3.176 \(\int \sqrt {a+b x} \sqrt {c+d x} \, dx\)

Optimal. Leaf size=116 \[ -\frac {(b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{3/2} d^{3/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 b d}+\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {50, 63, 217, 206} \[ -\frac {(b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{3/2} d^{3/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 b d}+\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]*Sqrt[c + d*x],x]

[Out]

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

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

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Mathematica [A]  time = 0.29, size = 118, normalized size = 1.02 \[ \frac {b \sqrt {d} \sqrt {a+b x} (c+d x) (a d+b (c+2 d x))-(b c-a d)^{5/2} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{4 b^2 d^{3/2} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 300, normalized size = 2.59 \[ \left [\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d^{2} x + b^{2} c d + a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b^{2} d^{2}}, \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (2 \, b^{2} d^{2} x + b^{2} c d + a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b^{2} d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.22, size = 232, normalized size = 2.00 \[ -\frac {\frac {4 \, {\left (\frac {{\left (b^{2} c - a b d\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d}} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a}\right )} a {\left | b \right |}}{b^{2}} - \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d}\right )} {\left | b \right |}}{b^{2}}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 305, normalized size = 2.63 \[ -\frac {\sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{8 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, b}+\frac {\sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a c \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{4 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}}-\frac {\sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b \,c^{2} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{8 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, d}+\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, a}{4 b}-\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, c}{4 d}+\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{2}}}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((b*x+a)^(1/2)*(d*x+c)^(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 details)Is a*d-b*c zero or nonzero?

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mupad [B]  time = 0.17, size = 88, normalized size = 0.76 \[ \left (\frac {x}{2}+\frac {a\,d+b\,c}{4\,b\,d}\right )\,\sqrt {a+b\,x}\,\sqrt {c+d\,x}-\frac {\ln \left (a\,d+b\,c+2\,b\,d\,x+2\,\sqrt {b}\,\sqrt {d}\,\sqrt {a+b\,x}\,\sqrt {c+d\,x}\right )\,{\left (a\,d-b\,c\right )}^2}{8\,b^{3/2}\,d^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b x} \sqrt {c + d x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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