3.320 \(\int \sqrt{a+b \sec (c+d x)} \tan (c+d x) \, dx\)

Optimal. Leaf size=51 \[ \frac{2 \sqrt{a+b \sec (c+d x)}}{d}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.0476281, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3885, 50, 63, 207} \[ \frac{2 \sqrt{a+b \sec (c+d x)}}{d}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x],x]

[Out]

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

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

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 207

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

Rubi steps

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

Mathematica [B]  time = 0.248529, size = 137, normalized size = 2.69 \[ \frac{\sqrt{a+b \sec (c+d x)} \left (2 \sqrt{a \cos (c+d x)+b}+\sqrt{a \cos (c+d x)} \log \left (1-\frac{\sqrt{a \cos (c+d x)+b}}{\sqrt{a \cos (c+d x)}}\right )-\sqrt{a \cos (c+d x)} \log \left (\frac{\sqrt{a \cos (c+d x)+b}}{\sqrt{a \cos (c+d x)}}+1\right )\right )}{d \sqrt{a \cos (c+d x)+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[b + a*Cos[c + d*x]] + Sqrt[a*Cos[c + d*x]]*Log[1 - Sqrt[b + a*Cos[c + d*x]]/Sqrt[a*Cos[c + d*x]]] - S
qrt[a*Cos[c + d*x]]*Log[1 + Sqrt[b + a*Cos[c + d*x]]/Sqrt[a*Cos[c + d*x]]])*Sqrt[a + b*Sec[c + d*x]])/(d*Sqrt[
b + a*Cos[c + d*x]])

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Maple [A]  time = 0.049, size = 42, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( 2\,\sqrt{a+b\sec \left ( dx+c \right ) }-2\,\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{a+b\sec \left ( dx+c \right ) }}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.00288, size = 489, normalized size = 9.59 \begin{align*} \left [\frac{\sqrt{a} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} + 4 \,{\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) + 4 \, \sqrt{\frac{a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{2 \, d}, \frac{\sqrt{-a} \arctan \left (\frac{2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) + 2 \, \sqrt{\frac{a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) - b^2 + 4*(2*a*cos(d*x + c)^2 + b*cos(d*x + c))*s
qrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))) + 4*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/d, (sqrt(-a)*arct
an(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) + 2*sqrt((a*cos(d*x
+ c) + b)/cos(d*x + c)))/d]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sec{\left (c + d x \right )}} \tan{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.38951, size = 250, normalized size = 4.9 \begin{align*} \frac{2 \,{\left (\frac{a \arctan \left (-\frac{\sqrt{a - b} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b} + \sqrt{a - b}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a}} - \frac{2 \, b}{\sqrt{a - b} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b} - \sqrt{a - b}}\right )} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a*arctan(-1/2*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^
4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b) + sqrt(a - b))/sqrt(-a))/sqrt(-a) - 2*b/(sqrt(a - b)*tan(1/2*d*x + 1/2
*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b) - sqrt(
a - b)))*sgn(cos(d*x + c))/d