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

Optimal. Leaf size=242 \[ -\frac{a^5}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac{a^4 \left (a^2+5 b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}+\frac{\sec ^4(c+d x) \left (a^2-2 a b \sin (c+d x)+b^2\right )}{4 d \left (a^2-b^2\right )^2}-\frac{\sec ^2(c+d x) \left (2 \left (3 a^2 b^2+2 a^4-b^4\right )-a b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{4 d \left (a^2-b^2\right )^3}-\frac{a (4 a+b) \log (1-\sin (c+d x))}{8 d (a+b)^4}-\frac{a (4 a-b) \log (\sin (c+d x)+1)}{8 d (a-b)^4} \]

[Out]

-(a*(4*a + b)*Log[1 - Sin[c + d*x]])/(8*(a + b)^4*d) - (a*(4*a - b)*Log[1 + Sin[c + d*x]])/(8*(a - b)^4*d) + (
a^4*(a^2 + 5*b^2)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^4*d) - a^5/((a^2 - b^2)^3*d*(a + b*Sin[c + d*x])) + (S
ec[c + d*x]^4*(a^2 + b^2 - 2*a*b*Sin[c + d*x]))/(4*(a^2 - b^2)^2*d) - (Sec[c + d*x]^2*(2*(2*a^4 + 3*a^2*b^2 -
b^4) - a*b*(9*a^2 - b^2)*Sin[c + d*x]))/(4*(a^2 - b^2)^3*d)

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Rubi [A]  time = 0.632621, antiderivative size = 242, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2721, 1647, 1629} \[ -\frac{a^5}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac{a^4 \left (a^2+5 b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}+\frac{\sec ^4(c+d x) \left (a^2-2 a b \sin (c+d x)+b^2\right )}{4 d \left (a^2-b^2\right )^2}-\frac{\sec ^2(c+d x) \left (2 \left (3 a^2 b^2+2 a^4-b^4\right )-a b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{4 d \left (a^2-b^2\right )^3}-\frac{a (4 a+b) \log (1-\sin (c+d x))}{8 d (a+b)^4}-\frac{a (4 a-b) \log (\sin (c+d x)+1)}{8 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(4*a + b)*Log[1 - Sin[c + d*x]])/(8*(a + b)^4*d) - (a*(4*a - b)*Log[1 + Sin[c + d*x]])/(8*(a - b)^4*d) + (
a^4*(a^2 + 5*b^2)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^4*d) - a^5/((a^2 - b^2)^3*d*(a + b*Sin[c + d*x])) + (S
ec[c + d*x]^4*(a^2 + b^2 - 2*a*b*Sin[c + d*x]))/(4*(a^2 - b^2)^2*d) - (Sec[c + d*x]^2*(2*(2*a^4 + 3*a^2*b^2 -
b^4) - a*b*(9*a^2 - b^2)*Sin[c + d*x]))/(4*(a^2 - b^2)^3*d)

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int \frac{\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5}{(a+x)^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^4(c+d x) \left (a^2+b^2-2 a b \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^2 d}+\frac{\operatorname{Subst}\left (\int \frac{\frac{2 a^3 b^6}{\left (a^2-b^2\right )^2}-\frac{4 a^4 b^4 x}{\left (a^2-b^2\right )^2}-\frac{6 a b^6 x^2}{\left (a^2-b^2\right )^2}-4 b^2 x^3}{(a+x)^2 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b^2 d}\\ &=\frac{\sec ^4(c+d x) \left (a^2+b^2-2 a b \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^2 d}-\frac{\sec ^2(c+d x) \left (2 \left (2 a^4+3 a^2 b^2-b^4\right )-a b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{2 a^3 b^6 \left (7 a^2+b^2\right )}{\left (a^2-b^2\right )^3}+\frac{4 a^2 b^4 \left (2 a^2+b^2\right ) x}{\left (a^2-b^2\right )^2}+\frac{2 a b^6 \left (9 a^2-b^2\right ) x^2}{\left (a^2-b^2\right )^3}}{(a+x)^2 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=\frac{\sec ^4(c+d x) \left (a^2+b^2-2 a b \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^2 d}-\frac{\sec ^2(c+d x) \left (2 \left (2 a^4+3 a^2 b^2-b^4\right )-a b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 d}+\frac{\operatorname{Subst}\left (\int \left (\frac{a b^4 (4 a+b)}{(a+b)^4 (b-x)}+\frac{8 a^5 b^4}{(a-b)^3 (a+b)^3 (a+x)^2}+\frac{8 a^4 b^4 \left (a^2+5 b^2\right )}{(a-b)^4 (a+b)^4 (a+x)}-\frac{a (4 a-b) b^4}{(a-b)^4 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=-\frac{a (4 a+b) \log (1-\sin (c+d x))}{8 (a+b)^4 d}-\frac{a (4 a-b) \log (1+\sin (c+d x))}{8 (a-b)^4 d}+\frac{a^4 \left (a^2+5 b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4 d}-\frac{a^5}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{\sec ^4(c+d x) \left (a^2+b^2-2 a b \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^2 d}-\frac{\sec ^2(c+d x) \left (2 \left (2 a^4+3 a^2 b^2-b^4\right )-a b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 d}\\ \end{align*}

