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

Optimal. Leaf size=157 \[ -\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{5/2}}+\frac {2 \sqrt {c+d x} \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right )}{\sqrt {a+b x}}+\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}}+\frac {2 (c+d x)^{3/2} (b c-a d)}{3 a b (a+b x)^{3/2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 157, 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, 150, 157, 63, 217, 206, 93, 208} \begin {gather*} \frac {2 \sqrt {c+d x} \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right )}{\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^{5/2}}+\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}}+\frac {2 (c+d x)^{3/2} (b c-a d)}{3 a b (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c^2/a^2 - d^2/b^2)*Sqrt[c + d*x])/Sqrt[a + b*x] + (2*(b*c - a*d)*(c + d*x)^(3/2))/(3*a*b*(a + b*x)^(3/2))
- (2*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/a^(5/2) + (2*d^(5/2)*ArcTanh[(Sqrt[d]*S
qrt[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 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(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 - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 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)^{5/2}} \, dx &=\frac {2 (b c-a d) (c+d x)^{3/2}}{3 a b (a+b x)^{3/2}}+\frac {2 \int \frac {\sqrt {c+d x} \left (\frac {3 b c^2}{2}+\frac {3}{2} a d^2 x\right )}{x (a+b x)^{3/2}} \, dx}{3 a b}\\ &=\frac {2 \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right ) \sqrt {c+d x}}{\sqrt {a+b x}}+\frac {2 (b c-a d) (c+d x)^{3/2}}{3 a b (a+b x)^{3/2}}-\frac {4 \int \frac {-\frac {3}{4} b^2 c^3-\frac {3}{4} a^2 d^3 x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{3 a^2 b^2}\\ &=\frac {2 \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right ) \sqrt {c+d x}}{\sqrt {a+b x}}+\frac {2 (b c-a d) (c+d x)^{3/2}}{3 a b (a+b x)^{3/2}}+\frac {c^3 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{a^2}+\frac {d^3 \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{b^2}\\ &=\frac {2 \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right ) \sqrt {c+d x}}{\sqrt {a+b x}}+\frac {2 (b c-a d) (c+d x)^{3/2}}{3 a b (a+b x)^{3/2}}+\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^2}+\frac {\left (2 d^3\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 {2 \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right ) \sqrt {c+d x}}{\sqrt {a+b x}}+\frac {2 (b c-a d) (c+d x)^{3/2}}{3 a b (a+b x)^{3/2}}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{5/2}}+\frac {\left (2 d^3\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 {2 \left (\frac {c^2}{a^2}-\frac {d^2}{b^2}\right ) \sqrt {c+d x}}{\sqrt {a+b x}}+\frac {2 (b c-a d) (c+d x)^{3/2}}{3 a b (a+b x)^{3/2}}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{5/2}}+\frac {2 d^{5/2} \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 = 0.40, size = 163, normalized size = 1.04 \begin {gather*} \frac {2 \left (-\frac {3 c^{5/2} (a+b x)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{5/2}}+\frac {c \sqrt {c+d x} (4 a c+a d x+3 b c x)}{a^2}+\frac {d \sqrt {c+d x} (a d-b c) \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {d (a+b x)}{a d-b c}\right )}{b^2 \sqrt {\frac {b (c+d x)}{b c-a d}}}\right )}{3 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((c*Sqrt[c + d*x]*(4*a*c + 3*b*c*x + a*d*x))/a^2 - (3*c^(5/2)*(a + b*x)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x
])/(Sqrt[a]*Sqrt[c + d*x])])/a^(5/2) + (d*(-(b*c) + a*d)*Sqrt[c + d*x]*Hypergeometric2F1[-3/2, -3/2, -1/2, (d*
(a + b*x))/(-(b*c) + a*d)])/(b^2*Sqrt[(b*(c + d*x))/(b*c - a*d)])))/(3*(a + b*x)^(3/2))

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IntegrateAlgebraic [B]  time = 3.08, size = 387, normalized size = 2.46 \begin {gather*} -\frac {2 c^{5/2} \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 )}{a^{5/2} \sqrt {b}}-\frac {2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}} \left (3 a^3 d^4 \sqrt {c+d x}+4 a^2 b d^3 (c+d x)^{3/2}-3 a^2 b c d^3 \sqrt {c+d x}-3 a b^2 c^2 d^2 \sqrt {c+d x}-a b^2 c d^2 (c+d x)^{3/2}+3 b^3 c^3 d \sqrt {c+d x}-3 b^3 c^2 d (c+d x)^{3/2}\right )}{3 a^2 b^2 (a d+b (c+d x)-b c)^2}-\frac {2 d^3 \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 )}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 5.93, size = 1361, normalized size = 8.67

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)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^2*b^2*d^2*x^2 + 2*a^3*b*d^2*x + a^4*d^2)*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) + 3*(b^4*c^
2*x^2 + 2*a*b^3*c^2*x + a^2*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*(4*a*b^2*c^2
- a^2*b*c*d - 3*a^3*d^2 + (3*b^3*c^2 + a*b^2*c*d - 4*a^2*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b^4*x^2
+ 2*a^3*b^3*x + a^4*b^2), -1/6*(6*(a^2*b^2*d^2*x^2 + 2*a^3*b*d^2*x + a^4*d^2)*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)) - 3*(b^4*c^2*x^2 +
2*a*b^3*c^2*x + a^2*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*(4*a*b^2*c^2 - a^2*
b*c*d - 3*a^3*d^2 + (3*b^3*c^2 + a*b^2*c*d - 4*a^2*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b^4*x^2 + 2*a^3
*b^3*x + a^4*b^2), 1/6*(6*(b^4*c^2*x^2 + 2*a*b^3*c^2*x + a^2*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)) + 3*(a^2*b^2*d^2*x^2 + 2
*a^3*b*d^2*x + a^4*d^2)*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*(4*a*b^2*c^2 - a^2*b*c*d - 3*a^3*d^
2 + (3*b^3*c^2 + a*b^2*c*d - 4*a^2*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b^4*x^2 + 2*a^3*b^3*x + a^4*b^2
), 1/3*(3*(b^4*c^2*x^2 + 2*a*b^3*c^2*x + a^2*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)) - 3*(a^2*b^2*d^2*x^2 + 2*a^3*b*d^2*x + a
^4*d^2)*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*(4*a*b^2*c^2 - a^2*b*c*d - 3*a^3*d^2 + (3*b^3*c^2 + a*b^2*c*d - 4*a^2*b*d^2)*x)*sqrt
(b*x + a)*sqrt(d*x + c))/(a^2*b^4*x^2 + 2*a^3*b^3*x + a^4*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)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 0.56

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maple [B]  time = 0.03, size = 566, normalized size = 3.61 \begin {gather*} \frac {\sqrt {d x +c}\, \left (3 \sqrt {a c}\, a^{2} b^{2} d^{3} x^{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 )-3 \sqrt {b d}\, b^{4} c^{3} x^{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 )+6 \sqrt {a c}\, a^{3} 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 )-6 \sqrt {b d}\, a \,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^{4} 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 )-3 \sqrt {b d}\, a^{2} 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 )-8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} b \,d^{2} x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a \,b^{2} c d x +6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, b^{3} c^{2} x -6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{3} d^{2}-2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} b c d +8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a \,b^{2} c^{2}\right )}{3 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, \left (b x +a \right )^{\frac {3}{2}} a^{2} 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)^(5/2),x)

[Out]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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