3.1361 \(\int \frac {\sin (c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=208 \[ -\frac {\left (15 a^2+21 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {\left (15 a^2-21 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {\sec ^2(c+d x) \left (4 b \left (3 a^2-2 b^2\right )-a \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}-\frac {a^6 \log (a+b \sin (c+d x))}{b d \left (a^2-b^2\right )^3} \]

[Out]

-1/16*(15*a^2+21*a*b+8*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d+1/16*(15*a^2-21*a*b+8*b^2)*ln(1+sin(d*x+c))/(a-b)^3/d-a
^6*ln(a+b*sin(d*x+c))/b/(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*(3*a
^2-2*b^2)-a*(9*a^2-5*b^2)*sin(d*x+c))/(a^2-b^2)^2/d

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Rubi [A]  time = 0.51, antiderivative size = 208, 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, 1629} \[ -\frac {a^6 \log (a+b \sin (c+d x))}{b d \left (a^2-b^2\right )^3}-\frac {\left (15 a^2+21 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {\left (15 a^2-21 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {\sec ^2(c+d x) \left (4 b \left (3 a^2-2 b^2\right )-a \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*a^2 + 21*a*b + 8*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) + ((15*a^2 - 21*a*b + 8*b^2)*Log[1 + Sin[c
+ d*x]])/(16*(a - b)^3*d) - (a^6*Log[a + b*Sin[c + d*x]])/(b*(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*(3*a^2 - 2*b^2) - a*(9*a^2 - 5*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 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]

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 {\sin (c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {x^6}{b^6 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{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 {\operatorname {Subst}\left (\int \frac {-\frac {a^2 b^6}{a^2-b^2}+\frac {3 a b^6 x}{a^2-b^2}-4 b^4 x^2-4 b^2 x^4}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b^3 d}\\ &=\frac {\sec ^2(c+d x) \left (4 b \left (3 a^2-2 b^2\right )-a \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 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^6 \left (7 a^2-3 b^2\right )}{\left (a^2-b^2\right )^2}-\frac {a b^6 \left (9 a^2-5 b^2\right ) x}{\left (a^2-b^2\right )^2}+8 b^4 x^2}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 b^5 d}\\ &=\frac {\sec ^2(c+d x) \left (4 b \left (3 a^2-2 b^2\right )-a \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 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 \left (\frac {b^5 \left (15 a^2+21 a b+8 b^2\right )}{2 (a+b)^3 (b-x)}-\frac {8 a^6 b^4}{(a-b)^3 (a+b)^3 (a+x)}+\frac {b^5 \left (15 a^2-21 a b+8 b^2\right )}{2 (a-b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^5 d}\\ &=-\frac {\left (15 a^2+21 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {\left (15 a^2-21 a b+8 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {a^6 \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right )^3 d}+\frac {\sec ^2(c+d x) \left (4 b \left (3 a^2-2 b^2\right )-a \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 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}\\ \end {align*}

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Mathematica [A]  time = 1.75, size = 187, normalized size = 0.90 \[ \frac {-\frac {16 a^6 \log (a+b \sin (c+d x))}{b (a-b)^3 (a+b)^3}-\frac {\left (15 a^2+21 a b+8 b^2\right ) \log (1-\sin (c+d x))}{(a+b)^3}+\frac {\left (15 a^2-21 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{(a-b)^3}+\frac {9 a+7 b}{(a+b)^2 (\sin (c+d x)-1)}+\frac {9 a-7 b}{(a-b)^2 (\sin (c+d x)+1)}+\frac {1}{(a+b) (\sin (c+d x)-1)^2}-\frac {1}{(a-b) (\sin (c+d x)+1)^2}}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(((15*a^2 + 21*a*b + 8*b^2)*Log[1 - Sin[c + d*x]])/(a + b)^3) + ((15*a^2 - 21*a*b + 8*b^2)*Log[1 + Sin[c + d
*x]])/(a - b)^3 - (16*a^6*Log[a + b*Sin[c + d*x]])/((a - b)^3*b*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^2)
+ (9*a + 7*b)/((a + b)^2*(-1 + Sin[c + d*x])) - 1/((a - b)*(1 + Sin[c + d*x])^2) + (9*a - 7*b)/((a - b)^2*(1
+ Sin[c + d*x])))/(16*d)

