3.1363 \(\int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=190 \[ -\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {\sec ^2(c+d x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}-\frac {a^4 b \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {a (3 a+b) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {a (3 a-b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]

[Out]

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

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Rubi [A]  time = 0.43, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2837, 12, 1647, 801} \[ -\frac {a^4 b \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {\sec ^2(c+d x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}-\frac {a (3 a+b) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {a (3 a-b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]*Tan[c + d*x]^4)/(a + b*Sin[c + d*x]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2837

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

Rubi steps

\begin {align*} \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {x^4}{b^4 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b \operatorname {Subst}\left (\int \frac {x^4}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\sec ^4(c+d x) \left (\frac {b}{a^2-b^2}-\frac {a \sin (c+d x)}{a^2-b^2}\right )}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {a^2 b^4}{a^2-b^2}+\frac {3 a b^4 x}{a^2-b^2}-4 b^2 x^2}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b d}\\ &=-\frac {\sec ^4(c+d x) \left (\frac {b}{a^2-b^2}-\frac {a \sin (c+d x)}{a^2-b^2}\right )}{4 d}+\frac {\sec ^2(c+d x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-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^2 b^4 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^2}-\frac {a b^4 \left (5 a^2-b^2\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^3 d}\\ &=-\frac {\sec ^4(c+d x) \left (\frac {b}{a^2-b^2}-\frac {a \sin (c+d x)}{a^2-b^2}\right )}{4 d}+\frac {\sec ^2(c+d x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac {\operatorname {Subst}\left (\int \left (\frac {a b^3 (3 a+b)}{2 (a+b)^3 (b-x)}-\frac {8 a^4 b^4}{(a-b)^3 (a+b)^3 (a+x)}+\frac {a (3 a-b) b^3}{2 (a-b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^3 d}\\ &=-\frac {a (3 a+b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {a (3 a-b) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {a^4 b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac {\sec ^4(c+d x) \left (\frac {b}{a^2-b^2}-\frac {a \sin (c+d x)}{a^2-b^2}\right )}{4 d}+\frac {\sec ^2(c+d x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-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 = 1.58, size = 169, normalized size = 0.89 \[ \frac {-\frac {16 a^4 b \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac {5 a+3 b}{(a+b)^2 (\sin (c+d x)-1)}+\frac {5 a-3 b}{(a-b)^2 (\sin (c+d x)+1)}+\frac {1}{(a+b) (\sin (c+d x)-1)^2}-\frac {1}{(a-b) (\sin (c+d x)+1)^2}-\frac {a (3 a+b) \log (1-\sin (c+d x))}{(a+b)^3}+\frac {a (3 a-b) \log (\sin (c+d x)+1)}{(a-b)^3}}{16 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]*Tan[c + d*x]^4)/(a + b*Sin[c + d*x]),x]

[Out]

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

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fricas [A]  time = 0.97, size = 261, normalized size = 1.37 \[ -\frac {16 \, a^{4} b \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (3 \, a^{5} + 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, a^{5} - 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{4} b - 8 \, a^{2} b^{3} + 4 \, b^{5} - 8 \, {\left (2 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} - 4 \, a^{3} b^{2} + 2 \, a b^{4} - {\left (5 \, a^{5} - 6 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/16*(16*a^4*b*cos(d*x + c)^4*log(b*sin(d*x + c) + a) - (3*a^5 + 8*a^4*b + 6*a^3*b^2 - a*b^4)*cos(d*x + c)^4*
log(sin(d*x + c) + 1) + (3*a^5 - 8*a^4*b + 6*a^3*b^2 - a*b^4)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a^4*b
- 8*a^2*b^3 + 4*b^5 - 8*(2*a^4*b - 3*a^2*b^3 + b^5)*cos(d*x + c)^2 - 2*(2*a^5 - 4*a^3*b^2 + 2*a*b^4 - (5*a^5 -
6*a^3*b^2 + a*b^4)*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 = 333, normalized size = 1.75 \[ -\frac {\frac {16 \, a^{4} b^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (3 \, a^{2} - a b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a^{2} + a b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (6 \, a^{4} b \sin \left (d x + c\right )^{4} - 5 \, a^{5} \sin \left (d x + c\right )^{3} + 6 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} - a b^{4} \sin \left (d x + c\right )^{3} - 4 \, a^{4} b \sin \left (d x + c\right )^{2} - 12 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} + 4 \, b^{5} \sin \left (d x + c\right )^{2} + 3 \, a^{5} \sin \left (d x + c\right ) - 2 \, a^{3} b^{2} \sin \left (d x + c\right ) - a b^{4} \sin \left (d x + c\right ) + 8 \, a^{2} b^{3} - 2 \, b^{5}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/16*(16*a^4*b^2*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - (3*a^2 - a*b)*log(abs(s
in(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a^2 + a*b)*log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b +
3*a*b^2 + b^3) + 2*(6*a^4*b*sin(d*x + c)^4 - 5*a^5*sin(d*x + c)^3 + 6*a^3*b^2*sin(d*x + c)^3 - a*b^4*sin(d*x +
c)^3 - 4*a^4*b*sin(d*x + c)^2 - 12*a^2*b^3*sin(d*x + c)^2 + 4*b^5*sin(d*x + c)^2 + 3*a^5*sin(d*x + c) - 2*a^3
*b^2*sin(d*x + c) - a*b^4*sin(d*x + c) + 8*a^2*b^3 - 2*b^5)/((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 = 0.40, size = 260, normalized size = 1.37 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {5 a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {3 b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) a^{2}}{16 d \left (a +b \right )^{3}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}-\frac {a^{4} b \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5 a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {3 b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) a^{2}}{16 d \left (a -b \right )^{3}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^4/(a+b*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.51, size = 276, normalized size = 1.45 \[ -\frac {\frac {16 \, a^{4} b \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (3 \, a^{2} - a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a^{2} + a b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (5 \, a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{3} + 6 \, a^{2} b - 2 \, b^{3} - 4 \, {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{2} - {\left (3 \, a^{3} + a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*(16*a^4*b*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - (3*a^2 - a*b)*log(sin(d*x + c) +
1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a^2 + a*b)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*
((5*a^3 - a*b^2)*sin(d*x + c)^3 + 6*a^2*b - 2*b^3 - 4*(2*a^2*b - b^3)*sin(d*x + c)^2 - (3*a^3 + a*b^2)*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 = 12.45, size = 507, normalized size = 2.67 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {5\,b}{8\,{\left (a-b\right )}^2}+\frac {3}{8\,\left (a-b\right )}\right )}{d}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (3\,a^3+a\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,a\,b^2-11\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (7\,a\,b^2-11\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2\,b-b^3\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^4-2\,a^2\,b^2+b^4}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2+b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {3}{8\,\left (a+b\right )}-\frac {5\,b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {a^4\,b\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^4/(cos(c + d*x)^5*(a + b*sin(c + d*x))),x)

[Out]

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

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

[Out]

Timed out

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