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

Optimal. Leaf size=177 \[ \frac {c \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{4 a^2 x}-\frac {\sqrt {c} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2}}+\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2} \]

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Rubi [A]  time = 0.12, antiderivative size = 177, 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, 149, 157, 63, 217, 206, 93, 208} \begin {gather*} -\frac {\sqrt {c} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2}}+\frac {c \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{4 a^2 x}+\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(3*b*c - 7*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*a^2*x) - (c*Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*a*x^2) - (Sqr
t[c]*(3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(5/2
)) + (2*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b]

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 149

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] && IntegerQ[m]

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^3 \sqrt {a+b x}} \, dx &=-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}-\frac {\int \frac {\sqrt {c+d x} \left (\frac {1}{2} c (3 b c-7 a d)-2 a d^2 x\right )}{x^2 \sqrt {a+b x}} \, dx}{2 a}\\ &=\frac {c (3 b c-7 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a^2 x}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}-\frac {\int \frac {-\frac {1}{4} c \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right )-2 a^2 d^3 x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 a^2}\\ &=\frac {c (3 b c-7 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a^2 x}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}+d^3 \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx+\frac {\left (c \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 a^2}\\ &=\frac {c (3 b c-7 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a^2 x}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^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}+\frac {\left (c \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{4 a^2}\\ &=\frac {c (3 b c-7 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a^2 x}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}-\frac {\sqrt {c} \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 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}\\ &=\frac {c (3 b c-7 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a^2 x}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{2 a x^2}-\frac {\sqrt {c} \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 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 )}{\sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 1.14, size = 189, normalized size = 1.07 \begin {gather*} -\frac {c \sqrt {a+b x} \sqrt {c+d x} (2 a c+9 a d x-3 b c x)}{4 a^2 x^2}-\frac {\sqrt {c} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2}}+\frac {2 d^{5/2} \sqrt {c+d x} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\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^3*Sqrt[a + b*x]),x]

[Out]

-1/4*(c*Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*c - 3*b*c*x + 9*a*d*x))/(a^2*x^2) + (2*d^(5/2)*Sqrt[c + d*x]*ArcSinh[
(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(Sqrt[b*c - a*d]*Sqrt[(b*(c + d*x))/(b*c - a*d)]) - (Sqrt[c]*(3*b^2*
c^2 - 10*a*b*c*d + 15*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(5/2))

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IntegrateAlgebraic [A]  time = 0.43, size = 248, normalized size = 1.40 \begin {gather*} \frac {\left (-15 a^2 \sqrt {c} d^2+10 a b c^{3/2} d-3 b^2 c^{5/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2}}-\frac {c \sqrt {a+b x} \left (9 a^3 d^2-\frac {7 a^2 c d^2 (a+b x)}{c+d x}-14 a^2 b c d-\frac {3 b^2 c^3 (a+b x)}{c+d x}+5 a b^2 c^2+\frac {10 a b c^2 d (a+b x)}{c+d x}\right )}{4 a^2 \sqrt {c+d x} \left (a-\frac {c (a+b x)}{c+d x}\right )^2}+\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(c*Sqrt[a + b*x]*(5*a*b^2*c^2 - 14*a^2*b*c*d + 9*a^3*d^2 - (3*b^2*c^3*(a + b*x))/(c + d*x) + (10*a*b*c^2*
d*(a + b*x))/(c + d*x) - (7*a^2*c*d^2*(a + b*x))/(c + d*x)))/(a^2*Sqrt[c + d*x]*(a - (c*(a + b*x))/(c + d*x))^
2) + ((-3*b^2*c^(5/2) + 10*a*b*c^(3/2)*d - 15*a^2*Sqrt[c]*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c
+ d*x])])/(4*a^(5/2)) + (2*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b]

