3.1360 \(\int \frac {\sin ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=221 \[ -\frac {\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {\left (24 a^2-37 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac {a^7 \log (a+b \sin (c+d x))}{b^2 d \left (a^2-b^2\right )^3}-\frac {\sin (c+d x)}{b d} \]

[Out]

-1/16*(24*a^2+37*a*b+15*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d-1/16*(24*a^2-37*a*b+15*b^2)*ln(1+sin(d*x+c))/(a-b)^3/d
+a^7*ln(a+b*sin(d*x+c))/b^2/(a^2-b^2)^3/d-sin(d*x+c)/b/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*(3*a^2-2*b^2)-b*(13*a^2-9*b^2)*sin(d*x+c))/(a^2-b^2)^2/d

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Rubi [A]  time = 0.54, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2837, 12, 1647, 1629} \[ \frac {a^7 \log (a+b \sin (c+d x))}{b^2 d \left (a^2-b^2\right )^3}-\frac {\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {\left (24 a^2-37 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \left (4 a \left (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}-\frac {\sin (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((24*a^2 + 37*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) - ((24*a^2 - 37*a*b + 15*b^2)*Log[1 + Sin
[c + d*x]])/(16*(a - b)^3*d) + (a^7*Log[a + b*Sin[c + d*x]])/(b^2*(a^2 - b^2)^3*d) - Sin[c + d*x]/(b*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*(3*a^2 - 2*b^2) - b*(13*a^2 - 9*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 ^2(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {x^7}{b^7 (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^7}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{b^2 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^8}{a^2-b^2}-\frac {b^6 \left (4 a^2-b^2\right ) x}{a^2-b^2}-4 b^4 x^3-4 b^2 x^5}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 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 (3 a^2-2 b^2\right )-b \left (13 a^2-9 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^8 \left (11 a^2-7 b^2\right )}{\left (a^2-b^2\right )^2}+\frac {b^6 \left (16 a^4-19 a^2 b^2+7 b^4\right ) x}{\left (a^2-b^2\right )^2}+8 b^4 x^3}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 b^6 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 (3 a^2-2 b^2\right )-b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac {\operatorname {Subst}\left (\int \left (-8 b^4+\frac {b^6 \left (24 a^2+37 a b+15 b^2\right )}{2 (a+b)^3 (b-x)}+\frac {8 a^7 b^4}{(a-b)^3 (a+b)^3 (a+x)}+\frac {b^6 \left (24 a^2-37 a b+15 b^2\right )}{2 (-a+b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^6 d}\\ &=-\frac {\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {\left (24 a^2-37 a b+15 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a^7 \log (a+b \sin (c+d x))}{b^2 \left (a^2-b^2\right )^3 d}-\frac {\sin (c+d x)}{b 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 (3 a^2-2 b^2\right )-b \left (13 a^2-9 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 = 2.50, size = 198, normalized size = 0.90 \[ \frac {\frac {16 a^7 \log (a+b \sin (c+d x))}{b^2 (a-b)^3 (a+b)^3}-\frac {\left (24 a^2+37 a b+15 b^2\right ) \log (1-\sin (c+d x))}{(a+b)^3}-\frac {\left (24 a^2-37 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{(a-b)^3}+\frac {11 a+9 b}{(a+b)^2 (\sin (c+d x)-1)}+\frac {9 b-11 a}{(a-b)^2 (\sin (c+d x)+1)}+\frac {1}{(a+b) (\sin (c+d x)-1)^2}+\frac {1}{(a-b) (\sin (c+d x)+1)^2}-\frac {16 \sin (c+d x)}{b}}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(((24*a^2 + 37*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(a + b)^3) - ((24*a^2 - 37*a*b + 15*b^2)*Log[1 + Sin[c +
d*x]])/(a - b)^3 + (16*a^7*Log[a + b*Sin[c + d*x]])/((a - b)^3*b^2*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x]
)^2) + (11*a + 9*b)/((a + b)^2*(-1 + Sin[c + d*x])) - (16*Sin[c + d*x])/b + 1/((a - b)*(1 + Sin[c + d*x])^2) +
(-11*a + 9*b)/((a - b)^2*(1 + Sin[c + d*x])))/(16*d)

