3.441 \(\int \frac{\tan ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=128 \[ \frac{\left (3 a^2+3 a b+b^2\right ) \sec ^2(c+d x)}{2 d (a+b)^3}-\frac{a^3 \log \left (a+b \sin ^2(c+d x)\right )}{2 d (a+b)^4}+\frac{a^3 \log (\cos (c+d x))}{d (a+b)^4}+\frac{\sec ^6(c+d x)}{6 d (a+b)}-\frac{(3 a+2 b) \sec ^4(c+d x)}{4 d (a+b)^2} \]

[Out]

(a^3*Log[Cos[c + d*x]])/((a + b)^4*d) - (a^3*Log[a + b*Sin[c + d*x]^2])/(2*(a + b)^4*d) + ((3*a^2 + 3*a*b + b^
2)*Sec[c + d*x]^2)/(2*(a + b)^3*d) - ((3*a + 2*b)*Sec[c + d*x]^4)/(4*(a + b)^2*d) + Sec[c + d*x]^6/(6*(a + b)*
d)

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Rubi [A]  time = 0.13169, antiderivative size = 128, 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} \[ \frac{\left (3 a^2+3 a b+b^2\right ) \sec ^2(c+d x)}{2 d (a+b)^3}-\frac{a^3 \log \left (a+b \sin ^2(c+d x)\right )}{2 d (a+b)^4}+\frac{a^3 \log (\cos (c+d x))}{d (a+b)^4}+\frac{\sec ^6(c+d x)}{6 d (a+b)}-\frac{(3 a+2 b) \sec ^4(c+d x)}{4 d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^7/(a + b*Sin[c + d*x]^2),x]

[Out]

(a^3*Log[Cos[c + d*x]])/((a + b)^4*d) - (a^3*Log[a + b*Sin[c + d*x]^2])/(2*(a + b)^4*d) + ((3*a^2 + 3*a*b + b^
2)*Sec[c + d*x]^2)/(2*(a + b)^3*d) - ((3*a + 2*b)*Sec[c + d*x]^4)/(4*(a + b)^2*d) + Sec[c + d*x]^6/(6*(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 ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{(1-x)^4 (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)^4}+\frac{3 a+2 b}{(a+b)^2 (-1+x)^3}+\frac{3 a^2+3 a b+b^2}{(a+b)^3 (-1+x)^2}+\frac{a^3}{(a+b)^4 (-1+x)}-\frac{a^3 b}{(a+b)^4 (a+b x)}\right ) \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac{a^3 \log (\cos (c+d x))}{(a+b)^4 d}-\frac{a^3 \log \left (a+b \sin ^2(c+d x)\right )}{2 (a+b)^4 d}+\frac{\left (3 a^2+3 a b+b^2\right ) \sec ^2(c+d x)}{2 (a+b)^3 d}-\frac{(3 a+2 b) \sec ^4(c+d x)}{4 (a+b)^2 d}+\frac{\sec ^6(c+d x)}{6 (a+b) d}\\ \end{align*}

Mathematica [A]  time = 0.286028, size = 113, normalized size = 0.88 \[ \frac{\frac{6 \left (3 a^2+3 a b+b^2\right ) \sec ^2(c+d x)}{(a+b)^3}-\frac{6 a^3 \log \left (a+b \sin ^2(c+d x)\right )}{(a+b)^4}+\frac{12 a^3 \log (\cos (c+d x))}{(a+b)^4}+\frac{2 \sec ^6(c+d x)}{a+b}-\frac{3 (3 a+2 b) \sec ^4(c+d x)}{(a+b)^2}}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^7/(a + b*Sin[c + d*x]^2),x]

[Out]

((12*a^3*Log[Cos[c + d*x]])/(a + b)^4 - (6*a^3*Log[a + b*Sin[c + d*x]^2])/(a + b)^4 + (6*(3*a^2 + 3*a*b + b^2)
*Sec[c + d*x]^2)/(a + b)^3 - (3*(3*a + 2*b)*Sec[c + d*x]^4)/(a + b)^2 + (2*Sec[c + d*x]^6)/(a + b))/(12*d)

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Maple [A]  time = 0.086, size = 170, normalized size = 1.3 \begin{align*}{\frac{{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{ \left ( a+b \right ) ^{4}d}}-{\frac{3\,a}{4\,d \left ( a+b \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{b}{2\,d \left ( a+b \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{2}}{2\,d \left ( a+b \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,ab}{2\,d \left ( a+b \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{2}}{2\,d \left ( a+b \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{1}{6\,d \left ( a+b \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}-{\frac{{a}^{3}\ln \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ) }{2\, \left ( a+b \right ) ^{4}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^7/(a+sin(d*x+c)^2*b),x)

