3.1362 \(\int \frac {\tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=204 \[ -\frac {\left (8 a^2+9 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {\left (8 a^2-9 a b+3 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac {a^5 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3} \]

[Out]

-1/16*(8*a^2+9*a*b+3*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d-1/16*(8*a^2-9*a*b+3*b^2)*ln(1+sin(d*x+c))/(a-b)^3/d+a^5*l
n(a+b*sin(d*x+c))/(a^2-b^2)^3/d+1/4*sec(d*x+c)^4*(a-b*sin(d*x+c))/(a^2-b^2)/d-1/8*sec(d*x+c)^2*(4*a*(2*a^2-b^2
)-b*(9*a^2-5*b^2)*sin(d*x+c))/(a^2-b^2)^2/d

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Rubi [A]  time = 0.35, antiderivative size = 204, 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 = {2721, 1647, 801} \[ \frac {a^5 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {\left (8 a^2+9 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {\left (8 a^2-9 a b+3 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^2 + 9*a*b + 3*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) - ((8*a^2 - 9*a*b + 3*b^2)*Log[1 + Sin[c + d
*x]])/(16*(a - b)^3*d) + (a^5*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^3*d) + (Sec[c + d*x]^4*(a - b*Sin[c + d*x]
))/(4*(a^2 - b^2)*d) - (Sec[c + d*x]^2*(4*a*(2*a^2 - b^2) - b*(9*a^2 - 5*b^2)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*
d)

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 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]

Rubi steps

\begin {align*} \int \frac {\tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}+\frac {\operatorname {Subst}\left (\int \frac {\frac {a b^6}{a^2-b^2}-\frac {b^4 \left (4 a^2-b^2\right ) x}{a^2-b^2}-4 b^2 x^3}{(a+x) \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) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {a b^6 \left (7 a^2-3 b^2\right )}{\left (a^2-b^2\right )^2}+\frac {b^4 \left (8 a^4-7 a^2 b^2+3 b^4\right ) x}{\left (a^2-b^2\right )^2}}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac {\operatorname {Subst}\left (\int \left (\frac {b^4 \left (8 a^2+9 a b+3 b^2\right )}{2 (a+b)^3 (b-x)}+\frac {8 a^5 b^4}{(a-b)^3 (a+b)^3 (a+x)}+\frac {b^4 \left (8 a^2-9 a b+3 b^2\right )}{2 (-a+b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=-\frac {\left (8 a^2+9 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {\left (8 a^2-9 a b+3 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a^5 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.40, size = 184, normalized size = 0.90 \[ \frac {\frac {16 a^5 \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}-\frac {\left (8 a^2+9 a b+3 b^2\right ) \log (1-\sin (c+d x))}{(a+b)^3}-\frac {\left (8 a^2-9 a b+3 b^2\right ) \log (\sin (c+d x)+1)}{(a-b)^3}+\frac {7 a+5 b}{(a+b)^2 (\sin (c+d x)-1)}+\frac {5 b-7 a}{(a-b)^2 (\sin (c+d x)+1)}+\frac {1}{(a+b) (\sin (c+d x)-1)^2}+\frac {1}{(a-b) (\sin (c+d x)+1)^2}}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(((8*a^2 + 9*a*b + 3*b^2)*Log[1 - Sin[c + d*x]])/(a + b)^3) - ((8*a^2 - 9*a*b + 3*b^2)*Log[1 + Sin[c + d*x]]
)/(a - b)^3 + (16*a^5*Log[a + b*Sin[c + d*x]])/((a - b)^3*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^2) + (7*
a + 5*b)/((a + b)^2*(-1 + Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x])^2) + (-7*a + 5*b)/((a - b)^2*(1 + Sin
[c + d*x])))/(16*d)

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fricas [A]  time = 0.95, size = 261, normalized size = 1.28 \[ \frac {16 \, a^{5} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (8 \, a^{5} + 15 \, a^{4} b - 10 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a^{5} - 15 \, a^{4} b + 10 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{5} - 8 \, a^{3} b^{2} + 4 \, a b^{4} - 8 \, {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - {\left (9 \, a^{4} b - 14 \, a^{2} b^{3} + 5 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*a^5*cos(d*x + c)^4*log(b*sin(d*x + c) + a) - (8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cos(d*x + c)^4*l
og(sin(d*x + c) + 1) - (8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a^5 -
8*a^3*b^2 + 4*a*b^4 - 8*(2*a^5 - 3*a^3*b^2 + a*b^4)*cos(d*x + c)^2 - 2*(2*a^4*b - 4*a^2*b^3 + 2*b^5 - (9*a^4*
b - 14*a^2*b^3 + 5*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^4)

