3.614 \(\int \frac {(a+b x)^{3/2} (c+d x)^{3/2}}{x^5} \, dx\)
Optimal. Leaf size=201 \[ -\frac {3 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{64 a^{5/2} c^{5/2}}+\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^3}{64 a^2 c^2 x}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d)}{8 c^2 x^3}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)^2}{32 a c^2 x^2}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4} \]
[Out]
-1/4*(b*x+a)^(3/2)*(d*x+c)^(5/2)/c/x^4-3/64*(-a*d+b*c)^4*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/
a^(5/2)/c^(5/2)-1/32*(-a*d+b*c)^2*(d*x+c)^(3/2)*(b*x+a)^(1/2)/a/c^2/x^2-1/8*(-a*d+b*c)*(d*x+c)^(5/2)*(b*x+a)^(
1/2)/c^2/x^3+3/64*(-a*d+b*c)^3*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2/c^2/x
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Rubi [A] time = 0.10, antiderivative size = 201, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used =
{94, 93, 208} \[ \frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^3}{64 a^2 c^2 x}-\frac {3 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{64 a^{5/2} c^{5/2}}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)^2}{32 a c^2 x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d)}{8 c^2 x^3}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*x)^(3/2)*(c + d*x)^(3/2))/x^5,x]
[Out]
(3*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*a^2*c^2*x) - ((b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(3/2))/(
32*a*c^2*x^2) - ((b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(5/2))/(8*c^2*x^3) - ((a + b*x)^(3/2)*(c + d*x)^(5/2))/(4
*c*x^4) - (3*(b*c - a*d)^4*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(64*a^(5/2)*c^(5/2))
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 94
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 + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && !(SumSimplerQ[p, 1] && !SumSimplerQ[m, 1])
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]
Rubi steps
\begin {align*} \int \frac {(a+b x)^{3/2} (c+d x)^{3/2}}{x^5} \, dx &=-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}+\frac {(3 (b c-a d)) \int \frac {\sqrt {a+b x} (c+d x)^{3/2}}{x^4} \, dx}{8 c}\\ &=-\frac {(b c-a d) \sqrt {a+b x} (c+d x)^{5/2}}{8 c^2 x^3}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}+\frac {(b c-a d)^2 \int \frac {(c+d x)^{3/2}}{x^3 \sqrt {a+b x}} \, dx}{16 c^2}\\ &=-\frac {(b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/2}}{32 a c^2 x^2}-\frac {(b c-a d) \sqrt {a+b x} (c+d x)^{5/2}}{8 c^2 x^3}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}-\frac {\left (3 (b c-a d)^3\right ) \int \frac {\sqrt {c+d x}}{x^2 \sqrt {a+b x}} \, dx}{64 a c^2}\\ &=\frac {3 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 a^2 c^2 x}-\frac {(b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/2}}{32 a c^2 x^2}-\frac {(b c-a d) \sqrt {a+b x} (c+d x)^{5/2}}{8 c^2 x^3}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}+\frac {\left (3 (b c-a d)^4\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{128 a^2 c^2}\\ &=\frac {3 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 a^2 c^2 x}-\frac {(b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/2}}{32 a c^2 x^2}-\frac {(b c-a d) \sqrt {a+b x} (c+d x)^{5/2}}{8 c^2 x^3}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}+\frac {\left (3 (b c-a d)^4\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{64 a^2 c^2}\\ &=\frac {3 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 a^2 c^2 x}-\frac {(b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/2}}{32 a c^2 x^2}-\frac {(b c-a d) \sqrt {a+b x} (c+d x)^{5/2}}{8 c^2 x^3}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}-\frac {3 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{64 a^{5/2} c^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.45, size = 173, normalized size = 0.86 \[ -\frac {x (b c-a d) \left (\frac {x (b c-a d) \left (3 x^2 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\sqrt {a} \sqrt {c} \sqrt {a+b x} \sqrt {c+d x} (2 a c+5 a d x-3 b c x)\right )}{a^{5/2} \sqrt {c}}+8 \sqrt {a+b x} (c+d x)^{5/2}\right )+16 c (a+b x)^{3/2} (c+d x)^{5/2}}{64 c^2 x^4} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*x)^(3/2)*(c + d*x)^(3/2))/x^5,x]
