3.7.62 \(\int \frac {(a+b x)^{5/2}}{x^3 (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=154 \[ -\frac {15 \sqrt {a} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 c^{7/2}}+\frac {15 \sqrt {a+b x} (b c-a d)^2}{4 c^3 \sqrt {c+d x}}-\frac {5 (a+b x)^{3/2} (b c-a d)}{4 c^2 x \sqrt {c+d x}}-\frac {(a+b x)^{5/2}}{2 c x^2 \sqrt {c+d x}} \]

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Rubi [A]  time = 0.06, antiderivative size = 154, 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} (b c-a d)}{4 c^2 x \sqrt {c+d x}}+\frac {15 \sqrt {a+b x} (b c-a d)^2}{4 c^3 \sqrt {c+d x}}-\frac {15 \sqrt {a} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 c^{7/2}}-\frac {(a+b x)^{5/2}}{2 c x^2 \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/(x^3*(c + d*x)^(3/2)),x]

[Out]

(15*(b*c - a*d)^2*Sqrt[a + b*x])/(4*c^3*Sqrt[c + d*x]) - (5*(b*c - a*d)*(a + b*x)^(3/2))/(4*c^2*x*Sqrt[c + d*x
]) - (a + b*x)^(5/2)/(2*c*x^2*Sqrt[c + d*x]) - (15*Sqrt[a]*(b*c - a*d)^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt
[a]*Sqrt[c + d*x])])/(4*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^3 (c+d x)^{3/2}} \, dx &=-\frac {(a+b x)^{5/2}}{2 c x^2 \sqrt {c+d x}}+\frac {(5 (b c-a d)) \int \frac {(a+b x)^{3/2}}{x^2 (c+d x)^{3/2}} \, dx}{4 c}\\ &=-\frac {5 (b c-a d) (a+b x)^{3/2}}{4 c^2 x \sqrt {c+d x}}-\frac {(a+b x)^{5/2}}{2 c x^2 \sqrt {c+d x}}+\frac {\left (15 (b c-a d)^2\right ) \int \frac {\sqrt {a+b x}}{x (c+d x)^{3/2}} \, dx}{8 c^2}\\ &=\frac {15 (b c-a d)^2 \sqrt {a+b x}}{4 c^3 \sqrt {c+d x}}-\frac {5 (b c-a d) (a+b x)^{3/2}}{4 c^2 x \sqrt {c+d x}}-\frac {(a+b x)^{5/2}}{2 c x^2 \sqrt {c+d x}}+\frac {\left (15 a (b c-a d)^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 c^3}\\ &=\frac {15 (b c-a d)^2 \sqrt {a+b x}}{4 c^3 \sqrt {c+d x}}-\frac {5 (b c-a d) (a+b x)^{3/2}}{4 c^2 x \sqrt {c+d x}}-\frac {(a+b x)^{5/2}}{2 c x^2 \sqrt {c+d x}}+\frac {\left (15 a (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{4 c^3}\\ &=\frac {15 (b c-a d)^2 \sqrt {a+b x}}{4 c^3 \sqrt {c+d x}}-\frac {5 (b c-a d) (a+b x)^{3/2}}{4 c^2 x \sqrt {c+d x}}-\frac {(a+b x)^{5/2}}{2 c x^2 \sqrt {c+d x}}-\frac {15 \sqrt {a} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 132, normalized size = 0.86 \begin {gather*} \frac {\sqrt {a+b x} \left (a^2 \left (-2 c^2+5 c d x+15 d^2 x^2\right )-a b c x (9 c+25 d x)+8 b^2 c^2 x^2\right )}{4 c^3 x^2 \sqrt {c+d x}}-\frac {15 \sqrt {a} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/(x^3*(c + d*x)^(3/2)),x]

[Out]

(Sqrt[a + b*x]*(8*b^2*c^2*x^2 - a*b*c*x*(9*c + 25*d*x) + a^2*(-2*c^2 + 5*c*d*x + 15*d^2*x^2)))/(4*c^3*x^2*Sqrt
[c + d*x]) - (15*Sqrt[a]*(b*c - a*d)^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*c^(7/2))

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IntegrateAlgebraic [A]  time = 0.25, size = 151, normalized size = 0.98 \begin {gather*} \frac {\sqrt {a+b x} (b c-a d)^2 \left (15 a^2+\frac {8 c^2 (a+b x)^2}{(c+d x)^2}-\frac {25 a c (a+b x)}{c+d x}\right )}{4 c^3 \sqrt {c+d x} \left (\frac {c (a+b x)}{c+d x}-a\right )^2}-\frac {15 \sqrt {a} (a d-b c)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(5/2)/(x^3*(c + d*x)^(3/2)),x]

[Out]

((b*c - a*d)^2*Sqrt[a + b*x]*(15*a^2 + (8*c^2*(a + b*x)^2)/(c + d*x)^2 - (25*a*c*(a + b*x))/(c + d*x)))/(4*c^3
*Sqrt[c + d*x]*(-a + (c*(a + b*x))/(c + d*x))^2) - (15*Sqrt[a]*(-(b*c) + a*d)^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x]
)/(Sqrt[a]*Sqrt[c + d*x])])/(4*c^(7/2))