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fricas [A]  time = 5.66, size = 1031, normalized size = 5.82 \begin {gather*} \left [\frac {8 \, a^{2} d^{2} x^{2} \sqrt {\frac {d}{b}} \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^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2} \sqrt {\frac {c}{a}} \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^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (2 \, a c^{2} - 3 \, {\left (b c^{2} - 3 \, a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, a^{2} x^{2}}, -\frac {16 \, a^{2} d^{2} x^{2} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) - {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2} \sqrt {\frac {c}{a}} \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^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (2 \, a c^{2} - 3 \, {\left (b c^{2} - 3 \, a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, a^{2} x^{2}}, \frac {4 \, a^{2} d^{2} x^{2} \sqrt {\frac {d}{b}} \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^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2} \sqrt {-\frac {c}{a}} \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 {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) - 2 \, {\left (2 \, a c^{2} - 3 \, {\left (b c^{2} - 3 \, a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, a^{2} x^{2}}, -\frac {8 \, a^{2} d^{2} x^{2} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) - {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2} \sqrt {-\frac {c}{a}} \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 {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) + 2 \, {\left (2 \, a c^{2} - 3 \, {\left (b c^{2} - 3 \, a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, a^{2} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^3/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(8*a^2*d^2*x^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^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*x^
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*(2*a*c^2 - 3*(b*c^2 - 3*a*c*d)*x)*sqrt(b*x + a)
*sqrt(d*x + c))/(a^2*x^2), -1/16*(16*a^2*d^2*x^2*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqr
t(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) - (3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*x^2*sqr
t(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)*sq
rt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(2*a*c^2 - 3*(b*c^2 - 3*a*c*d)*x)*sqrt(b*x + a)*sqrt
(d*x + c))/(a^2*x^2), 1/8*(4*a^2*d^2*x^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^2*c^2 - 10*a*b*
c*d + 15*a^2*d^2)*x^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)) - 2*(2*a*c^2 - 3*(b*c^2 - 3*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*
x^2), -1/8*(8*a^2*d^2*x^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^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*x^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)) + 2*(2*a*c
^2 - 3*(b*c^2 - 3*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*x^2)]

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giac [B]  time = 11.47, size = 1149, normalized size = 6.49

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^3/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/4*(4*sqrt(b*d)*d^2*abs(b)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/b + (3*sqr
t(b*d)*b^3*c^3*abs(b) - 10*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 15*sqrt(b*d)*a^2*b*c*d^2*abs(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)*a^2*b) - 2*(3*sqrt(b*d)*b^9*c^6*abs(b) - 21*sqrt(b*d)*a*b^8*c^5*d*abs(b) + 54*sqrt(b*d)*a^2*b^7*c^4*d^2*abs
(b) - 66*sqrt(b*d)*a^3*b^6*c^3*d^3*abs(b) + 39*sqrt(b*d)*a^4*b^5*c^2*d^4*abs(b) - 9*sqrt(b*d)*a^5*b^4*c*d^5*ab
s(b) - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^7*c^5*abs(b) + 32*sqrt(
b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^6*c^4*d*abs(b) - 10*sqrt(b*d)*(sqrt
(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^5*c^3*d^2*abs(b) - 40*sqrt(b*d)*(sqrt(b*d)*
sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^4*c^2*d^3*abs(b) + 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b
*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^3*c*d^4*abs(b) + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^5*c^4*abs(b) - 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^4*a*b^4*c^3*d*abs(b) - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b
*d - a*b*d))^4*a^2*b^3*c^2*d^2*abs(b) - 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a
*b*d))^4*a^3*b^2*c*d^3*abs(b) - 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*
b^3*c^3*abs(b) + 10*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^2*c^2*d*ab
s(b) + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b*c*d^2*abs(b))/((b^4
*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c -
2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b
^2*c + (b*x + a)*b*d - a*b*d))^4)^2*a^2))/b

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maple [B]  time = 0.02, size = 354, normalized size = 2.00 \begin {gather*} \frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (-15 \sqrt {b d}\, a^{2} c \,d^{2} 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 )+8 \sqrt {a c}\, a^{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 )+10 \sqrt {b d}\, a b \,c^{2} d \,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 )-3 \sqrt {b d}\, b^{2} 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 )-18 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a c d x +6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, b \,c^{2} x -4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a \,c^{2}\right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)/x^3/(b*x+a)^(1/2),x)

[Out]

1/8*(d*x+c)^(1/2)*(b*x+a)^(1/2)/a^2*(8*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*d^3*(a*c)^(1/2)-15*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*a^2*c*d^2*
(b*d)^(1/2)+10*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*a*b*c^2*d*(b*d)^(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^2*c^3*(b*d)^(1/2)-18*x*a*c*d*((b*x+a)*(d*x+
c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)+6*x*b*c^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-4*a*c^2*((b*x+a)*(
d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/x^2/(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((d*x+c)^(5/2)/x^3/(b*x+a)^(1/2),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^3\,\sqrt {a+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(x^3*(a + b*x)^(1/2)),x)

[Out]

int((c + d*x)^(5/2)/(x^3*(a + b*x)^(1/2)), 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^{3} \sqrt {a + b x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/2)/x**3/(b*x+a)**(1/2),x)

[Out]

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

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