3.4.37 \(\int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [337]

Optimal. Leaf size=54 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2}{a d \sqrt {a+b \sec (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 54, 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 = {3970, 53, 65, 213} \begin {gather*} \frac {2}{a d \sqrt {a+b \sec (c+d x)}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 3970

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]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(128\) vs. \(2(54)=108\).
time = 0.40, size = 128, normalized size = 2.37 \begin {gather*} \frac {\left (2 a \cos (c+d x)+\sqrt {a \cos (c+d x)} \sqrt {b+a \cos (c+d x)} \left (\log \left (1-\frac {\sqrt {b+a \cos (c+d x)}}{\sqrt {a \cos (c+d x)}}\right )-\log \left (1+\frac {\sqrt {b+a \cos (c+d x)}}{\sqrt {a \cos (c+d x)}}\right )\right )\right ) \sec (c+d x)}{a^2 d \sqrt {a+b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*Cos[c + d*x] + Sqrt[a*Cos[c + d*x]]*Sqrt[b + a*Cos[c + d*x]]*(Log[1 - Sqrt[b + a*Cos[c + d*x]]/Sqrt[a*Co
s[c + d*x]]] - Log[1 + Sqrt[b + a*Cos[c + d*x]]/Sqrt[a*Cos[c + d*x]]]))*Sec[c + d*x])/(a^2*d*Sqrt[a + b*Sec[c
+ d*x]])

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Maple [A]
time = 0.03, size = 45, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {-\frac {2 \arctanh \left (\frac {\sqrt {a +b \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}+\frac {2}{a \sqrt {a +b \sec \left (d x +c \right )}}}{d}\)	\(45\)
default	\(\frac {-\frac {2 \arctanh \left (\frac {\sqrt {a +b \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}+\frac {2}{a \sqrt {a +b \sec \left (d x +c \right )}}}{d}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.47, size = 70, normalized size = 1.30 \begin {gather*} \frac {\frac {\log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (46) = 92\).
time = 3.82, size = 260, normalized size = 4.81 \begin {gather*} \left [\frac {4 \, a \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right ) + b\right )} \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 )}{2 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{2} b d\right )}}, \frac {{\left (a \cos \left (d x + c\right ) + b\right )} \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 \, a \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a^{3} d \cos \left (d x + c\right ) + a^{2} b d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(4*a*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c) + (a*cos(d*x + c) + b)*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))*sqrt(a)*sqrt((a*cos(d*x + c) +
b)/cos(d*x + c))))/(a^3*d*cos(d*x + c) + a^2*b*d), ((a*cos(d*x + c) + b)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*co
s(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) + 2*a*sqrt((a*cos(d*x + c) + b)/cos(d*x + c
))*cos(d*x + c))/(a^3*d*cos(d*x + c) + a^2*b*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [B]
time = 1.95, size = 50, normalized size = 0.93 \begin {gather*} \frac {2}{a\,d\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}}-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}}{\sqrt {a}}\right )}{a^{3/2}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(a*d*(a + b/cos(c + d*x))^(1/2)) - (2*atanh((a + b/cos(c + d*x))^(1/2)/a^(1/2)))/(a^(3/2)*d)

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