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fricas [A]  time = 3.08, size = 479, normalized size = 3.11 \begin {gather*} \left [\frac {15 \, {\left ({\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x^{3} + {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{c}} \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^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} c^{2} - {\left (8 \, b^{2} c^{2} - 25 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2} + {\left (9 \, a b c^{2} - 5 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (c^{3} d x^{3} + c^{4} x^{2}\right )}}, \frac {15 \, {\left ({\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x^{3} + {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{c}} \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 {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) - 2 \, {\left (2 \, a^{2} c^{2} - {\left (8 \, b^{2} c^{2} - 25 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2} + {\left (9 \, a b c^{2} - 5 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (c^{3} d x^{3} + c^{4} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^3/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(15*((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x^3 + (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*x^2)*sqrt(a/c)*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^2 + (b*c^2 + a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*s
qrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a^2*c^2 - (8*b^2*c^2 - 25*a*b*c*d + 15*a^2*d^2)*x^2 + (9*a*b*c
^2 - 5*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^3*d*x^3 + c^4*x^2), 1/8*(15*((b^2*c^2*d - 2*a*b*c*d^2 + a^2
*d^3)*x^3 + (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*x^2)*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x +
a)*sqrt(d*x + c)*sqrt(-a/c)/(a*b*d*x^2 + a^2*c + (a*b*c + a^2*d)*x)) - 2*(2*a^2*c^2 - (8*b^2*c^2 - 25*a*b*c*d
+ 15*a^2*d^2)*x^2 + (9*a*b*c^2 - 5*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^3*d*x^3 + c^4*x^2)]

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giac [B]  time = 15.39, size = 1121, normalized size = 7.28 \begin {gather*} \frac {2 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \sqrt {b x + a}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} c^{3} {\left | b \right |}} - \frac {15 \, {\left (\sqrt {b d} a b^{4} c^{2} - 2 \, \sqrt {b d} a^{2} b^{3} c d + \sqrt {b d} a^{3} b^{2} d^{2}\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 )}{4 \, \sqrt {-a b c d} b c^{3} {\left | b \right |}} - \frac {9 \, \sqrt {b d} a b^{10} c^{5} - 43 \, \sqrt {b d} a^{2} b^{9} c^{4} d + 82 \, \sqrt {b d} a^{3} b^{8} c^{3} d^{2} - 78 \, \sqrt {b d} a^{4} b^{7} c^{2} d^{3} + 37 \, \sqrt {b d} a^{5} b^{6} c d^{4} - 7 \, \sqrt {b d} a^{6} b^{5} d^{5} - 27 \, \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^{8} c^{4} + 52 \, \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^{7} c^{3} d - 2 \, \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^{6} c^{2} d^{2} - 44 \, \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^{4} b^{5} c d^{3} + 21 \, \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^{5} b^{4} d^{4} + 27 \, \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^{6} c^{3} - 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^{2} b^{5} c^{2} d + 13 \, \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} b^{4} c d^{2} - 21 \, \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^{4} b^{3} d^{3} - 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 )}^{6} a b^{4} c^{2} - 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 )}^{6} a^{2} b^{3} c d + 7 \, \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 )}^{6} a^{3} b^{2} d^{2}}{2 \, {\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 )}^{2} c^{3} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^3/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

2*(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*sqrt(b*x + a)/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*c^3*abs(b)) - 15/4*
(sqrt(b*d)*a*b^4*c^2 - 2*sqrt(b*d)*a^2*b^3*c*d + sqrt(b*d)*a^3*b^2*d^2)*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)*b*c^3*abs(b)) -
1/2*(9*sqrt(b*d)*a*b^10*c^5 - 43*sqrt(b*d)*a^2*b^9*c^4*d + 82*sqrt(b*d)*a^3*b^8*c^3*d^2 - 78*sqrt(b*d)*a^4*b^7
*c^2*d^3 + 37*sqrt(b*d)*a^5*b^6*c*d^4 - 7*sqrt(b*d)*a^6*b^5*d^5 - 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*b^8*c^4 + 52*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^7*c^3*d - 2*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^6*c^2*d^2 - 44*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^5*c*d^3
+ 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^5*b^4*d^4 + 27*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^6*c^3 - 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^2*b^5*c^2*d + 13*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^4*c*d^2 - 21*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^3*d^3 - 9*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
^4*c^2 - 6*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^3*c*d + 7*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^2*d^2)/((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)^2*c^3*abs(b))

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maple [B]  time = 0.03, size = 507, normalized size = 3.29 \begin {gather*} -\frac {\sqrt {b x +a}\, \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 )-30 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 )+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 a^{3} c \,d^{2} 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^{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 )+15 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 )-30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} d^{2} x^{2}+50 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b c d \,x^{2}-16 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, b^{2} c^{2} x^{2}-10 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c d x +18 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b \,c^{2} x +4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c^{2}\right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, \sqrt {d x +c}\, c^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)/x^3/(d*x+c)^(3/2),x)

[Out]

-1/8*(b*x+a)^(1/2)*(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)-30*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)+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*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x
)*x^2*a^3*c*d^2-30*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a^2*b*c^2*d+15*ln((a*d*
x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a*b^2*c^3-30*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a
^2*d^2*x^2+50*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*b*c*d*x^2-16*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*b^2*c^2*x
^2-10*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*c*d*x+18*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*b*c^2*x+4*((b*x+a
)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*c^2)/c^3/((b*x+a)*(d*x+c))^(1/2)/x^2/(a*c)^(1/2)/(d*x+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^3/(d*x+c)^(3/2),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^3\,{\left (c+d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(5/2)/(x^3*(c + d*x)^(3/2)),x)

[Out]

int((a + b*x)^(5/2)/(x^3*(c + d*x)^(3/2)), 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**3/(d*x+c)**(3/2),x)

[Out]

Timed out

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