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giac [A]  time = 0.31, size = 343, normalized size = 1.68 \[ \frac {\frac {16 \, a^{5} b \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (8 \, a^{2} - 9 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (8 \, a^{2} + 9 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (6 \, a^{5} \sin \left (d x + c\right )^{4} - 9 \, a^{4} b \sin \left (d x + c\right )^{3} + 14 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} - 5 \, b^{5} \sin \left (d x + c\right )^{3} - 4 \, a^{5} \sin \left (d x + c\right )^{2} - 12 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 4 \, a b^{4} \sin \left (d x + c\right )^{2} + 7 \, a^{4} b \sin \left (d x + c\right ) - 10 \, a^{2} b^{3} \sin \left (d x + c\right ) + 3 \, b^{5} \sin \left (d x + c\right ) + 8 \, a^{3} b^{2} - 2 \, a b^{4}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*a^5*b*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - (8*a^2 - 9*a*b + 3*b^2)*lo
g(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (8*a^2 + 9*a*b + 3*b^2)*log(abs(sin(d*x + c) - 1))/
(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2*(6*a^5*sin(d*x + c)^4 - 9*a^4*b*sin(d*x + c)^3 + 14*a^2*b^3*sin(d*x + c)^3
- 5*b^5*sin(d*x + c)^3 - 4*a^5*sin(d*x + c)^2 - 12*a^3*b^2*sin(d*x + c)^2 + 4*a*b^4*sin(d*x + c)^2 + 7*a^4*b*
sin(d*x + c) - 10*a^2*b^3*sin(d*x + c) + 3*b^5*sin(d*x + c) + 8*a^3*b^2 - 2*a*b^4)/((a^6 - 3*a^4*b^2 + 3*a^2*b
^4 - b^6)*(sin(d*x + c)^2 - 1)^2))/d

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maple [A]  time = 0.45, size = 304, normalized size = 1.49 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {7 a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {5 b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a^{2}}{2 d \left (a +b \right )^{3}}-\frac {9 \ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) b^{2}}{16 d \left (a +b \right )^{3}}+\frac {a^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {7 a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {5 b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right ) a^{2}}{2 d \left (a -b \right )^{3}}+\frac {9 \ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) b^{2}}{16 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/(8*a+8*b)/(sin(d*x+c)-1)^2+7/16/d/(a+b)^2/(sin(d*x+c)-1)*a+5/16/d/(a+b)^2/(sin(d*x+c)-1)*b-1/2/d/(a+b)^3
*ln(sin(d*x+c)-1)*a^2-9/16/d/(a+b)^3*ln(sin(d*x+c)-1)*a*b-3/16/d/(a+b)^3*ln(sin(d*x+c)-1)*b^2+1/d*a^5/(a+b)^3/
(a-b)^3*ln(a+b*sin(d*x+c))+1/2/d/(8*a-8*b)/(1+sin(d*x+c))^2-7/16/d/(a-b)^2/(1+sin(d*x+c))*a+5/16/d/(a-b)^2/(1+
sin(d*x+c))*b-1/2/d/(a-b)^3*ln(1+sin(d*x+c))*a^2+9/16/d/(a-b)^3*ln(1+sin(d*x+c))*a*b-3/16/d/(a-b)^3*ln(1+sin(d
*x+c))*b^2

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maxima [A]  time = 0.33, size = 288, normalized size = 1.41 \[ \frac {\frac {16 \, a^{5} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (8 \, a^{2} - 9 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (8 \, a^{2} + 9 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (9 \, a^{2} b - 5 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 6 \, a^{3} - 2 \, a b^{2} - 4 \, {\left (2 \, a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (7 \, a^{2} b - 3 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*a^5*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - (8*a^2 - 9*a*b + 3*b^2)*log(sin(d*x
+ c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (8*a^2 + 9*a*b + 3*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*
a*b^2 + b^3) - 2*((9*a^2*b - 5*b^3)*sin(d*x + c)^3 + 6*a^3 - 2*a*b^2 - 4*(2*a^3 - a*b^2)*sin(d*x + c)^2 - (7*a
^2*b - 3*b^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b
^2 + b^4)*sin(d*x + c)^2))/d

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mupad [B]  time = 12.54, size = 498, normalized size = 2.44 \[ \frac {a^5\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {1}{a+b}-\frac {7\,b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {7\,b}{8\,{\left (a-b\right )}^2}+\frac {1}{a-b}\right )}{d}-\frac {\frac {2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a\,b^2-2\,a^3\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (7\,a^2\,b-3\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (15\,a^2\,b-11\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (15\,a^2\,b-11\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,a^2-3\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^5*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (log(
tan(c/2 + (d*x)/2) - 1)*(1/(a + b) - (7*b)/(8*(a + b)^2) + b^2/(4*(a + b)^3)))/d - (log(tan(c/2 + (d*x)/2) + 1
)*(b^2/(4*(a - b)^3) + (7*b)/(8*(a - b)^2) + 1/(a - b)))/d - ((2*a^3*tan(c/2 + (d*x)/2)^2)/(a^4 + b^4 - 2*a^2*
b^2) + (2*a^3*tan(c/2 + (d*x)/2)^6)/(a^4 + b^4 - 2*a^2*b^2) + (4*tan(c/2 + (d*x)/2)^4*(a*b^2 - 2*a^3))/(a^4 +
b^4 - 2*a^2*b^2) - (tan(c/2 + (d*x)/2)^7*(7*a^2*b - 3*b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^
3*(15*a^2*b - 11*b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(15*a^2*b - 11*b^3))/(4*(a^4 + b^4
- 2*a^2*b^2)) - (b*tan(c/2 + (d*x)/2)*(7*a^2 - 3*b^2))/(4*(a^4 + b^4 - 2*a^2*b^2)))/(d*(6*tan(c/2 + (d*x)/2)^4
- 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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