Mathematica [A]  time = 6.25186, size = 240, normalized size = 0.99 \[ -\frac{a^5}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac{a^4 \left (a^2+5 b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}-\frac{7 a+3 b}{16 d (a+b)^3 (1-\sin (c+d x))}-\frac{7 a-3 b}{16 d (a-b)^3 (\sin (c+d x)+1)}+\frac{1}{16 d (a+b)^2 (1-\sin (c+d x))^2}+\frac{1}{16 d (a-b)^2 (\sin (c+d x)+1)^2}-\frac{a (4 a+b) \log (1-\sin (c+d x))}{8 d (a+b)^4}-\frac{a (4 a-b) \log (\sin (c+d x)+1)}{8 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(4*a + b)*Log[1 - Sin[c + d*x]])/(8*(a + b)^4*d) - (a*(4*a - b)*Log[1 + Sin[c + d*x]])/(8*(a - b)^4*d) + (
a^4*(a^2 + 5*b^2)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^4*d) + 1/(16*(a + b)^2*d*(1 - Sin[c + d*x])^2) - (7*a
+ 3*b)/(16*(a + b)^3*d*(1 - Sin[c + d*x])) + 1/(16*(a - b)^2*d*(1 + Sin[c + d*x])^2) - (7*a - 3*b)/(16*(a - b)
^3*d*(1 + Sin[c + d*x])) - a^5/((a^2 - b^2)^3*d*(a + b*Sin[c + d*x]))

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Maple [A]  time = 0.093, size = 318, normalized size = 1.3 \begin{align*} -{\frac{{a}^{5}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3} \left ( a+b\sin \left ( dx+c \right ) \right ) }}+{\frac{{a}^{6}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}}+5\,{\frac{{a}^{4}\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{2}}{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}}+{\frac{1}{16\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{3\,b}{16\,d \left ( a+b \right ) ^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{7\,a}{16\,d \left ( a+b \right ) ^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{2\,d \left ( a+b \right ) ^{4}}}-{\frac{a\ln \left ( \sin \left ( dx+c \right ) -1 \right ) b}{8\,d \left ( a+b \right ) ^{4}}}+{\frac{1}{16\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,b}{16\,d \left ( a-b \right ) ^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{7\,a}{16\,d \left ( a-b \right ) ^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{{a}^{2}\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{2\,d \left ( a-b \right ) ^{4}}}+{\frac{a\ln \left ( 1+\sin \left ( dx+c \right ) \right ) b}{8\,d \left ( a-b \right ) ^{4}}} \end{align*}

