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

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Rubi [A]  time = 0.10, antiderivative size = 94, normalized size of antiderivative = 1.00, 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 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]))

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]

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*}

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Mathematica [A]  time = 0.26, 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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fricas [A]  time = 0.64, size = 118, normalized size = 1.26 \[ \frac {2 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 4 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) - 2 \, {\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}{4 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cos \left (d x + c\right )^{4}} \]

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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giac [B]  time = 2.75, size = 393, normalized size = 4.18 \[ \frac {\frac {6 \, a^{2} \log \left (a - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {4 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {12 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {25 \, a^{2} + \frac {124 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {24 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {246 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {144 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {48 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {124 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {24 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {25 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{4}}}{12 \, d} \]

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

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maple [A]  time = 0.46, size = 109, normalized size = 1.16 \[ \frac {a^{2} \ln \left (b \left (\cos ^{2}\left (d x +c \right )\right )-a -b \right )}{2 d \left (a +b \right )^{3}}-\frac {a}{d \left (a +b \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {b}{2 d \left (a +b \right )^{2} \cos \left (d x +c \right )^{2}}+\frac {1}{4 d \left (a +b \right ) \cos \left (d x +c \right )^{4}}-\frac {a^{2} \ln \left (\cos \left (d x +c \right )\right )}{\left (a +b \right )^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 159, normalized size = 1.69 \[ \frac {\frac {2 \, a^{2} \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, a^{2} \log \left (\sin \left (d x + c\right )^{2} - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (2 \, a + b\right )} \sin \left (d x + c\right )^{2} - 3 \, a - b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}}{4 \, d} \]

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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mupad [B]  time = 14.35, size = 90, normalized size = 0.96 \[ \frac {a^2\,\left (\frac {\ln \left (\left (a+b\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )}{2}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{4}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}-a\,b\,\left (\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{2}\right )}{d\,{\left (a+b\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{5}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \]

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]

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

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