[Out]

a^3*ln(cos(d*x+c))/(a+b)^4/d-3/4/d/(a+b)^2/cos(d*x+c)^4*a-1/2/d/(a+b)^2/cos(d*x+c)^4*b+3/2/d/(a+b)^3/cos(d*x+c
)^2*a^2+3/2/d/(a+b)^3/cos(d*x+c)^2*a*b+1/2/d/(a+b)^3/cos(d*x+c)^2*b^2+1/6/d/(a+b)/cos(d*x+c)^6-1/2/d*a^3/(a+b)
^4*ln(b*cos(d*x+c)^2-a-b)

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Maxima [B]  time = 1.0131, size = 369, normalized size = 2.88 \begin{align*} -\frac{\frac{6 \, a^{3} \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac{6 \, a^{3} \log \left (\sin \left (d x + c\right )^{2} - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac{6 \,{\left (3 \, a^{2} + 3 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{4} - 3 \,{\left (9 \, a^{2} + 7 \, a b + 2 \, b^{2}\right )} \sin \left (d x + c\right )^{2} + 11 \, a^{2} + 7 \, a b + 2 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )^{6} - 3 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )^{4} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3} + 3 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )^{2}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/12*(6*a^3*log(b*sin(d*x + c)^2 + a)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 6*a^3*log(sin(d*x + c)^2
- 1)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + (6*(3*a^2 + 3*a*b + b^2)*sin(d*x + c)^4 - 3*(9*a^2 + 7*a*b
+ 2*b^2)*sin(d*x + c)^2 + 11*a^2 + 7*a*b + 2*b^2)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sin(d*x + c)^6 - 3*(a^3 + 3
*a^2*b + 3*a*b^2 + b^3)*sin(d*x + c)^4 - a^3 - 3*a^2*b - 3*a*b^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sin
(d*x + c)^2))/d

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Fricas [A]  time = 4.12375, size = 420, normalized size = 3.28 \begin{align*} -\frac{6 \, a^{3} \cos \left (d x + c\right )^{6} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 12 \, a^{3} \cos \left (d x + c\right )^{6} \log \left (-\cos \left (d x + c\right )\right ) - 6 \,{\left (3 \, a^{3} + 6 \, a^{2} b + 4 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4} - 2 \, a^{3} - 6 \, a^{2} b - 6 \, a b^{2} - 2 \, b^{3} + 3 \,{\left (3 \, a^{3} + 8 \, a^{2} b + 7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}}{12 \,{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

-1/12*(6*a^3*cos(d*x + c)^6*log(-b*cos(d*x + c)^2 + a + b) - 12*a^3*cos(d*x + c)^6*log(-cos(d*x + c)) - 6*(3*a
^3 + 6*a^2*b + 4*a*b^2 + b^3)*cos(d*x + c)^4 - 2*a^3 - 6*a^2*b - 6*a*b^2 - 2*b^3 + 3*(3*a^3 + 8*a^2*b + 7*a*b^
2 + 2*b^3)*cos(d*x + c)^2)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*cos(d*x + c)^6)

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

[Out]

Timed out

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Giac [B]  time = 6.9696, size = 814, normalized size = 6.36 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^7/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

-1/60*(30*a^3*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^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 60*a^3*log(abs(-(co
s(d*x + c) - 1)/(cos(d*x + c) + 1) - 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + (147*a^3 + 1002*a^3*(co
s(d*x + c) - 1)/(cos(d*x + c) + 1) + 120*a^2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2925*a^3*(cos(d*x + c)
- 1)^2/(cos(d*x + c) + 1)^2 + 960*a^2*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 240*a*b^2*(cos(d*x + c) -
1)^2/(cos(d*x + c) + 1)^2 + 4780*a^3*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 3600*a^2*b*(cos(d*x + c) - 1)
^3/(cos(d*x + c) + 1)^3 + 2400*a*b^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 640*b^3*(cos(d*x + c) - 1)^3/
(cos(d*x + c) + 1)^3 + 2925*a^3*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 960*a^2*b*(cos(d*x + c) - 1)^4/(co
s(d*x + c) + 1)^4 + 240*a*b^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 1002*a^3*(cos(d*x + c) - 1)^5/(cos(d
*x + c) + 1)^5 + 120*a^2*b*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 147*a^3*(cos(d*x + c) - 1)^6/(cos(d*x +
c) + 1)^6)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^6))/d