3.2.81 \(\int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [181]

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

[Out]

-1/8*a*(4*a+b)*ln(1-sin(d*x+c))/(a+b)^4/d-1/8*a*(4*a-b)*ln(1+sin(d*x+c))/(a-b)^4/d+a^4*(a^2+5*b^2)*ln(a+b*sin(
d*x+c))/(a^2-b^2)^4/d-a^5/(a^2-b^2)^3/d/(a+b*sin(d*x+c))+1/4*sec(d*x+c)^4*(a^2+b^2-2*a*b*sin(d*x+c))/(a^2-b^2)
^2/d-1/4*sec(d*x+c)^2*(4*a^4+6*a^2*b^2-2*b^4-a*b*(9*a^2-b^2)*sin(d*x+c))/(a^2-b^2)^3/d

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Rubi [A]
time = 0.53, antiderivative size = 242, normalized size of antiderivative = 1.00, 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 = {2800, 1661, 1643} \begin {gather*} \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 {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 ^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 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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(a*(4*a + b)*Log[1 - Sin[c + d*x]])/((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])) +
(Sec[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 1643

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]

Rule 1661

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 2800

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]

Rubi steps

\begin {align*} \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {\text {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 {\text {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 {\text {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 {\text {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*}

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Mathematica [A]
time = 4.25, size = 204, normalized size = 0.84 \begin {gather*} \frac {-\frac {2 a (4 a+b) \log (1-\sin (c+d x))}{(a+b)^4}-\frac {2 a (4 a-b) \log (1+\sin (c+d x))}{(a-b)^4}+\frac {16 a^4 \left (a^2+5 b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4}+\frac {1}{(a+b)^2 (-1+\sin (c+d x))^2}+\frac {7 a+3 b}{(a+b)^3 (-1+\sin (c+d x))}+\frac {1}{(a-b)^2 (1+\sin (c+d x))^2}+\frac {-7 a+3 b}{(a-b)^3 (1+\sin (c+d x))}-\frac {16 a^5}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}}{16 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.59, size = 205, normalized size = 0.85 Too large to display

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (235) = 470\).
time = 0.40, size = 505, normalized size = 2.09 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 555 vs. \(2 (235) = 470\).
time = 0.55, size = 555, normalized size = 2.29 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{5}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 494 vs. \(2 (235) = 470\).
time = 7.52, size = 494, normalized size = 2.04 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 7.78, size = 755, normalized size = 3.12 \begin {gather*} \frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^6+5\,a^4\,b^2\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {1}{{\left (a+b\right )}^2}-\frac {7\,b}{4\,{\left (a+b\right )}^3}+\frac {3\,b^2}{4\,{\left (a+b\right )}^4}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {3\,b^2}{4\,{\left (a-b\right )}^4}+\frac {7\,b}{4\,{\left (a-b\right )}^3}+\frac {1}{{\left (a-b\right )}^2}\right )}{d}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^3+a\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a\,b^2-2\,a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a\,b^2-2\,a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (2\,a^3+a\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (11\,a^4+a^2\,b^2\right )}{2\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^4+a^2\,b^2\right )}{\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (2\,a^4+a^2\,b^2\right )}{\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (13\,a^4+31\,a^2\,b^2-8\,b^4\right )}{\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (11\,a^4+a^2\,b^2\right )}{2\,\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )} \end {gather*}

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]

(log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(a^6 + 5*a^4*b^2))/(d*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4
*b^4 - 4*a^6*b^2)) - (log(tan(c/2 + (d*x)/2) - 1)*(1/(a + b)^2 - (7*b)/(4*(a + b)^3) + (3*b^2)/(4*(a + b)^4)))
/d - (log(tan(c/2 + (d*x)/2) + 1)*((3*b^2)/(4*(a - b)^4) + (7*b)/(4*(a - b)^3) + 1/(a - b)^2))/d - ((tan(c/2 +
(d*x)/2)^2*(a*b^2 + 2*a^3))/(a^4 + b^4 - 2*a^2*b^2) + (3*tan(c/2 + (d*x)/2)^4*(a*b^2 - 2*a^3))/(a^4 + b^4 - 2
*a^2*b^2) + (3*tan(c/2 + (d*x)/2)^6*(a*b^2 - 2*a^3))/(a^4 + b^4 - 2*a^2*b^2) + (tan(c/2 + (d*x)/2)^8*(a*b^2 +
2*a^3))/(a^4 + b^4 - 2*a^2*b^2) - (b*tan(c/2 + (d*x)/2)^9*(11*a^4 + a^2*b^2))/(2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^
4*b^2)) + (8*b*tan(c/2 + (d*x)/2)^3*(2*a^4 + a^2*b^2))/((a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (8*b*tan(c/2 +
(d*x)/2)^7*(2*a^4 + a^2*b^2))/((a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b*tan(c/2 + (d*x)/2)^5*(13*a^4 - 8*b^4
+ 31*a^2*b^2))/((a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b*tan(c/2 + (d*x)/2)*(11*a^4 + a^2*b^2))/(2*(a^2 - b^2
)*(a^4 + b^4 - 2*a^2*b^2)))/(d*(a + 2*b*tan(c/2 + (d*x)/2) - 3*a*tan(c/2 + (d*x)/2)^2 + 2*a*tan(c/2 + (d*x)/2)
^4 + 2*a*tan(c/2 + (d*x)/2)^6 - 3*a*tan(c/2 + (d*x)/2)^8 + a*tan(c/2 + (d*x)/2)^10 - 8*b*tan(c/2 + (d*x)/2)^3
+ 12*b*tan(c/2 + (d*x)/2)^5 - 8*b*tan(c/2 + (d*x)/2)^7 + 2*b*tan(c/2 + (d*x)/2)^9))

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