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/d*a^5/(a+b)^3/(a-b)^3/(a+b*sin(d*x+c))+1/d*a^6/(a+b)^4/(a-b)^4*ln(a+b*sin(d*x+c))+5/d*a^4/(a+b)^4/(a-b)^4*l
n(a+b*sin(d*x+c))*b^2+1/16/d/(a+b)^2/(sin(d*x+c)-1)^2+3/16/d/(a+b)^3/(sin(d*x+c)-1)*b+7/16/d/(a+b)^3/(sin(d*x+
c)-1)*a-1/2/d*a^2/(a+b)^4*ln(sin(d*x+c)-1)-1/8/d*a/(a+b)^4*ln(sin(d*x+c)-1)*b+1/16/d/(a-b)^2/(1+sin(d*x+c))^2+
3/16/d/(a-b)^3/(1+sin(d*x+c))*b-7/16/d/(a-b)^3/(1+sin(d*x+c))*a-1/2/d*a^2/(a-b)^4*ln(1+sin(d*x+c))+1/8/d*a/(a-
b)^4*ln(1+sin(d*x+c))*b

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Maxima [B]  time = 1.8279, size = 682, normalized size = 2.82 \begin{align*} \frac{\frac{8 \,{\left (a^{6} + 5 \, a^{4} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac{{\left (4 \, a^{2} - a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac{{\left (4 \, a^{2} + a b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac{2 \,{\left (7 \, a^{5} + 6 \, a^{3} b^{2} - a b^{4} +{\left (4 \, a^{5} + 9 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{4} +{\left (5 \, a^{4} b - 7 \, a^{2} b^{3} + 2 \, b^{5}\right )} \sin \left (d x + c\right )^{3} -{\left (12 \, a^{5} + 13 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{2} -{\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )\right )}}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6} +{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )^{5} +{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \sin \left (d x + c\right )^{4} - 2 \,{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )^{3} - 2 \,{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \sin \left (d x + c\right )^{2} +{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )}}{8 \, 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]

1/8*(8*(a^6 + 5*a^4*b^2)*log(b*sin(d*x + c) + a)/(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) - (4*a^2 - a*
b)*log(sin(d*x + c) + 1)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (4*a^2 + a*b)*log(sin(d*x + c) - 1)/(a^
4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 2*(7*a^5 + 6*a^3*b^2 - a*b^4 + (4*a^5 + 9*a^3*b^2 - a*b^4)*sin(d*x
+ c)^4 + (5*a^4*b - 7*a^2*b^3 + 2*b^5)*sin(d*x + c)^3 - (12*a^5 + 13*a^3*b^2 - a*b^4)*sin(d*x + c)^2 - (4*a^4*
b - 5*a^2*b^3 + b^5)*sin(d*x + c))/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 + (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7
)*sin(d*x + c)^5 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*sin(d*x + c)^4 - 2*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b
^7)*sin(d*x + c)^3 - 2*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*sin(d*x + c)^2 + (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 -
b^7)*sin(d*x + c)))/d

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Fricas [B]  time = 3.38575, size = 1231, normalized size = 5.09 \begin{align*} \frac{2 \, a^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} - 2 \,{\left (4 \, a^{7} + 5 \, a^{5} b^{2} - 10 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (4 \, a^{7} - 9 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left ({\left (a^{6} b + 5 \, a^{4} b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) +{\left (a^{7} + 5 \, a^{5} b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) -{\left ({\left (4 \, a^{6} b + 15 \, a^{5} b^{2} + 20 \, a^{4} b^{3} + 10 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) +{\left (4 \, a^{7} + 15 \, a^{6} b + 20 \, a^{5} b^{2} + 10 \, a^{4} b^{3} - a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (4 \, a^{6} b - 15 \, a^{5} b^{2} + 20 \, a^{4} b^{3} - 10 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) +{\left (4 \, a^{7} - 15 \, a^{6} b + 20 \, a^{5} b^{2} - 10 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7} -{\left (5 \, a^{6} b - 12 \, a^{4} b^{3} + 9 \, a^{2} b^{5} - 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8 \,{\left ({\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) +{\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right )^{4}\right )}} \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]

