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

Optimal. Leaf size=275 \[ \frac {3 \sqrt {a+b x} \sqrt {c+d x} \left (a^2 d^2+6 a b c d+b^2 c^2\right )}{4 c}-\frac {3 \sqrt {a} \left (a^2 d^2+10 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {c}}+\frac {3 \sqrt {b} \left (5 a^2 d^2+10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {d}}-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{4 c x}+\frac {3 b \sqrt {a+b x} (c+d x)^{3/2} (a d+3 b c)}{4 c} \]

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Rubi [A]  time = 0.27, antiderivative size = 275, 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} \begin {gather*} \frac {3 \sqrt {a+b x} \sqrt {c+d x} \left (a^2 d^2+6 a b c d+b^2 c^2\right )}{4 c}-\frac {3 \sqrt {a} \left (a^2 d^2+10 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {c}}+\frac {3 \sqrt {b} \left (5 a^2 d^2+10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {d}}-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2} (3 a d+5 b c)}{4 c x}+\frac {3 b \sqrt {a+b x} (c+d x)^{3/2} (a d+3 b c)}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b^2*c^2 + 6*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*c) + (3*b*(3*b*c + a*d)*Sqrt[a + b*x]*(c +
d*x)^(3/2))/(4*c) - ((5*b*c + 3*a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(4*c*x) - ((a + b*x)^(5/2)*(c + d*x)^(3/
2))/(2*x^2) - (3*Sqrt[a]*(5*b^2*c^2 + 10*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c +
d*x])])/(4*Sqrt[c]) + (3*Sqrt[b]*(b^2*c^2 + 10*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*S
qrt[c + d*x])])/(4*Sqrt[d])

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)^{5/2} (c+d x)^{3/2}}{x^3} \, dx &=-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}+\frac {1}{2} \int \frac {(a+b x)^{3/2} \sqrt {c+d x} \left (\frac {1}{2} (5 b c+3 a d)+4 b d x\right )}{x^2} \, dx\\ &=-\frac {(5 b c+3 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}+\frac {\int \frac {\sqrt {a+b x} \sqrt {c+d x} \left (\frac {3}{4} \left (5 b^2 c^2+10 a b c d+a^2 d^2\right )+3 b d (3 b c+a d) x\right )}{x} \, dx}{2 c}\\ &=\frac {3 b (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(5 b c+3 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}+\frac {\int \frac {\sqrt {c+d x} \left (\frac {3}{2} a d \left (5 b^2 c^2+10 a b c d+a^2 d^2\right )+3 b d \left (b^2 c^2+6 a b c d+a^2 d^2\right ) x\right )}{x \sqrt {a+b x}} \, dx}{4 c d}\\ &=\frac {3 \left (b^2 c^2+6 a b c d+a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 c}+\frac {3 b (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(5 b c+3 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}+\frac {\int \frac {\frac {3}{2} a b c d \left (5 b^2 c^2+10 a b c d+a^2 d^2\right )+\frac {3}{2} b^2 c d \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{4 b c d}\\ &=\frac {3 \left (b^2 c^2+6 a b c d+a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 c}+\frac {3 b (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(5 b c+3 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}+\frac {1}{8} \left (3 a \left (5 b^2 c^2+10 a b c d+a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx+\frac {1}{8} \left (3 b \left (b^2 c^2+10 a b c d+5 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx\\ &=\frac {3 \left (b^2 c^2+6 a b c d+a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 c}+\frac {3 b (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(5 b c+3 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}+\frac {1}{4} \left (3 a \left (5 b^2 c^2+10 a b c d+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 )+\frac {1}{4} \left (3 \left (b^2 c^2+10 a b c d+5 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 )\\ &=\frac {3 \left (b^2 c^2+6 a b c d+a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 c}+\frac {3 b (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(5 b c+3 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}-\frac {3 \sqrt {a} \left (5 b^2 c^2+10 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {c}}+\frac {1}{4} \left (3 \left (b^2 c^2+10 a b c d+5 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 )\\ &=\frac {3 \left (b^2 c^2+6 a b c d+a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 c}+\frac {3 b (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(5 b c+3 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{2 x^2}-\frac {3 \sqrt {a} \left (5 b^2 c^2+10 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {c}}+\frac {3 \sqrt {b} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 2.09, size = 233, normalized size = 0.85 \begin {gather*} \frac {3 \sqrt {b c-a d} \left (5 a^2 d^2+10 a b c d+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 )}{4 \sqrt {d} \sqrt {c+d x}}-\frac {3 \sqrt {a} \left (a^2 d^2+10 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {c}}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (a^2 (2 c+5 d x)+9 a b x (c-d x)-b^2 x^2 (5 c+2 d x)\right )}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.79, size = 509, normalized size = 1.85 \begin {gather*} -\frac {3 \left (10 a^{3/2} b c d+a^{5/2} d^2+5 \sqrt {a} b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {c}}+\frac {3 \left (5 a^2 \sqrt {b} d^2+10 a b^{3/2} c d+b^{5/2} c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {d}}+\frac {\sqrt {a+b x} \left (-12 a^4 b^2 d^2-\frac {5 a^4 d^4 (a+b x)^2}{(c+d x)^2}+\frac {19 a^4 b d^3 (a+b x)}{c+d x}+\frac {15 a^3 b^2 c d^2 (a+b x)}{c+d x}+\frac {3 a^3 c d^4 (a+b x)^3}{(c+d x)^3}-\frac {26 a^3 b c d^3 (a+b x)^2}{(c+d x)^2}+12 a^2 b^4 c^2-\frac {15 a^2 b^3 c^2 d (a+b x)}{c+d x}+\frac {15 a^2 b c^2 d^3 (a+b x)^3}{(c+d x)^3}+\frac {5 b^4 c^4 (a+b x)^2}{(c+d x)^2}-\frac {19 a b^4 c^3 (a+b x)}{c+d x}-\frac {3 b^3 c^4 d (a+b x)^3}{(c+d x)^3}+\frac {26 a b^3 c^3 d (a+b x)^2}{(c+d x)^2}-\frac {15 a b^2 c^3 d^2 (a+b x)^3}{(c+d x)^3}\right )}{4 \sqrt {c+d x} \left (a-\frac {c (a+b x)}{c+d x}\right )^2 \left (\frac {d (a+b x)}{c+d x}-b\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(12*a^2*b^4*c^2 - 12*a^4*b^2*d^2 - (3*b^3*c^4*d*(a + b*x)^3)/(c + d*x)^3 - (15*a*b^2*c^3*d^2*(a
+ b*x)^3)/(c + d*x)^3 + (15*a^2*b*c^2*d^3*(a + b*x)^3)/(c + d*x)^3 + (3*a^3*c*d^4*(a + b*x)^3)/(c + d*x)^3 +
(5*b^4*c^4*(a + b*x)^2)/(c + d*x)^2 + (26*a*b^3*c^3*d*(a + b*x)^2)/(c + d*x)^2 - (26*a^3*b*c*d^3*(a + b*x)^2)/
(c + d*x)^2 - (5*a^4*d^4*(a + b*x)^2)/(c + d*x)^2 - (19*a*b^4*c^3*(a + b*x))/(c + d*x) - (15*a^2*b^3*c^2*d*(a
+ b*x))/(c + d*x) + (15*a^3*b^2*c*d^2*(a + b*x))/(c + d*x) + (19*a^4*b*d^3*(a + b*x))/(c + d*x)))/(4*Sqrt[c +
d*x]*(a - (c*(a + b*x))/(c + d*x))^2*(-b + (d*(a + b*x))/(c + d*x))^2) - (3*(5*Sqrt[a]*b^2*c^2 + 10*a^(3/2)*b*
c*d + a^(5/2)*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*Sqrt[c]) + (3*(b^(5/2)*c^2 + 1
0*a*b^(3/2)*c*d + 5*a^2*Sqrt[b]*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*Sqrt[d])

