3.6.93 \(\int \frac {x \sqrt {a+b x}}{(c+d x)^{5/2}} \, dx\) [593]

Optimal. Leaf size=102 \[ -\frac {2 c (a+b x)^{3/2}}{3 d (b c-a d) (c+d x)^{3/2}}-\frac {2 \sqrt {a+b x}}{d^2 \sqrt {c+d x}}+\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 49, 65, 223, 212} \begin {gather*} \frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{5/2}}-\frac {2 \sqrt {a+b x}}{d^2 \sqrt {c+d x}}-\frac {2 c (a+b x)^{3/2}}{3 d (c+d x)^{3/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 102, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {a+b x} \left (3 b c-3 a d+\frac {c d (a+b x)}{c+d x}\right )}{3 d^2 (-b c+a d) \sqrt {c+d x}}+\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{d^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(441\) vs. \(2(80)=160\).
time = 0.07, size = 442, normalized size = 4.33

method	result	size
default	\(\frac {\left (3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,d^{3} x^{2}-3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c \,d^{2} x^{2}+6 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c \,d^{2} x -6 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2} d x +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,c^{2} d -3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{3}-6 a \,d^{2} x \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+8 b c d x \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}-4 a c d \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+6 b \,c^{2} \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\right ) \sqrt {b x +a}}{3 \sqrt {b d}\, \left (a d -b c \right ) \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, d^{2} \left (d x +c \right )^{\frac {3}{2}}}\)	\(442\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b*d^3*x^2-3*ln(1/2*(2*b*d
*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^2*c*d^2*x^2+6*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x
+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b*c*d^2*x-6*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/
2)+a*d+b*c)/(b*d)^(1/2))*b^2*c^2*d*x+3*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1
/2))*a*b*c^2*d-3*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^2*c^3-6*a*d^2*x
*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+8*b*c*d*x*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)-4*a*c*d*(b*d)^(1/2)*((d*x+c
)*(b*x+a))^(1/2)+6*b*c^2*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2))*(b*x+a)^(1/2)/(b*d)^(1/2)/(a*d-b*c)/((d*x+c)*(b*
x+a))^(1/2)/d^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(x*(b*x+a)^(1/2)/(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 detail

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (80) = 160\).
time = 1.58, size = 469, normalized size = 4.60 \begin {gather*} \left [\frac {3 \, {\left (b c^{3} - a c^{2} d + {\left (b c d^{2} - a d^{3}\right )} x^{2} + 2 \, {\left (b c^{2} d - a c d^{2}\right )} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (3 \, b c^{2} - 2 \, a c d + {\left (4 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (b c^{3} d^{2} - a c^{2} d^{3} + {\left (b c d^{4} - a d^{5}\right )} x^{2} + 2 \, {\left (b c^{2} d^{3} - a c d^{4}\right )} x\right )}}, -\frac {3 \, {\left (b c^{3} - a c^{2} d + {\left (b c d^{2} - a d^{3}\right )} x^{2} + 2 \, {\left (b c^{2} d - a c d^{2}\right )} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, {\left (3 \, b c^{2} - 2 \, a c d + {\left (4 \, b c d - 3 \, a d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (b c^{3} d^{2} - a c^{2} d^{3} + {\left (b c d^{4} - a d^{5}\right )} x^{2} + 2 \, {\left (b c^{2} d^{3} - a c d^{4}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \sqrt {a + b x}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (80) = 160\).
time = 1.45, size = 188, normalized size = 1.84 \begin {gather*} -\frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (4 \, b^{4} c d^{2} {\left | b \right |} - 3 \, a b^{3} d^{3} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{3} c d^{3} - a b^{2} d^{4}} + \frac {3 \, {\left (b^{5} c^{2} d {\left | b \right |} - 2 \, a b^{4} c d^{2} {\left | b \right |} + a^{2} b^{3} d^{3} {\left | b \right |}\right )}}{b^{3} c d^{3} - a b^{2} d^{4}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {2 \, {\left | b \right |} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\sqrt {a+b\,x}}{{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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