[Out]
-1/64*(16*c*(a + b*x)^(3/2)*(c + d*x)^(5/2) + (b*c - a*d)*x*(8*Sqrt[a + b*x]*(c + d*x)^(5/2) + ((b*c - a*d)*x*
(Sqrt[a]*Sqrt[c]*Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*c - 3*b*c*x + 5*a*d*x) + 3*(b*c - a*d)^2*x^2*ArcTanh[(Sqrt[c
]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(a^(5/2)*Sqrt[c])))/(c^2*x^4)
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fricas [A] time = 4.77, size = 562, normalized size = 2.80 \[ \left [\frac {3 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {a c} x^{4} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (16 \, a^{4} c^{4} - {\left (3 \, a b^{3} c^{4} - 11 \, a^{2} b^{2} c^{3} d - 11 \, a^{3} b c^{2} d^{2} + 3 \, a^{4} c d^{3}\right )} x^{3} + 2 \, {\left (a^{2} b^{2} c^{4} + 22 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2}\right )} x^{2} + 24 \, {\left (a^{3} b c^{4} + a^{4} c^{3} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{256 \, a^{3} c^{3} x^{4}}, \frac {3 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-a c} x^{4} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (16 \, a^{4} c^{4} - {\left (3 \, a b^{3} c^{4} - 11 \, a^{2} b^{2} c^{3} d - 11 \, a^{3} b c^{2} d^{2} + 3 \, a^{4} c d^{3}\right )} x^{3} + 2 \, {\left (a^{2} b^{2} c^{4} + 22 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2}\right )} x^{2} + 24 \, {\left (a^{3} b c^{4} + a^{4} c^{3} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{128 \, a^{3} c^{3} x^{4}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(3/2)*(d*x+c)^(3/2)/x^5,x, algorithm="fricas")
[Out]
[1/256*(3*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*sqrt(a*c)*x^4*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 + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(
a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(16*a^4*c^4 - (3*a*b^3*c^4 - 11*a^2*b^2*c^3*d - 11*a^3*b*c^2*d^2 + 3*a^4*c*d^3)
*x^3 + 2*(a^2*b^2*c^4 + 22*a^3*b*c^3*d + a^4*c^2*d^2)*x^2 + 24*(a^3*b*c^4 + a^4*c^3*d)*x)*sqrt(b*x + a)*sqrt(d
*x + c))/(a^3*c^3*x^4), 1/128*(3*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*sqrt(
-a*c)*x^4*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (
a*b*c^2 + a^2*c*d)*x)) - 2*(16*a^4*c^4 - (3*a*b^3*c^4 - 11*a^2*b^2*c^3*d - 11*a^3*b*c^2*d^2 + 3*a^4*c*d^3)*x^3
+ 2*(a^2*b^2*c^4 + 22*a^3*b*c^3*d + a^4*c^2*d^2)*x^2 + 24*(a^3*b*c^4 + a^4*c^3*d)*x)*sqrt(b*x + a)*sqrt(d*x +
c))/(a^3*c^3*x^4)]
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giac [B] time = 16.84, size = 3832, normalized size = 19.06 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(3/2)*(d*x+c)^(3/2)/x^5,x, algorithm="giac")
[Out]
-1/64*(3*(sqrt(b*d)*b^5*c^4*abs(b) - 4*sqrt(b*d)*a*b^4*c^3*d*abs(b) + 6*sqrt(b*d)*a^2*b^3*c^2*d^2*abs(b) - 4*s
qrt(b*d)*a^3*b^2*c*d^3*abs(b) + sqrt(b*d)*a^4*b*d^4*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*c^2) - 2*(3*sqrt(b*d)*
b^19*c^11*abs(b) - 35*sqrt(b*d)*a*b^18*c^10*d*abs(b) + 161*sqrt(b*d)*a^2*b^17*c^9*d^2*abs(b) - 385*sqrt(b*d)*a
^3*b^16*c^8*d^3*abs(b) + 494*sqrt(b*d)*a^4*b^15*c^7*d^4*abs(b) - 238*sqrt(b*d)*a^5*b^14*c^6*d^5*abs(b) - 238*s
qrt(b*d)*a^6*b^13*c^5*d^6*abs(b) + 494*sqrt(b*d)*a^7*b^12*c^4*d^7*abs(b) - 385*sqrt(b*d)*a^8*b^11*c^3*d^8*abs(
b) + 161*sqrt(b*d)*a^9*b^10*c^2*d^9*abs(b) - 35*sqrt(b*d)*a^10*b^9*c*d^10*abs(b) + 3*sqrt(b*d)*a^11*b^8*d^11*a
bs(b) - 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^17*c^10*abs(b) + 178*
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^16*c^9*d*abs(b) - 561*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^15*c^8*d^2*abs(b) + 856*sqrt(b*d)*(s
qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^14*c^7*d^3*abs(b) - 698*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^13*c^6*d^4*abs(b) + 492*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^12*c^5*d^5*abs(b) - 698*sqrt(b*d)*(sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^6*b^11*c^4*d^6*abs(b) + 856*sqrt(b*d)*(sqrt(b*d)*sqrt(b*
x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^7*b^10*c^3*d^7*abs(b) - 561*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x +
a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^8*b^9*c^2*d^8*abs(b) + 178*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^9*b^8*c*d^9*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))^2*a^10*b^7*d^10*abs(b) + 63*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*
x + a)*b*d - a*b*d))^4*b^15*c^9*abs(b) - 377*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^14*c^8*d*abs(b) + 556*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^13*c^7*d^2*abs(b) - 36*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^12*c^6*d^3*abs(b) - 206*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^
11*c^5*d^4*abs(b) - 206*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^5*b^10*c
^4*d^5*abs(b) - 36*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^6*b^9*c^3*d^6
*abs(b) + 556*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^7*b^8*c^2*d^7*abs(
b) - 377*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^8*b^7*c*d^8*abs(b) + 63
*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^9*b^6*d^9*abs(b) - 105*sqrt(b*d
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^13*c^8*abs(b) + 440*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^12*c^7*d*abs(b) + 100*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^11*c^6*d^2*abs(b) - 632*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^10*c^5*d^3*abs(b) + 394*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^4*b^9*c^4*d^4*abs(b) - 632*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt
(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^5*b^8*c^3*d^5*abs(b) + 100*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*
c + (b*x + a)*b*d - a*b*d))^6*a^6*b^7*c^2*d^6*abs(b) + 440*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (
b*x + a)*b*d - a*b*d))^6*a^7*b^6*c*d^7*abs(b) - 105*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a
)*b*d - a*b*d))^6*a^8*b^5*d^8*abs(b) + 105*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a
*b*d))^8*b^11*c^7*abs(b) - 325*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a*b
^10*c^6*d*abs(b) - 551*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^2*b^9*c^5
*d^2*abs(b) - 253*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^3*b^8*c^4*d^3*
abs(b) - 253*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^4*b^7*c^3*d^4*abs(b
) - 551*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^5*b^6*c^2*d^5*abs(b) - 3
25*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^6*b^5*c*d^6*abs(b) + 105*sqrt
(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^7*b^4*d^7*abs(b) - 63*sqrt(b*d)*(sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^10*b^9*c^6*abs(b) + 170*sqrt(b*d)*(sqrt(b*d)*sqrt(
b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^10*a*b^8*c^5*d*abs(b) + 527*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^10*a^2*b^7*c^4*d^2*abs(b) + 780*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - s
qrt(b^2*c + (b*x + a)*b*d - a*b*d))^10*a^3*b^6*c^3*d^3*abs(b) + 527*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(
b^2*c + (b*x + a)*b*d - a*b*d))^10*a^4*b^5*c^2*d^4*abs(b) + 170*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*
c + (b*x + a)*b*d - a*b*d))^10*a^5*b^4*c*d^5*abs(b) - 63*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*
x + a)*b*d - a*b*d))^10*a^6*b^3*d^6*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))^12*b^7*c^5*abs(b) - 63*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^1
2*a*b^6*c^4*d*abs(b) - 342*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^12*a^2*b^
5*c^3*d^2*abs(b) - 342*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^12*a^3*b^4*c^
