3.7.100 \(\int \frac {x \sqrt {c+d x}}{\sqrt {a+b x}} \, dx\) [700]

Optimal. Leaf size=125 \[ -\frac {(b c+3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2 d}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b d}-\frac {(b c-a d) (b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{3/2}} \]

[Out]

-1/4*(-a*d+b*c)*(3*a*d+b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(5/2)/d^(3/2)+1/2*(d*x+c)^(
3/2)*(b*x+a)^(1/2)/b/d-1/4*(3*a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^2/d

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Rubi [A]
time = 0.04, antiderivative size = 125, 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 = {81, 52, 65, 223, 212} \begin {gather*} -\frac {(b c-a d) (3 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{3/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} (3 a d+b c)}{4 b^2 d}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

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 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 81

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

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

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Mathematica [A]
time = 0.31, size = 107, normalized size = 0.86 \begin {gather*} \frac {b \sqrt {a+b x} \sqrt {c+d x} (-3 a d+b (c+2 d x))+\sqrt {\frac {b}{d}} \left (b^2 c^2+2 a b c d-3 a^2 d^2\right ) \log \left (\sqrt {a+b x}-\sqrt {\frac {b}{d}} \sqrt {c+d x}\right )}{4 b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(250\) vs. \(2(99)=198\).
time = 0.07, size = 251, normalized size = 2.01

method	result	size
default	\(\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{2}-2 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c d -\ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2}+4 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b d x -6 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a d +2 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b c \right )}{8 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{2} d \sqrt {b d}}\)	\(251\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(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))
*a^2*d^2-2*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-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+4*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*b*d
*x-6*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*d+2*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*b*c)/((d*x+c)*(b*x+a))^(1/2
)/b^2/d/(b*d)^(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(x*(d*x+c)^(1/2)/(b*x+a)^(1/2),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 [A]
time = 0.69, size = 304, normalized size = 2.43 \begin {gather*} \left [-\frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\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 (2 \, b^{2} d^{2} x + b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b^{3} d^{2}}, \frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\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 (2 \, b^{2} d^{2} x + b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b^{3} d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*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*(2*b^2*d^2*x + b^2*c*d
- 3*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^2), 1/8*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*sqrt(-b*d)*arctan(
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*(2*b^2*d^2*x + b^2*c*d - 3*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^2)]

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Sympy [F(-1)] Timed out
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(x*(d*x+c)**(1/2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 1.06, size = 132, normalized size = 1.06 \begin {gather*} \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\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}\right )} {\left | b \right |}}{4 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)/d^2)*sqrt(b*x + a) + (b^3*c^2 + 2*a*
b^2*c*d - 3*a^2*b*d^2)*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))
*abs(b)/b^4

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Mupad [B]
time = 18.40, size = 584, normalized size = 4.67 \begin {gather*} \frac {\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (-\frac {3\,a^2\,b\,d^2}{2}+a\,b^2\,c\,d+\frac {b^3\,c^2}{2}\right )}{d^5\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {11\,a^2\,d^2}{2}+23\,a\,b\,c\,d+\frac {7\,b^2\,c^2}{2}\right )}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {8\,\sqrt {a}\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (-\frac {3\,a^2\,d^2}{2}+a\,b\,c\,d+\frac {b^2\,c^2}{2}\right )}{b^2\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (\frac {11\,a^2\,d^2}{2}+23\,a\,b\,c\,d+\frac {7\,b^2\,c^2}{2}\right )}{b\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}-\frac {\sqrt {a}\,\sqrt {c}\,\left (32\,a\,d+16\,b\,c\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {8\,\sqrt {a}\,b^2\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}+\frac {b^4}{d^4}-\frac {4\,b^3\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {6\,b^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {4\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}}+\frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )\,\left (a\,d-b\,c\right )\,\left (3\,a\,d+b\,c\right )}{2\,b^{5/2}\,d^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a + b*x)^(1/2) - a^(1/2))*((b^3*c^2)/2 - (3*a^2*b*d^2)/2 + a*b^2*c*d))/(d^5*((c + d*x)^(1/2) - c^(1/2))) +
(((a + b*x)^(1/2) - a^(1/2))^3*((11*a^2*d^2)/2 + (7*b^2*c^2)/2 + 23*a*b*c*d))/(d^4*((c + d*x)^(1/2) - c^(1/2)
)^3) - (8*a^(1/2)*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^6)/(d^2*((c + d*x)^(1/2) - c^(1/2))^6) + (((a + b*x)^(1/
2) - a^(1/2))^7*((b^2*c^2)/2 - (3*a^2*d^2)/2 + a*b*c*d))/(b^2*d^2*((c + d*x)^(1/2) - c^(1/2))^7) + (((a + b*x)
^(1/2) - a^(1/2))^5*((11*a^2*d^2)/2 + (7*b^2*c^2)/2 + 23*a*b*c*d))/(b*d^3*((c + d*x)^(1/2) - c^(1/2))^5) - (a^
(1/2)*c^(1/2)*(32*a*d + 16*b*c)*((a + b*x)^(1/2) - a^(1/2))^4)/(d^3*((c + d*x)^(1/2) - c^(1/2))^4) - (8*a^(1/2
)*b^2*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^2)/(d^4*((c + d*x)^(1/2) - c^(1/2))^2))/(((a + b*x)^(1/2) - a^(1/2))
^8/((c + d*x)^(1/2) - c^(1/2))^8 + b^4/d^4 - (4*b^3*((a + b*x)^(1/2) - a^(1/2))^2)/(d^3*((c + d*x)^(1/2) - c^(
1/2))^2) + (6*b^2*((a + b*x)^(1/2) - a^(1/2))^4)/(d^2*((c + d*x)^(1/2) - c^(1/2))^4) - (4*b*((a + b*x)^(1/2) -
a^(1/2))^6)/(d*((c + d*x)^(1/2) - c^(1/2))^6)) + (atanh((d^(1/2)*((a + b*x)^(1/2) - a^(1/2)))/(b^(1/2)*((c +
d*x)^(1/2) - c^(1/2))))*(a*d - b*c)*(3*a*d + b*c))/(2*b^(5/2)*d^(3/2))

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