3.563 \(\int \frac {\sqrt {a+b x} (c+d x)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=163 \[ -\frac {(b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} c^{3/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}{8 a^2 c x}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 c x^3}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}{12 a c x^2} \]

[Out]

-1/8*(-a*d+b*c)^3*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(5/2)/c^(3/2)-1/12*(-a*d+b*c)*(d*x+c)
^(3/2)*(b*x+a)^(1/2)/a/c/x^2-1/3*(d*x+c)^(5/2)*(b*x+a)^(1/2)/c/x^3+1/8*(-a*d+b*c)^2*(b*x+a)^(1/2)*(d*x+c)^(1/2
)/a^2/c/x

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Rubi [A]  time = 0.07, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {94, 93, 208} \[ -\frac {(b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} c^{3/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}{8 a^2 c x}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}{12 a c x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 c x^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x]*(c + d*x)^(3/2))/x^4,x]

[Out]

((b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*a^2*c*x) - ((b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(12*a*c*
x^2) - (Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*c*x^3) - ((b*c - a*d)^3*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqr
t[c + d*x])])/(8*a^(5/2)*c^(3/2))

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 {\sqrt {a+b x} (c+d x)^{3/2}}{x^4} \, dx &=-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 c x^3}+\frac {(b c-a d) \int \frac {(c+d x)^{3/2}}{x^3 \sqrt {a+b x}} \, dx}{6 c}\\ &=-\frac {(b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a c x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 c x^3}-\frac {(b c-a d)^2 \int \frac {\sqrt {c+d x}}{x^2 \sqrt {a+b x}} \, dx}{8 a c}\\ &=\frac {(b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 c x}-\frac {(b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a c x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 c x^3}+\frac {(b c-a d)^3 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 a^2 c}\\ &=\frac {(b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 c x}-\frac {(b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a c x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 c x^3}+\frac {(b c-a d)^3 \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}\\ &=\frac {(b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 c x}-\frac {(b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a c x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 c x^3}-\frac {(b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 141, normalized size = 0.87 \[ -\frac {\frac {x (b c-a d) \left (3 x^2 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\sqrt {a} \sqrt {c} \sqrt {a+b x} \sqrt {c+d x} (2 a c+5 a d x-3 b c x)\right )}{a^{5/2} \sqrt {c}}+8 \sqrt {a+b x} (c+d x)^{5/2}}{24 c x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x]*(c + d*x)^(3/2))/x^4,x]

[Out]

-1/24*(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)*S
qrt[c]))/(c*x^3)

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fricas [A]  time = 1.25, size = 438, normalized size = 2.69 \[ \left [-\frac {3 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a c} x^{3} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (8 \, a^{3} c^{3} - {\left (3 \, a b^{2} c^{3} - 8 \, a^{2} b c^{2} d - 3 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (a^{2} b c^{3} + 7 \, a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, a^{3} c^{2} x^{3}}, \frac {3 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-a c} x^{3} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (8 \, a^{3} c^{3} - {\left (3 \, a b^{2} c^{3} - 8 \, a^{2} b c^{2} d - 3 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (a^{2} b c^{3} + 7 \, a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, a^{3} c^{2} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)*(b*x+a)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[-1/96*(3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*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 - 8*a^2*b*c^2*d - 3*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 7*a^3*c^2*d)*x)*sqrt(
b*x + a)*sqrt(d*x + c))/(a^3*c^2*x^3), 1/48*(3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*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 - 8*a^2*b*c^2*d - 3*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 7*a^3*c^2*d
)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c^2*x^3)]

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giac [B]  time = 39.82, size = 2161, normalized size = 13.26 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)*(b*x+a)^(1/2)/x^4,x, algorithm="giac")

[Out]

-1/24*(3*(sqrt(b*d)*b^4*c^3*abs(b) - 3*sqrt(b*d)*a*b^3*c^2*d*abs(b) + 3*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*(3*sqrt(b*d)*b^14*c^8*abs(b) - 26*sqrt(b*d)*a*b^13*c
^7*d*abs(b) + 90*sqrt(b*d)*a^2*b^12*c^6*d^2*abs(b) - 162*sqrt(b*d)*a^3*b^11*c^5*d^3*abs(b) + 160*sqrt(b*d)*a^4
*b^10*c^4*d^4*abs(b) - 78*sqrt(b*d)*a^5*b^9*c^3*d^5*abs(b) + 6*sqrt(b*d)*a^6*b^8*c^2*d^6*abs(b) + 10*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) + 81*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) - 147*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) + 93*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) + 3*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) + 3*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) - 33*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*sqrt
(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) - 96*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) + 6*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) + 96*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) - 6*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^6*c^2*d^4*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) + 62*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) + 60*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*abs(b) + 98*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) + 30*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*s
qrt(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) - 30*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^5*c^3*d*abs(b) - 48*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^4*c^2*d^2*abs(b) - 114*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^3*c*d^3*abs(b) - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - s
qrt(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) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^10*b^4*c^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))^10*a*b^3*c^2*d*abs(b) + 39*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 + (sqrt(b*d)*sq
rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^3*a^2*c))/b

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maple [B]  time = 0.02, size = 485, normalized size = 2.98 \[ \frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (3 a^{3} d^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-9 a^{2} b c \,d^{2} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+9 a \,b^{2} c^{2} d \,x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-3 b^{3} c^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-6 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} d^{2} x^{2}-16 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b c d \,x^{2}+6 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{2} c^{2} x^{2}-28 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} c d x -4 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b \,c^{2} x -16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {a c}\, a^{2} c^{2}\right )}{48 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {a c}\, a^{2} c \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(3/2)*(b*x+a)^(1/2)/x^4,x)

[Out]

1/48*(d*x+c)^(1/2)*(b*x+a)^(1/2)/a^2/c*(3*a^3*d^3*x^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)-9*a^2*b*c*d^2*x^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)+9*a*
b^2*c^2*d*x^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)-3*b^3*c^3*x^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)-6*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^
2*d^2*x^2-16*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a*b*c*d*x^2+6*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(
1/2)*b^2*c^2*x^2-28*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*c*d*x-4*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a
*c)^(1/2)*a*b*c^2*x-16*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(a*c)^(1/2)*a^2*c^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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)*(b*x+a)^(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 [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^(1/2)*(c + d*x)^(3/2))/x^4,x)

[Out]

int(((a + b*x)^(1/2)*(c + d*x)^(3/2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b x} \left (c + d x\right )^{\frac {3}{2}}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(3/2)*(b*x+a)**(1/2)/x**4,x)

[Out]

Integral(sqrt(a + b*x)*(c + d*x)**(3/2)/x**4, x)

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