3.451 \(\int \frac{\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=74 \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{d (a+b)^{5/2}}+\frac{\tan ^3(c+d x)}{3 d (a+b)}-\frac{a \tan (c+d x)}{d (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.0976003, antiderivative size = 74, 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} \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{d (a+b)^{5/2}}+\frac{\tan ^3(c+d x)}{3 d (a+b)}-\frac{a \tan (c+d x)}{d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]

[Out]

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

Mathematica [A]  time = 0.309912, size = 75, normalized size = 1.01 \[ \frac{3 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )+\sqrt{a+b} \tan (c+d x) \left ((a+b) \sec ^2(c+d x)-4 a-b\right )}{3 d (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]

[Out]

(3*a^(3/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a + b]*(-4*a - b + (a + b)*Sec[c + d*x]^2)*Tan[c
+ d*x])/(3*(a + b)^(5/2)*d)

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Maple [A]  time = 0.109, size = 94, normalized size = 1.3 \begin{align*}{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\, \left ( a+b \right ) ^{2}d}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}b}{3\, \left ( a+b \right ) ^{2}d}}-{\frac{a\tan \left ( dx+c \right ) }{ \left ( a+b \right ) ^{2}d}}+{\frac{{a}^{2}}{ \left ( a+b \right ) ^{2}d}\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)^4/(a+sin(d*x+c)^2*b),x)

[Out]

1/3*a*tan(d*x+c)^3/(a+b)^2/d+1/3/d/(a+b)^2*tan(d*x+c)^3*b-a*tan(d*x+c)/(a+b)^2/d+1/d*a^2/(a+b)^2/(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)^4/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.91128, size = 879, normalized size = 11.88 \begin{align*} \left [\frac{3 \, a \sqrt{-\frac{a}{a + b}} \cos \left (d x + c\right )^{3} \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 (4 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sin \left (d x + c\right )}{12 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{3}}, -\frac{3 \, a \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 )^{3} + 2 \,{\left ({\left (4 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/12*(3*a*sqrt(-a/(a + b))*cos(d*x + c)^3*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*((4*a + b)*cos(d*x + c)^2 - a - b)*sin(d*x + c))/((a^2 + 2*a*b + b^2)*d*cos(d*x + c)^3), -1/6*(3*a*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)^3 + 2*((4*a + b)*cos(d*x + c)^2 - a - b)*sin(d*x + c))/((a^2 + 2*a*b + b^2)*d*cos(d*x + c)^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4/(a+b*sin(d*x+c)**2),x)

[Out]

Integral(tan(c + d*x)**4/(a + b*sin(c + d*x)**2), x)

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Giac [B]  time = 2.18947, size = 221, normalized size = 2.99 \begin{align*} \frac{\frac{3 \,{\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^{2}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a^{2} + a b}} + \frac{a^{2} \tan \left (d x + c\right )^{3} + 2 \, a b \tan \left (d x + c\right )^{3} + b^{2} \tan \left (d x + c\right )^{3} - 3 \, a^{2} \tan \left (d x + c\right ) - 3 \, a b \tan \left (d x + c\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

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