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

Optimal. Leaf size=142 \[ \frac {5 c^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{7/2}}-\frac {5 c \sqrt {c+d x} (b c-a d)}{a^3 \sqrt {a+b x}}-\frac {5 (c+d x)^{3/2} (b c-a d)}{3 a^2 (a+b x)^{3/2}}-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}} \]

[Out]

-5/3*(-a*d+b*c)*(d*x+c)^(3/2)/a^2/(b*x+a)^(3/2)-(d*x+c)^(5/2)/a/x/(b*x+a)^(3/2)+5*c^(3/2)*(-a*d+b*c)*arctanh(c
^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(7/2)-5*c*(-a*d+b*c)*(d*x+c)^(1/2)/a^3/(b*x+a)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 142, 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} \[ \frac {5 c^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{7/2}}-\frac {5 (c+d x)^{3/2} (b c-a d)}{3 a^2 (a+b x)^{3/2}}-\frac {5 c \sqrt {c+d x} (b c-a d)}{a^3 \sqrt {a+b x}}-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*c*(b*c - a*d)*Sqrt[c + d*x])/(a^3*Sqrt[a + b*x]) - (5*(b*c - a*d)*(c + d*x)^(3/2))/(3*a^2*(a + b*x)^(3/2))
- (c + d*x)^(5/2)/(a*x*(a + b*x)^(3/2)) + (5*c^(3/2)*(b*c - a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqr
t[c + d*x])])/a^(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 {(c+d x)^{5/2}}{x^2 (a+b x)^{5/2}} \, dx &=-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}}-\frac {(5 (b c-a d)) \int \frac {(c+d x)^{3/2}}{x (a+b x)^{5/2}} \, dx}{2 a}\\ &=-\frac {5 (b c-a d) (c+d x)^{3/2}}{3 a^2 (a+b x)^{3/2}}-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}}-\frac {(5 c (b c-a d)) \int \frac {\sqrt {c+d x}}{x (a+b x)^{3/2}} \, dx}{2 a^2}\\ &=-\frac {5 c (b c-a d) \sqrt {c+d x}}{a^3 \sqrt {a+b x}}-\frac {5 (b c-a d) (c+d x)^{3/2}}{3 a^2 (a+b x)^{3/2}}-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}}-\frac {\left (5 c^2 (b c-a d)\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 a^3}\\ &=-\frac {5 c (b c-a d) \sqrt {c+d x}}{a^3 \sqrt {a+b x}}-\frac {5 (b c-a d) (c+d x)^{3/2}}{3 a^2 (a+b x)^{3/2}}-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}}-\frac {\left (5 c^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 )}{a^3}\\ &=-\frac {5 c (b c-a d) \sqrt {c+d x}}{a^3 \sqrt {a+b x}}-\frac {5 (b c-a d) (c+d x)^{3/2}}{3 a^2 (a+b x)^{3/2}}-\frac {(c+d x)^{5/2}}{a x (a+b x)^{3/2}}+\frac {5 c^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 125, normalized size = 0.88 \[ -\frac {3 a^{5/2} (c+d x)^{5/2}+5 x (b c-a d) \left (\sqrt {a} \sqrt {c+d x} (4 a c+a d x+3 b c x)-3 c^{3/2} (a+b x)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )}{3 a^{7/2} x (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(3*a^(5/2)*(c + d*x)^(5/2) + 5*(b*c - a*d)*x*(Sqrt[a]*Sqrt[c + d*x]*(4*a*c + 3*b*c*x + a*d*x) - 3*c^(3/2)
*(a + b*x)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(a^(7/2)*x*(a + b*x)^(3/2))

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fricas [B]  time = 1.72, size = 505, normalized size = 3.56 \[ \left [-\frac {15 \, {\left ({\left (b^{3} c^{2} - a b^{2} c d\right )} x^{3} + 2 \, {\left (a b^{2} c^{2} - a^{2} b c d\right )} x^{2} + {\left (a^{2} b c^{2} - a^{3} c d\right )} x\right )} \sqrt {\frac {c}{a}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (3 \, a^{2} c^{2} + {\left (15 \, b^{2} c^{2} - 10 \, a b c d - 2 \, a^{2} d^{2}\right )} x^{2} + 2 \, {\left (10 \, a b c^{2} - 7 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{12 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}, -\frac {15 \, {\left ({\left (b^{3} c^{2} - a b^{2} c d\right )} x^{3} + 2 \, {\left (a b^{2} c^{2} - a^{2} b c d\right )} x^{2} + {\left (a^{2} b c^{2} - a^{3} c d\right )} x\right )} \sqrt {-\frac {c}{a}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) + 2 \, {\left (3 \, a^{2} c^{2} + {\left (15 \, b^{2} c^{2} - 10 \, a b c d - 2 \, a^{2} d^{2}\right )} x^{2} + 2 \, {\left (10 \, a b c^{2} - 7 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(15*((b^3*c^2 - a*b^2*c*d)*x^3 + 2*(a*b^2*c^2 - a^2*b*c*d)*x^2 + (a^2*b*c^2 - a^3*c*d)*x)*sqrt(c/a)*log
((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c
)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(3*a^2*c^2 + (15*b^2*c^2 - 10*a*b*c*d - 2*a^2*d^2)*x^2 + 2*(10
*a*b*c^2 - 7*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x), -1/6*(15*((b^3*c^2
- a*b^2*c*d)*x^3 + 2*(a*b^2*c^2 - a^2*b*c*d)*x^2 + (a^2*b*c^2 - a^3*c*d)*x)*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*
c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) + 2*(3*a^2*c^2 + (
15*b^2*c^2 - 10*a*b*c*d - 2*a^2*d^2)*x^2 + 2*(10*a*b*c^2 - 7*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b^2
*x^3 + 2*a^4*b*x^2 + a^5*x)]

