3.7.55 \(\int \frac {(a+b x)^{5/2}}{x^4 \sqrt {c+d x}} \, dx\)

Optimal. Leaf size=157 \[ -\frac {5 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{7/2}}-\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}{8 c^3 x}-\frac {5 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{12 c^2 x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 c x^3} \]

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Rubi [A]  time = 0.07, antiderivative size = 157, 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} \begin {gather*} -\frac {5 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{12 c^2 x^2}-\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}{8 c^3 x}-\frac {5 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{7/2}}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/(x^4*Sqrt[c + d*x]),x]

[Out]

(-5*(b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*c^3*x) - (5*(b*c - a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(12*c
^2*x^2) - ((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*c*x^3) - (5*(b*c - a*d)^3*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a
]*Sqrt[c + d*x])])/(8*Sqrt[a]*c^(7/2))

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)^{5/2}}{x^4 \sqrt {c+d x}} \, dx &=-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 c x^3}+\frac {(5 (b c-a d)) \int \frac {(a+b x)^{3/2}}{x^3 \sqrt {c+d x}} \, dx}{6 c}\\ &=-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 c^2 x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 c x^3}+\frac {\left (5 (b c-a d)^2\right ) \int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx}{8 c^2}\\ &=-\frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 c^3 x}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 c^2 x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 c x^3}+\frac {\left (5 (b c-a d)^3\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 c^3}\\ &=-\frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 c^3 x}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 c^2 x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 c x^3}+\frac {\left (5 (b c-a d)^3\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{8 c^3}\\ &=-\frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 c^3 x}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 c^2 x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 c x^3}-\frac {5 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 142, normalized size = 0.90 \begin {gather*} -\frac {\frac {5 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-3 a d x+5 b c x)\right )}{\sqrt {a} c^{5/2}}+8 (a+b x)^{5/2} \sqrt {c+d x}}{24 c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/(x^4*Sqrt[c + d*x]),x]

[Out]

-1/24*(8*(a + b*x)^(5/2)*Sqrt[c + d*x] + (5*(b*c - a*d)*x*(Sqrt[a]*Sqrt[c]*Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*c
+ 5*b*c*x - 3*a*d*x) + 3*(b*c - a*d)^2*x^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(Sqrt[a]
*c^(5/2)))/(c*x^3)

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IntegrateAlgebraic [A]  time = 0.21, size = 160, normalized size = 1.02 \begin {gather*} \frac {5 (a d-b c)^3 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{8 \sqrt {a} c^{7/2}}-\frac {(b c-a d)^3 \left (\frac {15 a^2 (c+d x)^{5/2}}{(a+b x)^{5/2}}+\frac {33 c^2 \sqrt {c+d x}}{\sqrt {a+b x}}-\frac {40 a c (c+d x)^{3/2}}{(a+b x)^{3/2}}\right )}{24 c^3 \left (c-\frac {a (c+d x)}{a+b x}\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(5/2)/(x^4*Sqrt[c + d*x]),x]

[Out]

-1/24*((b*c - a*d)^3*((33*c^2*Sqrt[c + d*x])/Sqrt[a + b*x] - (40*a*c*(c + d*x)^(3/2))/(a + b*x)^(3/2) + (15*a^
2*(c + d*x)^(5/2))/(a + b*x)^(5/2)))/(c^3*(c - (a*(c + d*x))/(a + b*x))^3) + (5*(-(b*c) + a*d)^3*ArcTanh[(Sqrt
[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(8*Sqrt[a]*c^(7/2))

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fricas [A]  time = 3.03, size = 438, normalized size = 2.79 \begin {gather*} \left [-\frac {15 \, {\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 (33 \, a b^{2} c^{3} - 40 \, a^{2} b c^{2} d + 15 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (13 \, a^{2} b c^{3} - 5 \, a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, a c^{4} x^{3}}, \frac {15 \, {\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 (33 \, a b^{2} c^{3} - 40 \, a^{2} b c^{2} d + 15 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (13 \, a^{2} b c^{3} - 5 \, a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, a c^{4} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^4/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(15*(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 + (33*a*b^2*c^3 - 40*a^2*b*c^2*d + 15*a^3*c*d^2)*x^2 + 2*(13*a^2*b*c^3 - 5*a^3*c^2*d)*x
)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^4*x^3), 1/48*(15*(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 + (33*a*b^2*c^3 - 40*a^2*b*c^2*d + 15*a^3*c*d^2)*x^2 + 2*(13*a^2*b*c^3 -
5*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^4*x^3)]

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giac [B]  time = 29.81, size = 2071, normalized size = 13.19

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^4/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

-1/24*b*(15*(sqrt(b*d)*b^4*c^3 - 3*sqrt(b*d)*a*b^3*c^2*d + 3*sqrt(b*d)*a^2*b^2*c*d^2 - sqrt(b*d)*a^3*b*d^3)*ar
ctan(-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*c^3) + 2*(33*sqrt(b*d)*b^14*c^8 - 238*sqrt(b*d)*a*b^13*c^7*d + 750*sqrt(b*d)*a^2*b^12*c^
6*d^2 - 1350*sqrt(b*d)*a^3*b^11*c^5*d^3 + 1520*sqrt(b*d)*a^4*b^10*c^4*d^4 - 1098*sqrt(b*d)*a^5*b^9*c^3*d^5 + 4
98*sqrt(b*d)*a^6*b^8*c^2*d^6 - 130*sqrt(b*d)*a^7*b^7*c*d^7 + 15*sqrt(b*d)*a^8*b^6*d^8 - 165*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 + 747*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 - 1233*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 + 687*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 + 417*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 - 783*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 + 405*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 - 75*sqr
t(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 + 330*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 - 768*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*b^9*c^5*d + 546*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 - 288*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 + 510*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 - 480*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
+ 150*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 - 330*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 + 202*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 - 108*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 - 180*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 + 310*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 - 150*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 + 165*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 + 102*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 - 150*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 + 75*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 - 33*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 - 45*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 + 45*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 - 1
5*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)/((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)*s
qrt(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)^3*c^3))/abs(b)

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maple [B]  time = 0.02, size = 405, normalized size = 2.58 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (15 a^{3} d^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-45 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 {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+45 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 {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-15 b^{3} c^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} d^{2} x^{2}+80 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b c d \,x^{2}-66 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, b^{2} c^{2} x^{2}+20 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c d x -52 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b \,c^{2} x -16 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, c^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)/x^4/(d*x+c)^(1/2),x)

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c^3*(15*a^3*d^3*x^3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/
2))/x)-45*a^2*b*c*d^2*x^3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)+45*a*b^2*c^2*d*x^3*l
n((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)-15*b^3*c^3*x^3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1
/2)*((b*x+a)*(d*x+c))^(1/2))/x)-30*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*d^2*x^2+80*((b*x+a)*(d*x+c))^(1/2)*
(a*c)^(1/2)*a*b*c*d*x^2-66*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*b^2*c^2*x^2+20*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1
/2)*a^2*c*d*x-52*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*b*c^2*x-16*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*c^2)
/((b*x+a)*(d*x+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)^(5/2)/x^4/(d*x+c)^(1/2),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 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{5/2}}{x^4\,\sqrt {c+d\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(5/2)/(x^4*(c + d*x)^(1/2)),x)

[Out]

int((a + b*x)^(5/2)/(x^4*(c + d*x)^(1/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)/x**4/(d*x+c)**(1/2),x)

[Out]

Timed out

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