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fricas [A]  time = 1.04, size = 303, normalized size = 1.46 \[ -\frac {16 \, a^{6} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, a^{4} b^{2} - 8 \, a^{2} b^{4} + 4 \, b^{6} - {\left (15 \, a^{5} b + 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} - 24 \, a^{2} b^{4} + 3 \, a b^{5} + 8 \, b^{6}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (15 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 3 \, a b^{5} - 8 \, b^{6}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, {\left (3 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} b - 4 \, a^{3} b^{3} + 2 \, a b^{5} - {\left (9 \, a^{5} b - 14 \, a^{3} b^{3} + 5 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/16*(16*a^6*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + 4*a^4*b^2 - 8*a^2*b^4 + 4*b^6 - (15*a^5*b + 24*a^4*b^2
- 10*a^3*b^3 - 24*a^2*b^4 + 3*a*b^5 + 8*b^6)*cos(d*x + c)^4*log(sin(d*x + c) + 1) + (15*a^5*b - 24*a^4*b^2 - 1
0*a^3*b^3 + 24*a^2*b^4 + 3*a*b^5 - 8*b^6)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 8*(3*a^4*b^2 - 5*a^2*b^4 + 2
*b^6)*cos(d*x + c)^2 - 2*(2*a^5*b - 4*a^3*b^3 + 2*a*b^5 - (9*a^5*b - 14*a^3*b^3 + 5*a*b^5)*cos(d*x + c)^2)*sin
(d*x + c))/((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x + c)^4)

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giac [A]  time = 0.29, size = 371, normalized size = 1.78 \[ -\frac {\frac {16 \, a^{6} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (15 \, a^{2} - 21 \, a b + 8 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (15 \, a^{2} + 21 \, a b + 8 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (18 \, a^{4} b \sin \left (d x + c\right )^{4} - 18 \, a^{2} b^{3} \sin \left (d x + c\right )^{4} + 6 \, b^{5} \sin \left (d x + c\right )^{4} - 9 \, a^{5} \sin \left (d x + c\right )^{3} + 14 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} - 5 \, a b^{4} \sin \left (d x + c\right )^{3} - 24 \, a^{4} b \sin \left (d x + c\right )^{2} + 16 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} - 4 \, b^{5} \sin \left (d x + c\right )^{2} + 7 \, a^{5} \sin \left (d x + c\right ) - 10 \, a^{3} b^{2} \sin \left (d x + c\right ) + 3 \, a b^{4} \sin \left (d x + c\right ) + 8 \, a^{4} b - 2 \, a^{2} b^{3}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/16*(16*a^6*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - (15*a^2 - 21*a*b + 8*b^2)*l
og(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (15*a^2 + 21*a*b + 8*b^2)*log(abs(sin(d*x + c) - 1
))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2*(18*a^4*b*sin(d*x + c)^4 - 18*a^2*b^3*sin(d*x + c)^4 + 6*b^5*sin(d*x +
c)^4 - 9*a^5*sin(d*x + c)^3 + 14*a^3*b^2*sin(d*x + c)^3 - 5*a*b^4*sin(d*x + c)^3 - 24*a^4*b*sin(d*x + c)^2 + 1
6*a^2*b^3*sin(d*x + c)^2 - 4*b^5*sin(d*x + c)^2 + 7*a^5*sin(d*x + c) - 10*a^3*b^2*sin(d*x + c) + 3*a*b^4*sin(d
*x + c) + 8*a^4*b - 2*a^2*b^3)/((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 = 0.48, size = 308, normalized size = 1.48 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {9 a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {7 b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right ) a^{2}}{16 d \left (a +b \right )^{3}}-\frac {21 \ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) b^{2}}{2 d \left (a +b \right )^{3}}-\frac {a^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3} b}-\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {9 a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {7 b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\sin \left (d x +c \right )\right ) a^{2}}{16 d \left (a -b \right )^{3}}-\frac {21 \ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) b^{2}}{2 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^6/(a+b*sin(d*x+c)),x)