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fricas [A]  time = 1.62, size = 351, normalized size = 1.59 \[ \frac {16 \, a^{7} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, a^{5} b^{2} - 8 \, a^{3} b^{4} + 4 \, a b^{6} - {\left (24 \, a^{5} b^{2} + 35 \, a^{4} b^{3} - 24 \, a^{3} b^{4} - 42 \, a^{2} b^{5} + 8 \, a b^{6} + 15 \, b^{7}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (24 \, a^{5} b^{2} - 35 \, a^{4} b^{3} - 24 \, a^{3} b^{4} + 42 \, a^{2} b^{5} + 8 \, a b^{6} - 15 \, b^{7}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, {\left (3 \, a^{5} b^{2} - 5 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{4} b^{3} - 4 \, a^{2} b^{5} + 2 \, b^{7} + 8 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} - {\left (13 \, a^{4} b^{3} - 22 \, a^{2} b^{5} + 9 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(16*a^7*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + 4*a^5*b^2 - 8*a^3*b^4 + 4*a*b^6 - (24*a^5*b^2 + 35*a^4*b
^3 - 24*a^3*b^4 - 42*a^2*b^5 + 8*a*b^6 + 15*b^7)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (24*a^5*b^2 - 35*a^4*b
^3 - 24*a^3*b^4 + 42*a^2*b^5 + 8*a*b^6 - 15*b^7)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 8*(3*a^5*b^2 - 5*a^3*
b^4 + 2*a*b^6)*cos(d*x + c)^2 - 2*(2*a^4*b^3 - 4*a^2*b^5 + 2*b^7 + 8*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos
(d*x + c)^4 - (13*a^4*b^3 - 22*a^2*b^5 + 9*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^
6 - b^8)*d*cos(d*x + c)^4)

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giac [A]  time = 0.31, size = 384, normalized size = 1.74 \[ \frac {\frac {16 \, a^{7} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}} - \frac {{\left (24 \, a^{2} - 37 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (24 \, a^{2} + 37 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {16 \, \sin \left (d x + c\right )}{b} + \frac {2 \, {\left (18 \, a^{5} \sin \left (d x + c\right )^{4} - 18 \, a^{3} b^{2} \sin \left (d x + c\right )^{4} + 6 \, a b^{4} \sin \left (d x + c\right )^{4} - 13 \, a^{4} b \sin \left (d x + c\right )^{3} + 22 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} - 9 \, b^{5} \sin \left (d x + c\right )^{3} - 24 \, a^{5} \sin \left (d x + c\right )^{2} + 16 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} - 4 \, a b^{4} \sin \left (d x + c\right )^{2} + 11 \, a^{4} b \sin \left (d x + c\right ) - 18 \, a^{2} b^{3} \sin \left (d x + c\right ) + 7 \, b^{5} \sin \left (d x + c\right ) + 8 \, a^{5} - 2 \, a^{3} b^{2}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*(16*a^7*log(abs(b*sin(d*x + c) + a))/(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8) - (24*a^2 - 37*a*b + 15*b^2)
*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (24*a^2 + 37*a*b + 15*b^2)*log(abs(sin(d*x + c)
- 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 16*sin(d*x + c)/b + 2*(18*a^5*sin(d*x + c)^4 - 18*a^3*b^2*sin(d*x + c)
^4 + 6*a*b^4*sin(d*x + c)^4 - 13*a^4*b*sin(d*x + c)^3 + 22*a^2*b^3*sin(d*x + c)^3 - 9*b^5*sin(d*x + c)^3 - 24*
a^5*sin(d*x + c)^2 + 16*a^3*b^2*sin(d*x + c)^2 - 4*a*b^4*sin(d*x + c)^2 + 11*a^4*b*sin(d*x + c) - 18*a^2*b^3*s
in(d*x + c) + 7*b^5*sin(d*x + c) + 8*a^5 - 2*a^3*b^2)/((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 = 0.49, size = 321, normalized size = 1.45 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {11 a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {9 b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) a^{2}}{2 d \left (a +b \right )^{3}}-\frac {37 \ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right ) b^{2}}{16 d \left (a +b \right )^{3}}-\frac {\sin \left (d x +c \right )}{b d}+\frac {a^{7} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{2} \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {11 a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {9 b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) a^{2}}{2 d \left (a -b \right )^{3}}+\frac {37 \ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}}-\frac {15 \ln \left (1+\sin \left (d x +c \right )\right ) b^{2}}{16 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x)

[Out]

