3.6.29 \(\int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^4} \, dx\)

Optimal. Leaf size=172 \[ -\frac {(a d+b c) (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} c^{5/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (\frac {b^2}{a^2}-\frac {d^2}{c^2}\right )}{8 x}+\frac {\sqrt {a+b x} (c+d x)^{3/2} (a d+b c)}{4 a c^2 x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 a c x^3} \]

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Rubi [A]  time = 0.09, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {96, 94, 93, 208} \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} \left (\frac {b^2}{a^2}-\frac {d^2}{c^2}\right )}{8 x}-\frac {(a d+b c) (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} c^{5/2}}+\frac {\sqrt {a+b x} (c+d x)^{3/2} (a d+b c)}{4 a c^2 x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x]*Sqrt[c + d*x])/x^4,x]

[Out]

((b^2/a^2 - d^2/c^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*x) + ((b*c + a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(4*a*c^2
*x^2) - ((a + b*x)^(3/2)*(c + d*x)^(3/2))/(3*a*c*x^3) - ((b*c - a*d)^2*(b*c + a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b
*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[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 {\sqrt {a+b x} \sqrt {c+d x}}{x^4} \, dx &=-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 a c x^3}-\frac {(b c+a d) \int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^3} \, dx}{2 a c}\\ &=\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 a c^2 x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 a c x^3}-\frac {\left (b^2-\frac {a^2 d^2}{c^2}\right ) \int \frac {\sqrt {c+d x}}{x^2 \sqrt {a+b x}} \, dx}{8 a}\\ &=\frac {\left (b^2-\frac {a^2 d^2}{c^2}\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 x}+\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 a c^2 x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 a c x^3}+\frac {\left ((b c-a d)^2 (b c+a d)\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 a^2 c^2}\\ &=\frac {\left (b^2-\frac {a^2 d^2}{c^2}\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 x}+\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 a c^2 x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 a c x^3}+\frac {\left ((b c-a d)^2 (b c+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 )}{8 a^2 c^2}\\ &=\frac {\left (b^2-\frac {a^2 d^2}{c^2}\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 x}+\frac {(b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 a c^2 x^2}-\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 a c x^3}-\frac {(b c-a d)^2 (b c+a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 142, normalized size = 0.83 \begin {gather*} -\frac {\frac {3 x (a d+b c) \left (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+a d x+b c x)\right )}{a^{3/2} c^{3/2}}+8 (a+b x)^{3/2} (c+d x)^{3/2}}{24 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x]*Sqrt[c + d*x])/x^4,x]

[Out]

-1/24*(8*(a + b*x)^(3/2)*(c + d*x)^(3/2) + (3*(b*c + a*d)*x*(-(Sqrt[a]*Sqrt[c]*Sqrt[a + b*x]*Sqrt[c + d*x]*(2*
a*c + b*c*x + a*d*x)) + (b*c - a*d)^2*x^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(a^(3/2)*
c^(3/2)))/(a*c*x^3)

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IntegrateAlgebraic [A]  time = 0.35, size = 213, normalized size = 1.24 \begin {gather*} \frac {\sqrt {c+d x} (a d-b c)^2 \left (\frac {3 a^3 d (c+d x)^2}{(a+b x)^2}+\frac {3 a^2 b c (c+d x)^2}{(a+b x)^2}-\frac {8 a^2 c d (c+d x)}{a+b x}+\frac {8 a b c^2 (c+d x)}{a+b x}-3 a c^2 d-3 b c^3\right )}{24 a^2 c^2 \sqrt {a+b x} \left (\frac {a (c+d x)}{a+b x}-c\right )^3}-\frac {(a d-b c)^2 (a d+b c) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{8 a^{5/2} c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[a + b*x]*Sqrt[c + d*x])/x^4,x]

[Out]

