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

Optimal. Leaf size=258 \[ -\frac {3 \sqrt {c} \left (5 a^2 d^2+10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a}}+\frac {3 \sqrt {d} \left (a^2 d^2+10 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b}}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{4 c x}+\frac {d \sqrt {a+b x} (c+d x)^{3/2} (5 a d+7 b c)}{4 c}+3 d \sqrt {a+b x} \sqrt {c+d x} (a d+b c) \]

[Out]

-1/2*(b*x+a)^(3/2)*(d*x+c)^(5/2)/x^2-3/4*(5*a^2*d^2+10*a*b*c*d+b^2*c^2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/
(d*x+c)^(1/2))*c^(1/2)/a^(1/2)+3/4*(a^2*d^2+10*a*b*c*d+5*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c
)^(1/2))*d^(1/2)/b^(1/2)+1/4*d*(5*a*d+7*b*c)*(d*x+c)^(3/2)*(b*x+a)^(1/2)/c-1/4*(5*a*d+3*b*c)*(d*x+c)^(5/2)*(b*
x+a)^(1/2)/c/x+3*d*(a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {97, 149, 154, 157, 63, 217, 206, 93, 208} \[ -\frac {3 \sqrt {c} \left (5 a^2 d^2+10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a}}+\frac {3 \sqrt {d} \left (a^2 d^2+10 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b}}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{4 c x}+\frac {d \sqrt {a+b x} (c+d x)^{3/2} (5 a d+7 b c)}{4 c}+3 d \sqrt {a+b x} \sqrt {c+d x} (a d+b c) \]

Antiderivative was successfully verified.

[In]

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

[Out]

3*d*(b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x] + (d*(7*b*c + 5*a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(4*c) - ((3*b*
c + 5*a*d)*Sqrt[a + b*x]*(c + d*x)^(5/2))/(4*c*x) - ((a + b*x)^(3/2)*(c + d*x)^(5/2))/(2*x^2) - (3*Sqrt[c]*(b^
2*c^2 + 10*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*Sqrt[a]) + (3*Sqr
t[d]*(5*b^2*c^2 + 10*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*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 97

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (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 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)^{3/2} (c+d x)^{5/2}}{x^3} \, dx &=-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}+\frac {1}{2} \int \frac {\sqrt {a+b x} (c+d x)^{3/2} \left (\frac {1}{2} (3 b c+5 a d)+4 b d x\right )}{x^2} \, dx\\ &=-\frac {(3 b c+5 a d) \sqrt {a+b x} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}+\frac {\int \frac {(c+d x)^{3/2} \left (\frac {3}{4} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right )+b d (7 b c+5 a d) x\right )}{x \sqrt {a+b x}} \, dx}{2 c}\\ &=\frac {d (7 b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(3 b c+5 a d) \sqrt {a+b x} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}+\frac {\int \frac {\sqrt {c+d x} \left (\frac {3}{2} b c \left (b^2 c^2+10 a b c d+5 a^2 d^2\right )+12 b^2 c d (b c+a d) x\right )}{x \sqrt {a+b x}} \, dx}{4 b c}\\ &=3 d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}+\frac {d (7 b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(3 b c+5 a d) \sqrt {a+b x} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}+\frac {\int \frac {\frac {3}{2} b^2 c^2 \left (b^2 c^2+10 a b c d+5 a^2 d^2\right )+\frac {3}{2} b^2 c d \left (5 b^2 c^2+10 a b c d+a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{4 b^2 c}\\ &=3 d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}+\frac {d (7 b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(3 b c+5 a d) \sqrt {a+b x} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}+\frac {1}{8} \left (3 d \left (5 b^2 c^2+10 a b c d+a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx+\frac {1}{8} \left (3 c \left (b^2 c^2+10 a b c d+5 a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx\\ &=3 d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}+\frac {d (7 b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(3 b c+5 a d) \sqrt {a+b x} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}+\frac {\left (3 d \left (5 b^2 c^2+10 a b c d+a^2 d^2\right )\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 )}{4 b}+\frac {1}{4} \left (3 c \left (b^2 c^2+10 a b c d+5 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 )\\ &=3 d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}+\frac {d (7 b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(3 b c+5 a d) \sqrt {a+b x} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}-\frac {3 \sqrt {c} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a}}+\frac {\left (3 d \left (5 b^2 c^2+10 a b c d+a^2 d^2\right )\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 )}{4 b}\\ &=3 d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}+\frac {d (7 b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(3 b c+5 a d) \sqrt {a+b x} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}-\frac {3 \sqrt {c} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a}}+\frac {3 \sqrt {d} \left (5 b^2 c^2+10 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 2.05, size = 234, normalized size = 0.91 \[ \frac {1}{4} \left (\frac {3 \sqrt {d} \sqrt {c+d x} \left (a^2 d^2+10 a b c d+5 b^2 c^2\right ) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}}}-\frac {3 \sqrt {c} \left (5 a^2 d^2+10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (a \left (-2 c^2-9 c d x+5 d^2 x^2\right )+b x \left (-5 c^2+9 c d x+2 d^2 x^2\right )\right )}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 5.65, size = 1185, normalized size = 4.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(5*b^2*c^2 + 10*a*b*c*d + 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 + 5*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*b*d^2*x^3 - 2*a*c^
2 + (9*b*c*d + 5*a*d^2)*x^2 - (5*b*c^2 + 9*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/x^2, -1/16*(6*(5*b^2*c^2 + 1
0*a*b*c*d + 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 + 5*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*b*d^2*x^3 - 2*a*c^2 + (9*b*c*d + 5*a*d^2)*x^2 - (5*b*c^2 + 9*a*c*d)*x)*sqr
t(b*x + a)*sqrt(d*x + c))/x^2, 1/16*(6*(b^2*c^2 + 10*a*b*c*d + 5*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)) + 3*(5*b^2*c^2 +
10*a*b*c*d + 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) + 4*(2*b*d^2*x^3 - 2*a*c^2 + (9*b*c*
d + 5*a*d^2)*x^2 - (5*b*c^2 + 9*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/x^2, 1/8*(3*(b^2*c^2 + 10*a*b*c*d + 5*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)) - 3*(5*b^2*c^2 + 10*a*b*c*d + 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)) + 2*(2*b*d^2*x^3 - 2*a*c^
2 + (9*b*c*d + 5*a*d^2)*x^2 - (5*b*c^2 + 9*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/x^2]

