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

Optimal. Leaf size=164 \[ -\frac {a^{3/2} (5 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}}-\frac {\sqrt {a+b x} (2 b c-3 a d) (b c-a d)}{c^2 d \sqrt {c+d x}}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}} \]

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Rubi [A]  time = 0.16, antiderivative size = 164, 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 {a^{3/2} (5 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}}-\frac {\sqrt {a+b x} (2 b c-3 a d) (b c-a d)}{c^2 d \sqrt {c+d x}}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*b*c - 3*a*d)*(b*c - a*d)*Sqrt[a + b*x])/(c^2*d*Sqrt[c + d*x])) - (a*(a + b*x)^(3/2))/(c*x*Sqrt[c + d*x])
- (a^(3/2)*(5*b*c - 3*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/c^(5/2) + (2*b^(5/2)*Arc
Tanh[(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 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 {(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx &=-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}}-\frac {\int \frac {\sqrt {a+b x} \left (-\frac {1}{2} a (5 b c-3 a d)-b^2 c x\right )}{x (c+d x)^{3/2}} \, dx}{c}\\ &=-\frac {(2 b c-3 a d) (b c-a d) \sqrt {a+b x}}{c^2 d \sqrt {c+d x}}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}}+\frac {2 \int \frac {\frac {1}{4} a^2 d (5 b c-3 a d)+\frac {1}{2} b^3 c^2 x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{c^2 d}\\ &=-\frac {(2 b c-3 a d) (b c-a d) \sqrt {a+b x}}{c^2 d \sqrt {c+d x}}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}}+\frac {b^3 \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{d}+\frac {\left (a^2 (5 b c-3 a d)\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 c^2}\\ &=-\frac {(2 b c-3 a d) (b c-a d) \sqrt {a+b x}}{c^2 d \sqrt {c+d x}}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}}+\frac {\left (2 b^2\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 )}{d}+\frac {\left (a^2 (5 b c-3 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{c^2}\\ &=-\frac {(2 b c-3 a d) (b c-a d) \sqrt {a+b x}}{c^2 d \sqrt {c+d x}}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}}-\frac {a^{3/2} (5 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {\left (2 b^2\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 )}{d}\\ &=-\frac {(2 b c-3 a d) (b c-a d) \sqrt {a+b x}}{c^2 d \sqrt {c+d x}}-\frac {a (a+b x)^{3/2}}{c x \sqrt {c+d x}}-\frac {a^{3/2} (5 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}+\frac {2 b^{5/2} \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 [C]  time = 4.92, size = 333, normalized size = 2.03 \begin {gather*} \frac {10 a^{5/2} \sqrt {c+d x} (3 a d-5 b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {5 b c^{3/2} \sqrt {b c-a d} \left (9 a^2 d^2-8 a b c d+3 b^2 c^2\right ) \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^{5/2}}-\frac {5 \sqrt {c} \sqrt {a+b x} \left (2 a^3 d^2 (c+3 d x)-2 a^2 b c d^2 x-a b^2 c d x (7 c+3 d x)+b^3 c^2 x (3 c+d x)\right )}{d^2 x}+\frac {2 c^{3/2} (a+b x)^{5/2} (b c-3 a d) \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}}{10 a c^{5/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-5*Sqrt[c]*Sqrt[a + b*x]*(-2*a^2*b*c*d^2*x + b^3*c^2*x*(3*c + d*x) + 2*a^3*d^2*(c + 3*d*x) - a*b^2*c*d*x*(7*
c + 3*d*x)))/(d^2*x) + (5*b*c^(3/2)*Sqrt[b*c - a*d]*(3*b^2*c^2 - 8*a*b*c*d + 9*a^2*d^2)*Sqrt[(b*(c + d*x))/(b*
c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/d^(5/2) + 10*a^(5/2)*(-5*b*c + 3*a*d)*Sqrt[c + d*x
]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + (2*c^(3/2)*(b*c - 3*a*d)*(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))/(10*a*c^(
5/2)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.35, size = 230, normalized size = 1.40 \begin {gather*} \frac {\left (3 a^{5/2} d-5 a^{3/2} b c\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{c^{5/2}}-\frac {\sqrt {a+b x} \left (-3 a^3 d^2+\frac {2 a^2 c d^2 (a+b x)}{c+d x}+5 a^2 b c d+\frac {2 b^2 c^3 (a+b x)}{c+d x}-2 a b^2 c^2-\frac {4 a b c^2 d (a+b x)}{c+d x}\right )}{c^2 d \sqrt {c+d x} \left (\frac {c (a+b x)}{c+d x}-a\right )}+\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 5.65, size = 1233, normalized size = 7.52

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 5.90, size = 560, normalized size = 3.41 \begin {gather*} -\frac {\sqrt {b d} b^{3} \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 {{\left (5 \, \sqrt {b d} a^{2} b^{3} c - 3 \, \sqrt {b d} a^{3} b^{2} d\right )} \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} b c^{2} {\left | b \right |}} - \frac {2 \, {\left (\sqrt {b d} a^{2} b^{5} c^{2} - 2 \, \sqrt {b d} a^{3} b^{4} c d + \sqrt {b d} a^{4} b^{3} d^{2} - \sqrt {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} a^{2} b^{3} c - \sqrt {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} a^{3} b^{2} d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} 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 )}^{4}\right )} c^{2} {\left | b \right |}} - \frac {2 \, {\left (b^{4} c^{2} {\left | b \right |} - 2 \, a b^{3} c d {\left | b \right |} + a^{2} b^{2} d^{2} {\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^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(b*d)*b^3*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)) - (5*sqrt(b
*d)*a^2*b^3*c - 3*sqrt(b*d)*a^3*b^2*d)*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)*b*c^2*abs(b)) - 2*(sqrt(b*d)*a^2*b^5*c^2 - 2*sqrt
(b*d)*a^3*b^4*c*d + sqrt(b*d)*a^4*b^3*d^2 - 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^3*c - 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^2*d)/(
(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) - sq
rt(b^2*c + (b*x + a)*b*d - a*b*d))^4)*c^2*abs(b)) - 2*(b^4*c^2*abs(b) - 2*a*b^3*c*d*abs(b) + a^2*b^2*d^2*abs(b
))*sqrt(b*x + a)/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*b^2*c^2*d)

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maple [B]  time = 0.02, size = 502, normalized size = 3.06 \begin {gather*} \frac {\sqrt {b x +a}\, \left (3 \sqrt {b d}\, a^{3} d^{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 )-5 \sqrt {b d}\, a^{2} b 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 )+2 \sqrt {a c}\, b^{3} c^{2} d \,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}\, a^{3} c \,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 )-5 \sqrt {b d}\, a^{2} b \,c^{2} d 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 )+2 \sqrt {a c}\, b^{3} c^{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}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2} x +8 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d x -4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2} x -2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} c d \right )}{2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, \sqrt {d x +c}\, c^{2} d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

Timed out

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