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

Optimal. Leaf size=127 \[ -\frac {(3 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{5/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (3 b c-a d)}{d^2 (b c-a d)}-\frac {2 c (a+b x)^{3/2}}{d \sqrt {c+d x} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 127, 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 = {78, 50, 63, 217, 206} \[ \frac {\sqrt {a+b x} \sqrt {c+d x} (3 b c-a d)}{d^2 (b c-a d)}-\frac {(3 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{5/2}}-\frac {2 c (a+b x)^{3/2}}{d \sqrt {c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

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

Rule 63

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 78

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] || LtQ
[p, n]))))

Rule 206

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

Rule 217

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

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Mathematica [A]  time = 0.42, size = 126, normalized size = 0.99 \[ \frac {\sqrt {d} \sqrt {a+b x} (3 c+d x)-\frac {\left (a^2 d^2-4 a b c d+3 b^2 c^2\right ) \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{b \sqrt {b c-a d}}}{d^{5/2} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d]*Sqrt[a + b*x]*(3*c + d*x) - ((3*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSin
h[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(b*Sqrt[b*c - a*d]))/(d^(5/2)*Sqrt[c + d*x])

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fricas [A]  time = 1.10, size = 308, normalized size = 2.43 \[ \left [-\frac {{\left (3 \, b c^{2} - a c d + {\left (3 \, b c d - a d^{2}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (b d^{2} x + 3 \, b c d\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, {\left (b d^{4} x + b c d^{3}\right )}}, \frac {{\left (3 \, b c^{2} - a c d + {\left (3 \, b c d - a d^{2}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (b d^{2} x + 3 \, b c d\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b d^{4} x + b c d^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)^(1/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*((3*b*c^2 - a*c*d + (3*b*c*d - a*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*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(b*d^2*x + 3*b*c*d
)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^4*x + b*c*d^3), 1/2*((3*b*c^2 - a*c*d + (3*b*c*d - a*d^2)*x)*sqrt(-b*d)*ar
ctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*
d^2)*x)) + 2*(b*d^2*x + 3*b*c*d)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^4*x + b*c*d^3)]

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giac [A]  time = 1.28, size = 134, normalized size = 1.06 \[ \frac {\sqrt {b x + a} {\left (\frac {{\left (b x + a\right )} {\left | b \right |}}{b d} + \frac {3 \, b^{2} c d {\left | b \right |} - a b d^{2} {\left | b \right |}}{b^{2} d^{3}}\right )}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {{\left (3 \, b c {\left | b \right |} - a d {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)^(1/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.02, size = 264, normalized size = 2.08 \[ \frac {\sqrt {b x +a}\, \left (a \,d^{2} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-3 b c d x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+a c d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-3 b \,c^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, d x +6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, c \right )}{2 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {d x +c}\, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x+a)^(1/2)/(d*x+c)^(3/2),x)

[Out]

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

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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(x*(b*x+a)^(1/2)/(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 \[ \int \frac {x\,\sqrt {a+b\,x}}{{\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*x)^(1/2))/(c + d*x)^(3/2),x)

[Out]

int((x*(a + b*x)^(1/2))/(c + d*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sqrt {a + b x}}{\left (c + d x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)**(1/2)/(d*x+c)**(3/2),x)

[Out]

Integral(x*sqrt(a + b*x)/(c + d*x)**(3/2), x)

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