3.442 \(\int \frac{\tan ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=94 \[ \frac{a^2 \log \left (a+b \sin ^2(c+d x)\right )}{2 d (a+b)^3}-\frac{a^2 \log (\cos (c+d x))}{d (a+b)^3}+\frac{\sec ^4(c+d x)}{4 d (a+b)}-\frac{(2 a+b) \sec ^2(c+d x)}{2 d (a+b)^2} \]

[Out]

-((a^2*Log[Cos[c + d*x]])/((a + b)^3*d)) + (a^2*Log[a + b*Sin[c + d*x]^2])/(2*(a + b)^3*d) - ((2*a + b)*Sec[c
+ d*x]^2)/(2*(a + b)^2*d) + Sec[c + d*x]^4/(4*(a + b)*d)

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Rubi [A]  time = 0.102406, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3194, 88} \[ \frac{a^2 \log \left (a+b \sin ^2(c+d x)\right )}{2 d (a+b)^3}-\frac{a^2 \log (\cos (c+d x))}{d (a+b)^3}+\frac{\sec ^4(c+d x)}{4 d (a+b)}-\frac{(2 a+b) \sec ^2(c+d x)}{2 d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + b*Sin[c + d*x]^2),x]

[Out]

-((a^2*Log[Cos[c + d*x]])/((a + b)^3*d)) + (a^2*Log[a + b*Sin[c + d*x]^2])/(2*(a + b)^3*d) - ((2*a + b)*Sec[c
+ d*x]^2)/(2*(a + b)^2*d) + Sec[c + d*x]^4/(4*(a + b)*d)

Rule 3194

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

Mathematica [A]  time = 0.275684, size = 78, normalized size = 0.83 \[ \frac{-2 \left (2 a^2+3 a b+b^2\right ) \sec ^2(c+d x)+2 a^2 \left (\log \left (a+b \sin ^2(c+d x)\right )-2 \log (\cos (c+d x))\right )+(a+b)^2 \sec ^4(c+d x)}{4 d (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + b*Sin[c + d*x]^2),x]

[Out]

(2*a^2*(-2*Log[Cos[c + d*x]] + Log[a + b*Sin[c + d*x]^2]) - 2*(2*a^2 + 3*a*b + b^2)*Sec[c + d*x]^2 + (a + b)^2
*Sec[c + d*x]^4)/(4*(a + b)^3*d)

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Maple [A]  time = 0.085, size = 109, normalized size = 1.2 \begin{align*} -{\frac{a}{d \left ( a+b \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{b}{2\,d \left ( a+b \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{1}{4\,d \left ( a+b \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{ \left ( a+b \right ) ^{3}d}}+{\frac{{a}^{2}\ln \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ) }{2\, \left ( a+b \right ) ^{3}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+sin(d*x+c)^2*b),x)

[Out]

-1/d/(a+b)^2/cos(d*x+c)^2*a-1/2/d/(a+b)^2/cos(d*x+c)^2*b+1/4/d/(a+b)/cos(d*x+c)^4-a^2*ln(cos(d*x+c))/(a+b)^3/d
+1/2/d*a^2/(a+b)^3*ln(b*cos(d*x+c)^2-a-b)

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Maxima [A]  time = 1.03609, size = 215, normalized size = 2.29 \begin{align*} \frac{\frac{2 \, a^{2} \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{2 \, a^{2} \log \left (\sin \left (d x + c\right )^{2} - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{2 \,{\left (2 \, a + b\right )} \sin \left (d x + c\right )^{2} - 3 \, a - b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{4} - 2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

1/4*(2*a^2*log(b*sin(d*x + c)^2 + a)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*a^2*log(sin(d*x + c)^2 - 1)/(a^3 + 3*
a^2*b + 3*a*b^2 + b^3) + (2*(2*a + b)*sin(d*x + c)^2 - 3*a - b)/((a^2 + 2*a*b + b^2)*sin(d*x + c)^4 - 2*(a^2 +
2*a*b + b^2)*sin(d*x + c)^2 + a^2 + 2*a*b + b^2))/d

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Fricas [A]  time = 2.96964, size = 288, normalized size = 3.06 \begin{align*} \frac{2 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 4 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) - 2 \,{\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}{4 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

1/4*(2*a^2*cos(d*x + c)^4*log(-b*cos(d*x + c)^2 + a + b) - 4*a^2*cos(d*x + c)^4*log(-cos(d*x + c)) - 2*(2*a^2
+ 3*a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cos(d*x + c)^4)

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

[Out]

Timed out

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Giac [B]  time = 3.14534, size = 531, normalized size = 5.65 \begin{align*} \frac{\frac{6 \, a^{2} \log \left (a - \frac{2 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{4 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{12 \, a^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{25 \, a^{2} + \frac{124 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{24 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{246 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{144 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{48 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{124 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{24 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{25 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/12*(6*a^2*log(a - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a*
(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 12*a^2*log(abs(-(cos(d*x + c) - 1
)/(cos(d*x + c) + 1) - 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + (25*a^2 + 124*a^2*(cos(d*x + c) - 1)/(cos(d*x + c
) + 1) + 24*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 246*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 14
4*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 48*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 124*a^2*(
cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 24*a*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 25*a^2*(cos(d*x
+ c) - 1)^4/(cos(d*x + c) + 1)^4)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)
^4))/d