3.5.52 \(\int \frac {\tan ^2(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) [452]

Optimal. Leaf size=53 \[ -\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{3/2} d}+\frac {\tan (c+d x)}{(a+b) d} \]

[Out]

-arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))*a^(1/2)/(a+b)^(3/2)/d+tan(d*x+c)/(a+b)/d

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Rubi [A]
time = 0.05, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3274, 327, 211} \begin {gather*} \frac {\tan (c+d x)}{d (a+b)}-\frac {\sqrt {a} \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{d (a+b)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 3274

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{ff
= FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(p
+ 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m}, x] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 53, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{3/2} d}+\frac {\tan (c+d x)}{(a+b) d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.50, size = 51, normalized size = 0.96

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (d x +c \right )}{a +b}-\frac {a \arctan \left (\frac {\tan \left (d x +c \right ) \left (a +b \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right ) \sqrt {a \left (a +b \right )}}}{d}\)	\(51\)
default	\(\frac {\frac {\tan \left (d x +c \right )}{a +b}-\frac {a \arctan \left (\frac {\tan \left (d x +c \right ) \left (a +b \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right ) \sqrt {a \left (a +b \right )}}}{d}\)	\(51\)
risch	\(\frac {2 i}{d \left (a +b \right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right )^{2} d}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right )^{2} d}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 51, normalized size = 0.96 \begin {gather*} -\frac {\frac {a \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a + b\right )}} - \frac {\tan \left (d x + c\right )}{a + b}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a)*(a + b)) - tan(d*x + c)/(a + b))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (45) = 90\).
time = 0.41, size = 300, normalized size = 5.66 \begin {gather*} \left [\frac {\sqrt {-\frac {a}{a + b}} \cos \left (d x + c\right ) \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) + 4 \, \sin \left (d x + c\right )}{4 \, {\left (a + b\right )} d \cos \left (d x + c\right )}, \frac {\sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right )}{2 \, {\left (a + b\right )} d \cos \left (d x + c\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(-a/(a + b))*cos(d*x + c)*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*
x + c)^2 + 4*((2*a^2 + 3*a*b + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-a/(a + b))*sin(d*
x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)) + 4*sin(d
*x + c))/((a + b)*d*cos(d*x + c)), 1/2*(sqrt(a/(a + b))*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a/(
a + b))/(a*cos(d*x + c)*sin(d*x + c)))*cos(d*x + c) + 2*sin(d*x + c))/((a + b)*d*cos(d*x + c))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.61, size = 86, normalized size = 1.62 \begin {gather*} -\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} a}{\sqrt {a^{2} + a b} {\left (a + b\right )}} - \frac {\tan \left (d x + c\right )}{a + b}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 14.92, size = 53, normalized size = 1.00 \begin {gather*} \frac {\mathrm {tan}\left (c+d\,x\right )}{d\,\left (a+b\right )}-\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )}{2\,\sqrt {a}\,\sqrt {a+b}}\right )}{d\,{\left (a+b\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c + d*x)/(d*(a + b)) - (a^(1/2)*atan((tan(c + d*x)*(2*a + 2*b))/(2*a^(1/2)*(a + b)^(1/2))))/(d*(a + b)^(3/
2))

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