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

Optimal. Leaf size=148 \[ -\frac {(b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{7/2}}-\frac {d \sqrt {a+b x} (13 b c-15 a d)}{3 c^3 \sqrt {c+d x} (b c-a d)}-\frac {5 d \sqrt {a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{c x (c+d x)^{3/2}} \]

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Rubi [A]  time = 0.13, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {99, 152, 12, 93, 208} \begin {gather*} -\frac {d \sqrt {a+b x} (13 b c-15 a d)}{3 c^3 \sqrt {c+d x} (b c-a d)}-\frac {5 d \sqrt {a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac {(b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{7/2}}-\frac {\sqrt {a+b x}}{c x (c+d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*d*Sqrt[a + b*x])/(3*c^2*(c + d*x)^(3/2)) - Sqrt[a + b*x]/(c*x*(c + d*x)^(3/2)) - (d*(13*b*c - 15*a*d)*Sqrt
[a + b*x])/(3*c^3*(b*c - a*d)*Sqrt[c + d*x]) - ((b*c - 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c
+ d*x])])/(Sqrt[a]*c^(7/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 99

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[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

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}}{x^2 (c+d x)^{5/2}} \, dx &=-\frac {\sqrt {a+b x}}{c x (c+d x)^{3/2}}+\frac {\int \frac {\frac {1}{2} (b c-5 a d)-2 b d x}{x \sqrt {a+b x} (c+d x)^{5/2}} \, dx}{c}\\ &=-\frac {5 d \sqrt {a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{c x (c+d x)^{3/2}}-\frac {2 \int \frac {-\frac {3}{4} (b c-5 a d) (b c-a d)+\frac {5}{2} b d (b c-a d) x}{x \sqrt {a+b x} (c+d x)^{3/2}} \, dx}{3 c^2 (b c-a d)}\\ &=-\frac {5 d \sqrt {a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{c x (c+d x)^{3/2}}-\frac {d (13 b c-15 a d) \sqrt {a+b x}}{3 c^3 (b c-a d) \sqrt {c+d x}}+\frac {4 \int \frac {3 (b c-5 a d) (b c-a d)^2}{8 x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{3 c^3 (b c-a d)^2}\\ &=-\frac {5 d \sqrt {a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{c x (c+d x)^{3/2}}-\frac {d (13 b c-15 a d) \sqrt {a+b x}}{3 c^3 (b c-a d) \sqrt {c+d x}}+\frac {(b c-5 a d) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 c^3}\\ &=-\frac {5 d \sqrt {a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{c x (c+d x)^{3/2}}-\frac {d (13 b c-15 a d) \sqrt {a+b x}}{3 c^3 (b c-a d) \sqrt {c+d x}}+\frac {(b c-5 a d) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{c^3}\\ &=-\frac {5 d \sqrt {a+b x}}{3 c^2 (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{c x (c+d x)^{3/2}}-\frac {d (13 b c-15 a d) \sqrt {a+b x}}{3 c^3 (b c-a d) \sqrt {c+d x}}-\frac {(b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 150, normalized size = 1.01 \begin {gather*} \frac {\frac {3 (c+d x) (b c-5 a d) \left (\sqrt {c} \sqrt {a+b x}-\sqrt {a} \sqrt {c+d x} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )}{c^{5/2}}-\frac {d (a+b x)^{3/2} (3 b c-5 a d)}{c (b c-a d)}-\frac {3 (a+b x)^{3/2}}{x}}{3 a c (c+d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.26, size = 189, normalized size = 1.28 \begin {gather*} \frac {(5 a d-b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{7/2}}-\frac {\sqrt {a+b x} \left (15 a^2 d^2-\frac {2 c^2 d^2 (a+b x)^2}{(c+d x)^2}+\frac {12 b c^2 d (a+b x)}{c+d x}-\frac {10 a c d^2 (a+b x)}{c+d x}-18 a b c d+3 b^2 c^2\right )}{3 c^3 \sqrt {c+d x} (b c-a d) \left (\frac {c (a+b x)}{c+d x}-a\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Sqrt[a + b*x]*(3*b^2*c^2 - 18*a*b*c*d + 15*a^2*d^2 - (2*c^2*d^2*(a + b*x)^2)/(c + d*x)^2 + (12*b*c^2*d*(
a + b*x))/(c + d*x) - (10*a*c*d^2*(a + b*x))/(c + d*x)))/(c^3*(b*c - a*d)*Sqrt[c + d*x]*(-a + (c*(a + b*x))/(c
+ d*x))) + ((-(b*c) + 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(Sqrt[a]*c^(7/2))