2*d^3*abs(b) - 63*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^12*a^4*b^3*c*d^4*a
bs(b) + 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^12*a^5*b^2*d^5*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))^14*b^5*c^4*abs(b) + 12*sqrt(b*d)*(s
qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^14*a*b^4*c^3*d*abs(b) + 110*sqrt(b*d)*(sqrt(b*d)
*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^14*a^2*b^3*c^2*d^2*abs(b) + 12*sqrt(b*d)*(sqrt(b*d)*sqrt
(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^14*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))^14*a^4*b*d^4*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)^4*a^2*c^2
))/b
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maple [B] time = 0.02, size = 705, normalized size = 3.51 \[ -\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (3 a^{4} d^{4} x^{4} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-12 a^{3} b c \,d^{3} x^{4} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+18 a^{2} b^{2} c^{2} d^{2} x^{4} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-12 a \,b^{3} c^{3} d \,x^{4} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+3 b^{4} c^{4} x^{4} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-6 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{3} d^{3} x^{3}+22 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} b c \,d^{2} x^{3}+22 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a \,b^{2} c^{2} d \,x^{3}-6 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{3} c^{3} x^{3}+4 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{3} c \,d^{2} x^{2}+88 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} b \,c^{2} d \,x^{2}+4 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a \,b^{2} c^{3} x^{2}+48 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{3} c^{2} d x +48 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} b \,c^{3} x +32 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {a c}\, a^{3} c^{3}\right )}{128 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {a c}\, a^{2} c^{2} x^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+a)^(3/2)*(d*x+c)^(3/2)/x^5,x)
[Out]
-1/128*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2/c^2*(3*a^4*d^4*x^4*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)-12*a^3*b*c*d^3*x^4*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)
+18*a^2*b^2*c^2*d^2*x^4*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)-12*a*b^3*c^3*d
*x^4*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^4*c^4*x^4*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)-6*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^3*d^3*x^3
+22*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*b*c*d^2*x^3+22*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)
*a*b^2*c^2*d*x^3-6*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*b^3*c^3*x^3+4*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+
a*c)^(1/2)*a^3*c*d^2*x^2+88*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*b*c^2*d*x^2+4*(a*c)^(1/2)*(b*d*x^2
+a*d*x+b*c*x+a*c)^(1/2)*a*b^2*c^3*x^2+48*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^3*c^2*d*x+48*(a*c)^(1/2
)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*b*c^3*x+32*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(a*c)^(1/2)*a^3*c^3)/(b*d*x^2
+a*d*x+b*c*x+a*c)^(1/2)/x^4/(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)^(3/2)/x^5,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 \[ \int \frac {{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{3/2}}{x^5} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((a + b*x)^(3/2)*(c + d*x)^(3/2))/x^5,x)
[Out]
int(((a + b*x)^(3/2)*(c + d*x)^(3/2))/x^5, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {3}{2}}}{x^{5}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)**(3/2)*(d*x+c)**(3/2)/x**5,x)
[Out]
Integral((a + b*x)**(3/2)*(c + d*x)**(3/2)/x**5, x)
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