sqrt {b d}\, a^{3} d^{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 )+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 )-3 \sqrt {a c}\, b^{3} c^{2} 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 )-2 \sqrt {b d}\, a^{3} c \,d^{2} \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 )+5 \sqrt {a c}\, a \,b^{2} c^{2} d \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 )-3 \sqrt {a c}\, b^{3} c^{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 )+2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c d x +4 \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 +6 \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 {d x +c}\, c \,d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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