1/8*(2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 - 2*(4*a^7 + 5*a^5*b^2 - 10*a^3*b^4 + a*b^6)*cos(d*x + c)^4 - 2*(
4*a^7 - 9*a^5*b^2 + 6*a^3*b^4 - a*b^6)*cos(d*x + c)^2 + 8*((a^6*b + 5*a^4*b^3)*cos(d*x + c)^4*sin(d*x + c) + (
a^7 + 5*a^5*b^2)*cos(d*x + c)^4)*log(b*sin(d*x + c) + a) - ((4*a^6*b + 15*a^5*b^2 + 20*a^4*b^3 + 10*a^3*b^4 -
a*b^6)*cos(d*x + c)^4*sin(d*x + c) + (4*a^7 + 15*a^6*b + 20*a^5*b^2 + 10*a^4*b^3 - a^2*b^5)*cos(d*x + c)^4)*lo
g(sin(d*x + c) + 1) - ((4*a^6*b - 15*a^5*b^2 + 20*a^4*b^3 - 10*a^3*b^4 + a*b^6)*cos(d*x + c)^4*sin(d*x + c) +
(4*a^7 - 15*a^6*b + 20*a^5*b^2 - 10*a^4*b^3 + a^2*b^5)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 2*(a^6*b - 3*a
^4*b^3 + 3*a^2*b^5 - b^7 - (5*a^6*b - 12*a^4*b^3 + 9*a^2*b^5 - 2*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b -
4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^4*sin(d*x + c) + (a^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*
b^6 + a*b^8)*d*cos(d*x + c)^4)

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

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

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Giac [B]  time = 4.67247, size = 667, normalized size = 2.76 \begin{align*} \frac{\frac{8 \,{\left (a^{6} b + 5 \, a^{4} b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}} - \frac{{\left (4 \, a^{2} - a b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac{{\left (4 \, a^{2} + a b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac{8 \,{\left (a^{6} b \sin \left (d x + c\right ) + 5 \, a^{4} b^{3} \sin \left (d x + c\right ) + 2 \, a^{7} + 4 \, a^{5} b^{2}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}} + \frac{2 \,{\left (3 \, a^{6} \sin \left (d x + c\right )^{4} + 15 \, a^{4} b^{2} \sin \left (d x + c\right )^{4} - 9 \, a^{5} b \sin \left (d x + c\right )^{3} + 10 \, a^{3} b^{3} \sin \left (d x + c\right )^{3} - a b^{5} \sin \left (d x + c\right )^{3} - 2 \, a^{6} \sin \left (d x + c\right )^{2} - 28 \, a^{4} b^{2} \sin \left (d x + c\right )^{2} - 8 \, a^{2} b^{4} \sin \left (d x + c\right )^{2} + 2 \, b^{6} \sin \left (d x + c\right )^{2} + 7 \, a^{5} b \sin \left (d x + c\right ) - 6 \, a^{3} b^{3} \sin \left (d x + c\right ) - a b^{5} \sin \left (d x + c\right ) + 12 \, a^{4} b^{2} + 7 \, a^{2} b^{4} - b^{6}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{8 \, 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]

1/8*(8*(a^6*b + 5*a^4*b^3)*log(abs(b*sin(d*x + c) + a))/(a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9) - (4
*a^2 - a*b)*log(abs(sin(d*x + c) + 1))/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (4*a^2 + a*b)*log(abs(sin
(d*x + c) - 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 8*(a^6*b*sin(d*x + c) + 5*a^4*b^3*sin(d*x + c) +
2*a^7 + 4*a^5*b^2)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(b*sin(d*x + c) + a)) + 2*(3*a^6*sin(d*x
+ c)^4 + 15*a^4*b^2*sin(d*x + c)^4 - 9*a^5*b*sin(d*x + c)^3 + 10*a^3*b^3*sin(d*x + c)^3 - a*b^5*sin(d*x + c)^3
- 2*a^6*sin(d*x + c)^2 - 28*a^4*b^2*sin(d*x + c)^2 - 8*a^2*b^4*sin(d*x + c)^2 + 2*b^6*sin(d*x + c)^2 + 7*a^5*
b*sin(d*x + c) - 6*a^3*b^3*sin(d*x + c) - a*b^5*sin(d*x + c) + 12*a^4*b^2 + 7*a^2*b^4 - b^6)/((a^8 - 4*a^6*b^2
+ 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(sin(d*x + c)^2 - 1)^2))/d