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fricas [A]  time = 5.96, size = 1185, normalized size = 4.31

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 17.77, size = 1239, normalized size = 4.51

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(3/2)/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*abs(b) + (5*b*c*d^2*abs(b) + 7*a*d^3*abs(b))/d^2)*sq
rt(b*x + a) - 3*(sqrt(b*d)*b^2*c^2*abs(b) + 10*sqrt(b*d)*a*b*c*d*abs(b) + 5*sqrt(b*d)*a^2*d^2*abs(b))*log((sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/d - 6*(5*sqrt(b*d)*a*b^3*c^2*abs(b) + 10*sqrt(b
*d)*a^2*b^2*c*d*abs(b) + sqrt(b*d)*a^3*b*d^2*abs(b))*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - s
qrt(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*(9*sqrt(b*d)*a*b^9*c^5*abs(b
) - 31*sqrt(b*d)*a^2*b^8*c^4*d*abs(b) + 34*sqrt(b*d)*a^3*b^7*c^3*d^2*abs(b) - 6*sqrt(b*d)*a^4*b^6*c^2*d^3*abs(
b) - 11*sqrt(b*d)*a^5*b^5*c*d^4*abs(b) + 5*sqrt(b*d)*a^6*b^4*d^5*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*b^7*c^4*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^2*b^6*c^3*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^3*b^5*c^2*d^2*abs(b) - 8*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^4*c*d^3*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*a^5*b^3*d^4*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*
b^5*c^3*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^4*c^2*d*
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^3*b^3*c*d^2*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*a^4*b^2*d^3*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))^6*a*b^3*c^2*abs(b) - 18*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^2*c*d*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*a^3*b*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 = 2.36 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-3 \sqrt {b d}\, a^{3} 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 )-30 \sqrt {b d}\, a^{2} b c 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 )+15 \sqrt {a c}\, a^{2} b \,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 )-15 \sqrt {b d}\, a \,b^{2} c^{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 )+30 \sqrt {a c}\, a \,b^{2} c 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 )+3 \sqrt {a c}\, b^{3} c^{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 )+4 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} d \,x^{3}+18 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b d \,x^{2}+10 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{2} c \,x^{2}-10 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} d x -18 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b c x -4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} c \right )}{8 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x+a)^(1/2)*(d*x+c)^(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*a^2*b*d^2*(a*c)^(1/2)+30*(a*c)^(1/2)*a*b^2*c*d*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))+3*(a*c)^(1/2)*b^3*c^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))-3*(b*d)^(1/2)*a^3*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^2*b*c*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)-15*(b*d)^(1/2)*a*b^2*c^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)+4*x^3*b^2*d*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)+18*x^2*a*b*d*(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)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*b^2*c*x^2-10*(b
*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*d*x-18*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+
a*c)^(1/2)*a*b*c*x-4*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*c)/(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 \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)*(d*x+c)^(3/2)/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 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{3/2}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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