1/2/d/(8*a+8*b)/(sin(d*x+c)-1)^2+11/16/d/(a+b)^2/(sin(d*x+c)-1)*a+9/16/d/(a+b)^2/(sin(d*x+c)-1)*b-3/2/d/(a+b)^
3*ln(sin(d*x+c)-1)*a^2-37/16/d/(a+b)^3*ln(sin(d*x+c)-1)*a*b-15/16/d/(a+b)^3*ln(sin(d*x+c)-1)*b^2-sin(d*x+c)/b/
d+1/d/b^2*a^7/(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))^2-11/16/d/(a-b)^2/(1+sin(d*x+c
))*a+9/16/d/(a-b)^2/(1+sin(d*x+c))*b-3/2/d/(a-b)^3*ln(1+sin(d*x+c))*a^2+37/16/d/(a-b)^3*ln(1+sin(d*x+c))*a*b-1
5/16/d/(a-b)^3*ln(1+sin(d*x+c))*b^2

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maxima [A]  time = 0.33, size = 303, normalized size = 1.37 \[ \frac {\frac {16 \, a^{7} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}} - \frac {{\left (24 \, a^{2} - 37 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (24 \, a^{2} + 37 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (13 \, a^{2} b - 9 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 10 \, a^{3} - 6 \, a b^{2} - 4 \, {\left (3 \, a^{3} - 2 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (11 \, a^{2} b - 7 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}} - \frac {16 \, \sin \left (d x + c\right )}{b}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/16*(16*a^7*log(b*sin(d*x + c) + a)/(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8) - (24*a^2 - 37*a*b + 15*b^2)*log(
sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (24*a^2 + 37*a*b + 15*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*
a^2*b + 3*a*b^2 + b^3) - 2*((13*a^2*b - 9*b^3)*sin(d*x + c)^3 + 10*a^3 - 6*a*b^2 - 4*(3*a^3 - 2*a*b^2)*sin(d*x
+ c)^2 - (11*a^2*b - 7*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) - 16*sin(d*x + c)/b)/d

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mupad [B]  time = 14.07, size = 627, normalized size = 2.84 \[ -\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 {11\,b}{8\,{\left (a-b\right )}^2}+\frac {3}{a-b}\right )}{d}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a\,b^2-4\,a^3\right )}{{\left (a^2-b^2\right )}^2}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a\,b^2-2\,a^3\right )}{{\left (a^2-b^2\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (3\,a\,b^2-4\,a^3\right )}{{\left (a^2-b^2\right )}^2}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\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 )}^5\,\left (24\,a^4-29\,a^2\,b^2+9\,b^4\right )}{2\,b\,{\left (a^2-b^2\right )}^2}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a^2-5\,b^2\right )}{b\,\left (a^2-b^2\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (4\,a^2-5\,b^2\right )}{b\,\left (a^2-b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (8\,a^4-27\,a^2\,b^2+15\,b^4\right )}{4\,b\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (8\,a^4-27\,a^2\,b^2+15\,b^4\right )}{4\,b\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,{\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}{a+b}-\frac {11\,b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b^2\,d}-\frac {a^7\,\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^2+3\,a^4\,b^4-3\,a^2\,b^6+b^8\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^7/(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) + (11*b)/(8*(a - b)^2) + 3/(a - b)))/d - ((2*tan(c/2 + (d*x)
/2)^4*(3*a*b^2 - 4*a^3))/(a^2 - b^2)^2 - (2*tan(c/2 + (d*x)/2)^2*(a*b^2 - 2*a^3))/(a^2 - b^2)^2 + (2*tan(c/2 +
(d*x)/2)^6*(3*a*b^2 - 4*a^3))/(a^2 - b^2)^2 - (2*tan(c/2 + (d*x)/2)^8*(a*b^2 - 2*a^3))/(a^4 + b^4 - 2*a^2*b^2
) + (tan(c/2 + (d*x)/2)^5*(24*a^4 + 9*b^4 - 29*a^2*b^2))/(2*b*(a^2 - b^2)^2) - (2*tan(c/2 + (d*x)/2)^3*(4*a^2
- 5*b^2))/(b*(a^2 - b^2)) - (2*tan(c/2 + (d*x)/2)^7*(4*a^2 - 5*b^2))/(b*(a^2 - b^2)) + (tan(c/2 + (d*x)/2)*(8*
a^4 + 15*b^4 - 27*a^2*b^2))/(4*b*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^9*(8*a^4 + 15*b^4 - 27*a^2*b^2
))/(4*b*(a^4 + b^4 - 2*a^2*b^2)))/(d*(2*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^6
- 3*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (log(tan(c/2 + (d*x)/2) - 1)*(3/(a + b) - (11*b)/(8*
(a + b)^2) + b^2/(4*(a + b)^3)))/d - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/(b^2*d) - (a^7*log(a + 2*b*tan(c/2 + (d
*x)/2) + a*tan(c/2 + (d*x)/2)^2))/(d*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*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)**7/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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