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fricas [B]  time = 2.89, size = 670, normalized size = 4.53 \begin {gather*} \left [-\frac {3 \, {\left ({\left (b^{2} c^{2} d^{2} - 6 \, a b c d^{3} + 5 \, a^{2} d^{4}\right )} x^{3} + 2 \, {\left (b^{2} c^{3} d - 6 \, a b c^{2} d^{2} + 5 \, a^{2} c d^{3}\right )} x^{2} + {\left (b^{2} c^{4} - 6 \, a b c^{3} d + 5 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt {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 + {\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 (3 \, a b c^{4} - 3 \, a^{2} c^{3} d + {\left (13 \, a b c^{2} d^{2} - 15 \, a^{2} c d^{3}\right )} x^{2} + 2 \, {\left (9 \, a b c^{3} d - 10 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{12 \, {\left ({\left (a b c^{5} d^{2} - a^{2} c^{4} d^{3}\right )} x^{3} + 2 \, {\left (a b c^{6} d - a^{2} c^{5} d^{2}\right )} x^{2} + {\left (a b c^{7} - a^{2} c^{6} d\right )} x\right )}}, \frac {3 \, {\left ({\left (b^{2} c^{2} d^{2} - 6 \, a b c d^{3} + 5 \, a^{2} d^{4}\right )} x^{3} + 2 \, {\left (b^{2} c^{3} d - 6 \, a b c^{2} d^{2} + 5 \, a^{2} c d^{3}\right )} x^{2} + {\left (b^{2} c^{4} - 6 \, a b c^{3} d + 5 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt {-a c} \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 (3 \, a b c^{4} - 3 \, a^{2} c^{3} d + {\left (13 \, a b c^{2} d^{2} - 15 \, a^{2} c d^{3}\right )} x^{2} + 2 \, {\left (9 \, a b c^{3} d - 10 \, a^{2} c^{2} d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left ({\left (a b c^{5} d^{2} - a^{2} c^{4} d^{3}\right )} x^{3} + 2 \, {\left (a b c^{6} d - a^{2} c^{5} d^{2}\right )} x^{2} + {\left (a b c^{7} - a^{2} c^{6} d\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*((b^2*c^2*d^2 - 6*a*b*c*d^3 + 5*a^2*d^4)*x^3 + 2*(b^2*c^3*d - 6*a*b*c^2*d^2 + 5*a^2*c*d^3)*x^2 + (b^
2*c^4 - 6*a*b*c^3*d + 5*a^2*c^2*d^2)*x)*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 + (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*(3*a*b*c^4 - 3*
a^2*c^3*d + (13*a*b*c^2*d^2 - 15*a^2*c*d^3)*x^2 + 2*(9*a*b*c^3*d - 10*a^2*c^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x +
c))/((a*b*c^5*d^2 - a^2*c^4*d^3)*x^3 + 2*(a*b*c^6*d - a^2*c^5*d^2)*x^2 + (a*b*c^7 - a^2*c^6*d)*x), 1/6*(3*((b
^2*c^2*d^2 - 6*a*b*c*d^3 + 5*a^2*d^4)*x^3 + 2*(b^2*c^3*d - 6*a*b*c^2*d^2 + 5*a^2*c*d^3)*x^2 + (b^2*c^4 - 6*a*b
*c^3*d + 5*a^2*c^2*d^2)*x)*sqrt(-a*c)*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*(3*a*b*c^4 - 3*a^2*c^3*d + (13*a*b*c^2*d^2 - 15*a^2*c*d
^3)*x^2 + 2*(9*a*b*c^3*d - 10*a^2*c^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/((a*b*c^5*d^2 - a^2*c^4*d^3)*x^3 +
2*(a*b*c^6*d - a^2*c^5*d^2)*x^2 + (a*b*c^7 - a^2*c^6*d)*x)]

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giac [B]  time = 6.99, size = 571, normalized size = 3.86 \begin {gather*} -\frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (5 \, b^{4} c^{4} d^{3} {\left | b \right |} - 6 \, a b^{3} c^{3} d^{4} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{3} c^{7} d - a b^{2} c^{6} d^{2}} + \frac {6 \, {\left (b^{5} c^{5} d^{2} {\left | b \right |} - 2 \, a b^{4} c^{4} d^{3} {\left | b \right |} + a^{2} b^{3} c^{3} d^{4} {\left | b \right |}\right )}}{b^{3} c^{7} d - a b^{2} c^{6} d^{2}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {{\left (\sqrt {b d} b^{3} c - 5 \, \sqrt {b d} a b^{2} d\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} b c^{3} {\left | b \right |}} - \frac {2 \, {\left (\sqrt {b d} b^{5} c^{2} - 2 \, \sqrt {b d} a b^{4} c d + \sqrt {b d} a^{2} b^{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} b^{3} c - \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^{2} d\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 )} c^{3} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 653, normalized size = 4.41 \begin {gather*} \frac {\left (15 a^{2} d^{4} 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 )-18 a b c \,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 )+3 b^{2} c^{2} 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 )+30 a^{2} c \,d^{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 )-36 a b \,c^{2} 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 )+6 b^{2} c^{3} 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^{2} c^{2} d^{2} 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 )-18 a b \,c^{3} 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 )+3 b^{2} c^{4} 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 )-30 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,d^{3} x^{2}+26 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b c \,d^{2} x^{2}-40 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a c \,d^{2} x +36 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b \,c^{2} d x -6 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,c^{2} d +6 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b \,c^{3}\right ) \sqrt {b x +a}}{6 \left (a d -b c \right ) \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \left (d x +c \right )^{\frac {3}{2}} c^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(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^3*a^2*d^4-18*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^3*a*b*c*d^3+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)*x^3*b^2*c^2*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*c*d^3-36
*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*c^2*d^2+6*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*b^2*c^3*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*a^2*c^2*d^2-18*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*a*b*c^3*d+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)*x*b^2*c^4-30*x^2*a*d^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^
(1/2)+26*x^2*b*c*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-40*x*a*c*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+36*x
*b*c^2*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-6*a*c^2*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+6*b*c^3*(a*c)^(1/2)
*((b*x+a)*(d*x+c))^(1/2))/c^3*(b*x+a)^(1/2)/(a*d-b*c)/(a*c)^(1/2)/x/((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(3/2)

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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)/x^2/(d*x+c)^(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 \begin {gather*} \int \frac {\sqrt {a+b\,x}}{x^2\,{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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