3.2.37 \(\int \sqrt {a+a \sec (c+d x)} \tan (c+d x) \, dx\) [137]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3965, 52, 65, 213} \begin {gather*} \frac {2 \sqrt {a \sec (c+d x)+a}}{d}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 213

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

Rule 3965

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 60, normalized size = 1.18 \begin {gather*} \frac {\sqrt {a (1+\sec (c+d x))} \left (-2 \tanh ^{-1}\left (\sqrt {1+\sec (c+d x)}\right )+2 \sqrt {1+\sec (c+d x)}\right )}{d \sqrt {1+\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Sec[c + d*x])]*(-2*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + 2*Sqrt[1 + Sec[c + d*x]]))/(d*Sqrt[1 + Sec[c
+ d*x]])

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Maple [A]
time = 0.05, size = 42, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {2 \sqrt {a +a \sec \left (d x +c \right )}-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d}\)	\(42\)
default	\(\frac {2 \sqrt {a +a \sec \left (d x +c \right )}-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 67, normalized size = 1.31 \begin {gather*} \frac {\sqrt {a} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 2 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(a)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) + sqrt(a))) + 2*sqrt(a + a/cos(d*x
+ c)))/d

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Fricas [A]
time = 3.98, size = 184, normalized size = 3.61 \begin {gather*} \left [\frac {\sqrt {a} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\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 ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \tan {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*tan(c + d*x), x)

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Giac [A]
time = 0.82, size = 75, normalized size = 1.47 \begin {gather*} \frac {\sqrt {2} {\left (\frac {\sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, a}{\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.48, size = 47, normalized size = 0.92 \begin {gather*} \frac {2\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{d}-\frac {2\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{\sqrt {a}}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a + a/cos(c + d*x))^(1/2))/d - (2*a^(1/2)*atanh((a + a/cos(c + d*x))^(1/2)/a^(1/2)))/d

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