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giac [B]  time = 18.59, size = 990, normalized size = 6.97 \[ \frac {5 \, {\left (\sqrt {b d} b c^{3} {\left | b \right |} - \sqrt {b d} a c^{2} d {\left | b \right |}\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} a^{3} b} - \frac {2 \, {\left (\sqrt {b d} b^{3} c^{4} {\left | b \right |} - 2 \, \sqrt {b d} a b^{2} c^{3} d {\left | b \right |} + \sqrt {b d} a^{2} b c^{2} d^{2} {\left | b \right |} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b c^{3} {\left | b \right |} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a c^{2} d {\left | b \right |}\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}\right )} a^{3}} - \frac {4 \, {\left (6 \, \sqrt {b d} b^{7} c^{5} {\left | b \right |} - 23 \, \sqrt {b d} a b^{6} c^{4} d {\left | b \right |} + 32 \, \sqrt {b d} a^{2} b^{5} c^{3} d^{2} {\left | b \right |} - 18 \, \sqrt {b d} a^{3} b^{4} c^{2} d^{3} {\left | b \right |} + 2 \, \sqrt {b d} a^{4} b^{3} c d^{4} {\left | b \right |} + \sqrt {b d} a^{5} b^{2} d^{5} {\left | b \right |} - 12 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{5} c^{4} {\left | b \right |} + 36 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{4} c^{3} d {\left | b \right |} - 36 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{3} c^{2} d^{2} {\left | b \right |} + 12 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{3} b^{2} c d^{3} {\left | b \right |} + 6 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} b^{3} c^{3} {\left | b \right |} - 9 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a b^{2} c^{2} d {\left | b \right |} + 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a^{3} d^{3} {\left | b \right |}\right )}}{3 \, {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{3} a^{3} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*(sqrt(b*d)*b*c^3*abs(b) - sqrt(b*d)*a*c^2*d*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^3*b) - 2*(sqrt(b*d)*b^3*c^4*abs(
b) - 2*sqrt(b*d)*a*b^2*c^3*d*abs(b) + sqrt(b*d)*a^2*b*c^2*d^2*abs(b) - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq
rt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b*c^3*abs(b) - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a
)*b*d - a*b*d))^2*a*c^2*d*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)*a^3) - 4/3*(6*sqrt(b*d)*b^7*c^5*abs(b
) - 23*sqrt(b*d)*a*b^6*c^4*d*abs(b) + 32*sqrt(b*d)*a^2*b^5*c^3*d^2*abs(b) - 18*sqrt(b*d)*a^3*b^4*c^2*d^3*abs(b
) + 2*sqrt(b*d)*a^4*b^3*c*d^4*abs(b) + sqrt(b*d)*a^5*b^2*d^5*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))^2*b^5*c^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))^2*a*b^4*c^3*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))^2*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))^2*a^3*b^2*c*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*
b^3*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))^4*a*b^2*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))^4*a^3*d^3*abs(b))/((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)^3*a^3*b^2)

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maple [B]  time = 0.03, size = 502, normalized size = 3.54 \[ -\frac {\sqrt {d x +c}\, \left (15 a \,b^{2} c^{2} d \,x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-15 b^{3} c^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+30 a^{2} b \,c^{2} d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-30 a \,b^{2} c^{3} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+15 a^{3} c^{2} d x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-15 a^{2} b \,c^{3} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} d^{2} x^{2}-20 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b c d \,x^{2}+30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, b^{2} c^{2} x^{2}-28 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c d x +40 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b \,c^{2} x +6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c^{2}\right )}{6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, \left (b x +a \right )^{\frac {3}{2}} a^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(d*x+c)^(1/2)*(15*a*b^2*c^2*d*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)-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*a^2*b*c^2*d*x^2*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*a*b^2*c^3*x^2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*
x+c))^(1/2))/x)+15*a^3*c^2*d*x*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)-15*a^2*b*c^3*x*
ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)-4*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*d^2*
x^2-20*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*b*c*d*x^2+30*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*b^2*c^2*x^2-28*(
(b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*c*d*x+40*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*b*c^2*x+6*((b*x+a)*(d*x+
c))^(1/2)*(a*c)^(1/2)*a^2*c^2)/a^3/((b*x+a)*(d*x+c))^(1/2)/x/(a*c)^(1/2)/(b*x+a)^(3/2)

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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)^(5/2)/x^2/(b*x+a)^(5/2),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 \[ \int \frac {{\left (c+d\,x\right )}^{5/2}}{x^2\,{\left (a+b\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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