((-(b*c) + a*d)^2*Sqrt[c + d*x]*(-3*b*c^3 - 3*a*c^2*d + (8*a*b*c^2*(c + d*x))/(a + b*x) - (8*a^2*c*d*(c + d*x)
)/(a + b*x) + (3*a^2*b*c*(c + d*x)^2)/(a + b*x)^2 + (3*a^3*d*(c + d*x)^2)/(a + b*x)^2))/(24*a^2*c^2*Sqrt[a + b
*x]*(-c + (a*(c + d*x))/(a + b*x))^3) - ((-(b*c) + a*d)^2*(b*c + a*d)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]
*Sqrt[a + b*x])])/(8*a^(5/2)*c^(5/2))

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fricas [A]  time = 2.35, size = 434, normalized size = 2.52 \begin {gather*} \left [\frac {3 \, {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {a c} x^{3} \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 (8 \, a^{3} c^{3} - {\left (3 \, a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + 3 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (a^{2} b c^{3} + a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, a^{3} c^{3} x^{3}}, \frac {3 \, {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {-a c} x^{3} \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 (8 \, a^{3} c^{3} - {\left (3 \, a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + 3 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (a^{2} b c^{3} + a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, a^{3} c^{3} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(d*x+c)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[1/96*(3*(b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*sqrt(a*c)*x^3*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*(8*a^3*c^3 - (3*a*b^2*c^3 - 2*a^2*b*c^2*d + 3*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x + a
)*sqrt(d*x + c))/(a^3*c^3*x^3), 1/48*(3*(b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*sqrt(-a*c)*x^3*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*(8*a^3*c^3 - (3*a*b^2*c^3 - 2*a^2*b*c^2*d + 3*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x
+ a)*sqrt(d*x + c))/(a^3*c^3*x^3)]

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giac [B]  time = 39.91, size = 2104, normalized size = 12.23

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(d*x+c)^(1/2)/x^4,x, algorithm="giac")

[Out]

-1/24*(3*(sqrt(b*d)*b^4*c^3*abs(b) - sqrt(b*d)*a*b^3*c^2*d*abs(b) - sqrt(b*d)*a^2*b^2*c*d^2*abs(b) + sqrt(b*d)
*a^3*b*d^3*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^14*c^8*abs(b) - 20*sqrt(b*d)*a*b^13*c^7
*d*abs(b) + 60*sqrt(b*d)*a^2*b^12*c^6*d^2*abs(b) - 108*sqrt(b*d)*a^3*b^11*c^5*d^3*abs(b) + 130*sqrt(b*d)*a^4*b
^10*c^4*d^4*abs(b) - 108*sqrt(b*d)*a^5*b^9*c^3*d^5*abs(b) + 60*sqrt(b*d)*a^6*b^8*c^2*d^6*abs(b) - 20*sqrt(b*d)
*a^7*b^7*c*d^7*abs(b) + 3*sqrt(b*d)*a^8*b^6*d^8*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^12*c^7*abs(b) + 51*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^11*c^6*d*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)
)^2*a^2*b^10*c^5*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))^2*a
^3*b^9*c^4*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^8
*c^3*d^4*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))^2*a^5*b^7*c^2*d
^5*abs(b) + 51*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^6*b^6*c*d^6*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^7*b^5*d^7*abs(b) + 30*sqr
t(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^10*c^6*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*b^9*c^5*d*abs(b) - 30*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^8*c^4*d^2*abs(b) + 72*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^7*c^3*d^3*abs(b) - 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - s
qrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^4*b^6*c^2*d^4*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^5*b^5*c*d^5*abs(b) + 30*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^4*d^6*abs(b) - 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*
d - a*b*d))^6*b^8*c^5*abs(b) + 2*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^7*c^4*d*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))^6*a^2*b^6*c^3
*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))^6*a^3*b^5*c^2*d^3*a
bs(b) + 2*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^4*c*d^4*abs(b) - 3
0*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^3*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))^8*b^6*c^4*abs(b) + 66*sqrt(b*d)*(sqrt(b*d)*s
qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^2*b^4*c^2*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))^8*a^4*b^2*d^4*abs(b) - 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq
rt(b^2*c + (b*x + a)*b*d - a*b*d))^10*b^4*c^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))^10*a*b^3*c^2*d*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))^10*a^2*b^2*c*d^2*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
))^10*a^3*b*d^3*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 + (sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^3*a^2*c^2))/b