[Out]

1/2/d/(8*a+8*b)/(sin(d*x+c)-1)^2+9/16/d/(a+b)^2/(sin(d*x+c)-1)*a+7/16/d/(a+b)^2/(sin(d*x+c)-1)*b-15/16/d/(a+b)
^3*ln(sin(d*x+c)-1)*a^2-21/16/d/(a+b)^3*ln(sin(d*x+c)-1)*a*b-1/2/d/(a+b)^3*ln(sin(d*x+c)-1)*b^2-1/d*a^6/(a+b)^
3/(a-b)^3/b*ln(a+b*sin(d*x+c))-1/2/d/(8*a-8*b)/(1+sin(d*x+c))^2+9/16/d/(a-b)^2/(1+sin(d*x+c))*a-7/16/d/(a-b)^2
/(1+sin(d*x+c))*b+15/16/d/(a-b)^3*ln(1+sin(d*x+c))*a^2-21/16/d/(a-b)^3*ln(1+sin(d*x+c))*a*b+1/2/d/(a-b)^3*ln(1
+sin(d*x+c))*b^2

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maxima [A]  time = 0.34, size = 289, normalized size = 1.39 \[ -\frac {\frac {16 \, a^{6} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (15 \, a^{2} - 21 \, a b + 8 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (15 \, a^{2} + 21 \, a b + 8 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (9 \, a^{3} - 5 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} + 10 \, a^{2} b - 6 \, b^{3} - 4 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*(16*a^6*log(b*sin(d*x + c) + a)/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - (15*a^2 - 21*a*b + 8*b^2)*log(si
n(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (15*a^2 + 21*a*b + 8*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2
*b + 3*a*b^2 + b^3) - 2*((9*a^3 - 5*a*b^2)*sin(d*x + c)^3 + 10*a^2*b - 6*b^3 - 4*(3*a^2*b - 2*b^3)*sin(d*x + c
)^2 - (7*a^3 - 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 = 13.38, size = 549, normalized size = 2.64 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {9\,b}{8\,{\left (a-b\right )}^2}+\frac {15}{8\,\left (a-b\right )}\right )}{d}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (11\,a\,b^2-15\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (3\,a\,b^2-7\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (11\,a\,b^2-15\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2\,b-b^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^2\,b-2\,b^3\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (2\,a^2\,b-b^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,a^2-3\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {15}{8\,\left (a+b\right )}-\frac {9\,b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {a^6\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6\,b-3\,a^4\,b^3+3\,a^2\,b^5-b^7\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^6/(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) + (9*b)/(8*(a - b)^2) + 15/(8*(a - b))))/d - ((tan(c/2 + (d*x)
/2)^3*(11*a*b^2 - 15*a^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) - (tan(c/2 + (d*x)/2)^7*(3*a*b^2 - 7*a^3))/(4*(a^4 + b^
4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(11*a*b^2 - 15*a^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) - (2*tan(c/2 + (d*x)/
2)^2*(2*a^2*b - b^3))/(a^4 + b^4 - 2*a^2*b^2) + (4*tan(c/2 + (d*x)/2)^4*(3*a^2*b - 2*b^3))/(a^4 + b^4 - 2*a^2*
b^2) - (2*tan(c/2 + (d*x)/2)^6*(2*a^2*b - b^3))/(a^4 + b^4 - 2*a^2*b^2) + (a*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)) + log(tan(c/2 + (d*x)/2)^2 + 1)/(b*d) - (log(tan(c/2 + (d*x)/2) - 1)*(15/(8*(a + b
)) - (9*b)/(8*(a + b)^2) + b^2/(4*(a + b)^3)))/d - (a^6*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^
2))/(d*(a^6*b - b^7 + 3*a^2*b^5 - 3*a^4*b^3))

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

[Out]

Timed out

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