3.450 \(\int \frac{\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=97 \[ -\frac{a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{d (a+b)^{7/2}}+\frac{a^2 \tan (c+d x)}{d (a+b)^3}+\frac{\tan ^5(c+d x)}{5 d (a+b)}-\frac{a \tan ^3(c+d x)}{3 d (a+b)^2} \]

[Out]

-((a^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/((a + b)^(7/2)*d)) + (a^2*Tan[c + d*x])/((a + b)^3*d) -
(a*Tan[c + d*x]^3)/(3*(a + b)^2*d) + Tan[c + d*x]^5/(5*(a + b)*d)

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Rubi [A]  time = 0.106833, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3195, 302, 205} \[ -\frac{a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{d (a+b)^{7/2}}+\frac{a^2 \tan (c+d x)}{d (a+b)^3}+\frac{\tan ^5(c+d x)}{5 d (a+b)}-\frac{a \tan ^3(c+d x)}{3 d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]

[Out]

-((a^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/((a + b)^(7/2)*d)) + (a^2*Tan[c + d*x])/((a + b)^3*d) -
(a*Tan[c + d*x]^3)/(3*(a + b)^2*d) + Tan[c + d*x]^5/(5*(a + b)*d)

Rule 3195

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]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2}{(a+b)^3}-\frac{a x^2}{(a+b)^2}+\frac{x^4}{a+b}-\frac{a^3}{(a+b)^3 \left (a+(a+b) x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{a^2 \tan (c+d x)}{(a+b)^3 d}-\frac{a \tan ^3(c+d x)}{3 (a+b)^2 d}+\frac{\tan ^5(c+d x)}{5 (a+b) d}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{(a+b)^3 d}\\ &=-\frac{a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{(a+b)^{7/2} d}+\frac{a^2 \tan (c+d x)}{(a+b)^3 d}-\frac{a \tan ^3(c+d x)}{3 (a+b)^2 d}+\frac{\tan ^5(c+d x)}{5 (a+b) d}\\ \end{align*}

Mathematica [A]  time = 0.784555, size = 111, normalized size = 1.14 \[ \frac{\sqrt{a+b} \tan (c+d x) \left (-\left (11 a^2+17 a b+6 b^2\right ) \sec ^2(c+d x)+23 a^2+3 (a+b)^2 \sec ^4(c+d x)+11 a b+3 b^2\right )-15 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{15 d (a+b)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]

[Out]

(-15*a^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a + b]*(23*a^2 + 11*a*b + 3*b^2 - (11*a^2 + 17*
a*b + 6*b^2)*Sec[c + d*x]^2 + 3*(a + b)^2*Sec[c + d*x]^4)*Tan[c + d*x])/(15*(a + b)^(7/2)*d)

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Maple [A]  time = 0.121, size = 161, normalized size = 1.7 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}{a}^{2}}{5\,d \left ( a+b \right ) ^{3}}}+{\frac{2\, \left ( \tan \left ( dx+c \right ) \right ) ^{5}ab}{5\,d \left ( a+b \right ) ^{3}}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}{b}^{2}}{5\,d \left ( a+b \right ) ^{3}}}-{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d \left ( a+b \right ) ^{3}}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}ab}{3\,d \left ( a+b \right ) ^{3}}}+{\frac{{a}^{2}\tan \left ( dx+c \right ) }{d \left ( a+b \right ) ^{3}}}-{\frac{{a}^{3}}{d \left ( a+b \right ) ^{3}}\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^6/(a+sin(d*x+c)^2*b),x)

[Out]

1/5/d/(a+b)^3*tan(d*x+c)^5*a^2+2/5/d/(a+b)^3*tan(d*x+c)^5*a*b+1/5/d/(a+b)^3*tan(d*x+c)^5*b^2-1/3*a^2*tan(d*x+c
)^3/(a+b)^3/d-1/3/d/(a+b)^3*tan(d*x+c)^3*a*b+a^2*tan(d*x+c)/(a+b)^3/d-1/d*a^3/(a+b)^3/(a*(a+b))^(1/2)*arctan((
a+b)*tan(d*x+c)/(a*(a+b))^(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(tan(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.07295, size = 1122, normalized size = 11.57 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{-\frac{a}{a + b}} \cos \left (d x + c\right )^{5} \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 \,{\left ({\left (23 \, a^{2} + 11 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} -{\left (11 \, a^{2} + 17 \, a b + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{60 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cos \left (d x + c\right )^{5}}, \frac{15 \, a^{2} \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 )^{5} + 2 \,{\left ({\left (23 \, a^{2} + 11 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} -{\left (11 \, a^{2} + 17 \, a b + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cos \left (d x + c\right )^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/60*(15*a^2*sqrt(-a/(a + b))*cos(d*x + c)^5*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*((23*a^2 + 11*a*b + 3*b^2)*cos(d*x + c)^4 - (11*a^2 + 17*a*b + 6*b^2)*cos(d*x + c)^2 + 3*a^2 + 6*a*b + 3*
b^2)*sin(d*x + c))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cos(d*x + c)^5), 1/30*(15*a^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)^5 + 2*((23*a^2
+ 11*a*b + 3*b^2)*cos(d*x + c)^4 - (11*a^2 + 17*a*b + 6*b^2)*cos(d*x + c)^2 + 3*a^2 + 6*a*b + 3*b^2)*sin(d*x
+ c))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cos(d*x + c)^5)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**6/(a+b*sin(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 4.58227, size = 400, normalized size = 4.12 \begin{align*} -\frac{\frac{15 \,{\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^{3}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt{a^{2} + a b}} - \frac{3 \, a^{4} \tan \left (d x + c\right )^{5} + 12 \, a^{3} b \tan \left (d x + c\right )^{5} + 18 \, a^{2} b^{2} \tan \left (d x + c\right )^{5} + 12 \, a b^{3} \tan \left (d x + c\right )^{5} + 3 \, b^{4} \tan \left (d x + c\right )^{5} - 5 \, a^{4} \tan \left (d x + c\right )^{3} - 15 \, a^{3} b \tan \left (d x + c\right )^{3} - 15 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} - 5 \, a b^{3} \tan \left (d x + c\right )^{3} + 15 \, a^{4} \tan \left (d x + c\right ) + 30 \, a^{3} b \tan \left (d x + c\right ) + 15 \, a^{2} b^{2} \tan \left (d x + c\right )}{a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

-1/15*(15*(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^3/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a^2 + a*b)) - (3*a^4*tan(d*x + c)^5 + 12*a^3*b*tan(d*x + c)^5
+ 18*a^2*b^2*tan(d*x + c)^5 + 12*a*b^3*tan(d*x + c)^5 + 3*b^4*tan(d*x + c)^5 - 5*a^4*tan(d*x + c)^3 - 15*a^3*b
*tan(d*x + c)^3 - 15*a^2*b^2*tan(d*x + c)^3 - 5*a*b^3*tan(d*x + c)^3 + 15*a^4*tan(d*x + c) + 30*a^3*b*tan(d*x
+ c) + 15*a^2*b^2*tan(d*x + c))/(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5))/d