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maple [B]  time = 0.02, size = 485, normalized size = 2.82 \begin {gather*} -\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (3 a^{3} d^{3} x^{3} \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 a^{2} b c \,d^{2} x^{3} \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 a \,b^{2} c^{2} d \,x^{3} \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^{3} c^{3} x^{3} \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^{2} d^{2} x^{2}+4 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b c d \,x^{2}-6 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{2} c^{2} x^{2}+4 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} c d x +4 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b \,c^{2} x +16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {a c}\, a^{2} c^{2}\right )}{48 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {a c}\, a^{2} c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)*(d*x+c)^(1/2)/x^4,x)

[Out]

-1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2/c^2*(3*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)*x^3*a^3*d^3-3*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)*x^3*a^2*b*c*d^2-3
*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)*x^3*a*b^2*c^2*d+3*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)*x^3*b^3*c^3-6*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)
*x^2*a^2*d^2+4*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x^2*a*b*c*d-6*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)*x^2*b^2*c^2+4*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^2*c*d+4*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+
a*c)^(1/2)*x*a*b*c^2+16*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*c^2*(a*c)^(1/2))/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/x
^3/(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)^(1/2)*(d*x+c)^(1/2)/x^4,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 [B]  time = 93.34, size = 1459, normalized size = 8.48

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^(1/2)*(c + d*x)^(1/2))/x^4,x)

[Out]