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giac [B]  time = 18.23, size = 1247, normalized size = 4.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*d^2*abs(b)/b + 3*(3*b*c*d^3*abs(b) + a*d^4*abs(b))/(b*
d^2))*sqrt(b*x + a) - 3*(5*sqrt(b*d)*b^2*c^2*abs(b) + 10*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b*d)*a^2*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 - 6*(sqrt(b*d)*b^3*c^3*abs(b) + 10*sq
rt(b*d)*a*b^2*c^2*d*abs(b) + 5*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)*b) - 4*(5*sqrt(b*d)*b^9*c^6
*abs(b) - 11*sqrt(b*d)*a*b^8*c^5*d*abs(b) - 6*sqrt(b*d)*a^2*b^7*c^4*d^2*abs(b) + 34*sqrt(b*d)*a^3*b^6*c^3*d^3*
abs(b) - 31*sqrt(b*d)*a^4*b^5*c^2*d^4*abs(b) + 9*sqrt(b*d)*a^5*b^4*c*d^5*abs(b) - 15*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) - 8*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr
t(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^6*c^4*d*abs(b) + 34*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) + 16*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) + 15*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) + 37*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) + 33*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*ab
s(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) - 5
*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) - 18*sqrt(b*d)*(sq
rt(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*abs(b) - 9*sqrt(b*d)*(sqrt(b*d)*sqr
t(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)/b

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maple [B]  time = 0.02, size = 650, normalized size = 2.52 \[ \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-15 \sqrt {b d}\, a^{2} c \,d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+3 \sqrt {a c}\, a^{2} d^{3} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-30 \sqrt {b d}\, a b \,c^{2} d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+30 \sqrt {a c}\, a b c \,d^{2} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-3 \sqrt {b d}\, b^{2} c^{3} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+15 \sqrt {a c}\, b^{2} c^{2} d \,x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+4 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b \,d^{2} x^{3}+10 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a \,d^{2} x^{2}+18 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b c d \,x^{2}-18 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a c d x -10 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b \,c^{2} x -4 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a \,c^{2}\right )}{8 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(3*(a*c)^(1/2)*a^2*d^3*x^2*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))+30*(a*c)^(1/2)*a*b*c*d^2*x^2*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a
*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))+15*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/
(b*d)^(1/2))*x^2*b^2*c^2*d*(a*c)^(1/2)-15*(b*d)^(1/2)*a^2*c*d^2*x^2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x
^2+a*d*x+b*c*x+a*c)^(1/2))/x)-30*(b*d)^(1/2)*a*b*c^2*d*x^2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+
b*c*x+a*c)^(1/2))/x)-3*(b*d)^(1/2)*b^2*c^3*x^2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(
1/2))/x)+4*x^3*b*d^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)+10*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2
)*(b*d)^(1/2)*(a*c)^(1/2)*a*d^2*x^2+18*x^2*b*c*d*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-18*(b
*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a*c*d*x-10*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(
a*c)^(1/2)*b*c^2*x-4*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a*c^2)/(b*d*x^2+a*d*x+b*c*x+a*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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(d*x+c)^(5/2)/x^3,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 = -0.00 \[ \int \frac {{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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