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((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/(a+b)^(7/2)/d+a^2*tan(d*x+c)/(a+b)^3/d-1/3*a*tan(d*x+c)^3/(a+b
)^2/d+1/5*tan(d*x+c)^5/(a+b)/d

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Rubi [A]  time = 0.11, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.75, 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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fricas [B]  time = 0.51, size = 472, normalized size = 4.87 \[ \left [\frac {15 \, a^{2} \sqrt {-\frac {a}{a + b}} \cos \left (d x + c\right )^{5} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) + 4 \, {\left ({\left (23 \, a^{2} + 11 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - {\left (11 \, a^{2} + 17 \, a b + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cos \left (d x + c\right )^{5}}, \frac {15 \, a^{2} \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{5} + 2 \, {\left ({\left (23 \, a^{2} + 11 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - {\left (11 \, a^{2} + 17 \, a b + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cos \left (d x + c\right )^{5}}\right ] \]

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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giac [B]  time = 4.32, size = 296, normalized size = 3.05 \[ -\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} a^{3}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a^{2} + a b}} - \frac {3 \, a^{4} \tan \left (d x + c\right )^{5} + 12 \, a^{3} b \tan \left (d x + c\right )^{5} + 18 \, a^{2} b^{2} \tan \left (d x + c\right )^{5} + 12 \, a b^{3} \tan \left (d x + c\right )^{5} + 3 \, b^{4} \tan \left (d x + c\right )^{5} - 5 \, a^{4} \tan \left (d x + c\right )^{3} - 15 \, a^{3} b \tan \left (d x + c\right )^{3} - 15 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} - 5 \, a b^{3} \tan \left (d x + c\right )^{3} + 15 \, a^{4} \tan \left (d x + c\right ) + 30 \, a^{3} b \tan \left (d x + c\right ) + 15 \, a^{2} b^{2} \tan \left (d x + c\right )}{a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}}}{15 \, d} \]

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

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maple [A]  time = 0.54, size = 161, normalized size = 1.66 \[ \frac {\left (\tan ^{5}\left (d x +c \right )\right ) a^{2}}{5 d \left (a +b \right )^{3}}+\frac {2 \left (\tan ^{5}\left (d x +c \right )\right ) a b}{5 d \left (a +b \right )^{3}}+\frac {\left (\tan ^{5}\left (d x +c \right )\right ) b^{2}}{5 d \left (a +b \right )^{3}}-\frac {a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3 \left (a +b \right )^{3} d}-\frac {\left (\tan ^{3}\left (d x +c \right )\right ) b a}{3 d \left (a +b \right )^{3}}+\frac {a^{2} \tan \left (d x +c \right )}{\left (a +b \right )^{3} d}-\frac {a^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{d \left (a +b \right )^{3} \sqrt {a \left (a +b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^6/(a+b*sin(d*x+c)^2),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*b*a+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 [A]  time = 0.42, size = 130, normalized size = 1.34 \[ -\frac {\frac {15 \, a^{3} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{5} - 5 \, {\left (a^{2} + a b\right )} \tan \left (d x + c\right )^{3} + 15 \, a^{2} \tan \left (d x + c\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{15 \, d} \]

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]

-1/15*(15*a^3*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt((a + b)*a)) -
(3*(a^2 + 2*a*b + b^2)*tan(d*x + c)^5 - 5*(a^2 + a*b)*tan(d*x + c)^3 + 15*a^2*tan(d*x + c))/(a^3 + 3*a^2*b +
3*a*b^2 + b^3))/d

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mupad [B]  time = 15.11, size = 112, normalized size = 1.15 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,d\,\left (a+b\right )}-\frac {a^{5/2}\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{7/2}}\right )}{d\,{\left (a+b\right )}^{7/2}}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d\,{\left (a+b\right )}^2}+\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )}{d\,{\left (a+b\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c + d*x)^5/(5*d*(a + b)) - (a^(5/2)*atan((tan(c + d*x)*(2*a + 2*b)*(3*a*b^2 + 3*a^2*b + a^3 + b^3))/(2*a^(
1/2)*(a + b)^(7/2))))/(d*(a + b)^(7/2)) - (a*tan(c + d*x)^3)/(3*d*(a + b)^2) + (a^2*tan(c + d*x))/(d*(a + b)^3
)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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