((((a + b*x)^(1/2) - a^(1/2))^4*((5*b^6*c^4)/64 + (5*a^4*b^2*d^4)/64 + (9*a^3*b^3*c*d^3)/32 - (3*a^2*b^4*c^2*d
^2)/32 + (9*a*b^5*c^3*d)/32))/(a^3*c^3*d^3*((c + d*x)^(1/2) - c^(1/2))^4) - (((a + b*x)^(1/2) - a^(1/2))^3*((1
7*b^6*c^3)/192 + (17*a^3*b^3*d^3)/192))/(a^(5/2)*c^(5/2)*d^3*((c + d*x)^(1/2) - c^(1/2))^3) - b^6/(192*a*c*d^3
) - (((a + b*x)^(1/2) - a^(1/2))^5*((13*a^4*b^2*c*d^4)/32 - (a^5*b*d^5)/64 - (b^6*c^5)/64 + (a^2*b^4*c^3*d^2)/
16 + (a^3*b^3*c^2*d^3)/16 + (13*a*b^5*c^4*d)/32))/(a^(7/2)*c^(7/2)*d^3*((c + d*x)^(1/2) - c^(1/2))^5) + (((a +
b*x)^(1/2) - a^(1/2))^2*((b^6*c^2)/64 + (a^2*b^4*d^2)/64 - (a*b^5*c*d)/16))/(a^2*c^2*d^3*((c + d*x)^(1/2) - c
^(1/2))^2) + (((a + b*x)^(1/2) - a^(1/2))^7*((3*a^5*d^5)/64 + (3*b^5*c^5)/64 + (3*a^2*b^3*c^3*d^2)/64 + (3*a^3
*b^2*c^2*d^3)/64 - (15*a*b^4*c^4*d)/64 - (15*a^4*b*c*d^4)/64))/(a^(7/2)*c^(7/2)*d^2*((c + d*x)^(1/2) - c^(1/2)
)^7) + (((b^6*c)/64 + (a*b^5*d)/64)*((a + b*x)^(1/2) - a^(1/2)))/(a^(3/2)*c^(3/2)*d^3*((c + d*x)^(1/2) - c^(1/
2))) + (((a + b*x)^(1/2) - a^(1/2))^6*((15*a^2*b^4*c^4*d^2)/32 - (5*b^6*c^6)/192 - (5*a^6*d^6)/192 - (3*a^3*b^
3*c^3*d^3)/8 + (15*a^4*b^2*c^2*d^4)/32 + (3*a*b^5*c^5*d)/32 + (3*a^5*b*c*d^5)/32))/(a^4*c^4*d^3*((c + d*x)^(1/
2) - c^(1/2))^6) - (((a + b*x)^(1/2) - a^(1/2))^8*((a^4*d^4)/64 + (b^4*c^4)/64 + (7*a^2*b^2*c^2*d^2)/64 - (3*a
*b^3*c^3*d)/32 - (3*a^3*b*c*d^3)/32))/(a^3*c^3*d*((c + d*x)^(1/2) - c^(1/2))^8))/(((a + b*x)^(1/2) - a^(1/2))^
9/((c + d*x)^(1/2) - c^(1/2))^9 + (b^3*((a + b*x)^(1/2) - a^(1/2))^3)/(d^3*((c + d*x)^(1/2) - c^(1/2))^3) + ((
(a + b*x)^(1/2) - a^(1/2))^7*(3*a^2*d^2 + 3*b^2*c^2 + 9*a*b*c*d))/(a*c*d^2*((c + d*x)^(1/2) - c^(1/2))^7) - ((
(a + b*x)^(1/2) - a^(1/2))^6*(a^3*d^3 + b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d^2))/(a^(3/2)*c^(3/2)*d^3*((c + d
*x)^(1/2) - c^(1/2))^6) - ((3*a*d + 3*b*c)*((a + b*x)^(1/2) - a^(1/2))^8)/(a^(1/2)*c^(1/2)*d*((c + d*x)^(1/2)
- c^(1/2))^8) - ((3*b^3*c + 3*a*b^2*d)*((a + b*x)^(1/2) - a^(1/2))^4)/(a^(1/2)*c^(1/2)*d^3*((c + d*x)^(1/2) -
c^(1/2))^4) + (((a + b*x)^(1/2) - a^(1/2))^5*(3*b^3*c^2 + 3*a^2*b*d^2 + 9*a*b^2*c*d))/(a*c*d^3*((c + d*x)^(1/2
) - c^(1/2))^5)) + (((d*(3*a^2*d^2 + 3*b^2*c^2 + 8*a*b*c*d))/(32*a^2*c^2) - (4*a^2*d^7 + 4*b^2*c^2*d^5 + 17*a*
b*c*d^6)/(64*a^2*c^2*d^4))*((a + b*x)^(1/2) - a^(1/2)))/((c + d*x)^(1/2) - c^(1/2)) + (log(((c^(1/2)*(a + b*x)
^(1/2) - a^(1/2)*(c + d*x)^(1/2))*(b*c^(1/2) - (a^(1/2)*d*((a + b*x)^(1/2) - a^(1/2)))/((c + d*x)^(1/2) - c^(1
/2))))/((c + d*x)^(1/2) - c^(1/2)))*(a^(1/2)*b^3*c^(7/2) + a^(7/2)*c^(1/2)*d^3 - a^(3/2)*b^2*c^(5/2)*d - a^(5/
2)*b*c^(3/2)*d^2))/(16*a^3*c^3) - (log(((a + b*x)^(1/2) - a^(1/2))/((c + d*x)^(1/2) - c^(1/2)))*(a^(1/2)*b^3*c
^(7/2) + a^(7/2)*c^(1/2)*d^3 - a^(3/2)*b^2*c^(5/2)*d - a^(5/2)*b*c^(3/2)*d^2))/(16*a^3*c^3) - (d^3*((a + b*x)^
(1/2) - a^(1/2))^3)/(192*a*c*((c + d*x)^(1/2) - c^(1/2))^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x} \sqrt {c + d x}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)*(d*x+c)**(1/2)/x**4,x)

[Out]

Integral(sqrt(a + b*x)*sqrt(c + d*x)/x**4, x)

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