3.1501 \(\int (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=202 \[ -\frac {b \left (24 a^2+35 b^2\right ) \sin (c+d x)}{8 d}-\frac {(a+b) \left (8 a^2+37 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}-\frac {(a-b) \left (8 a^2-37 a b+35 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}-\frac {b^3 \sin ^3(c+d x)}{3 d} \]

[Out]

-1/16*(a+b)*(8*a^2+37*a*b+35*b^2)*ln(1-sin(d*x+c))/d-1/16*(a-b)*(8*a^2-37*a*b+35*b^2)*ln(1+sin(d*x+c))/d-1/8*b
*(24*a^2+35*b^2)*sin(d*x+c)/d-3/2*a*b^2*sin(d*x+c)^2/d-1/3*b^3*sin(d*x+c)^3/d+1/4*sec(d*x+c)^4*(a+b*sin(d*x+c)
)^3/d-1/8*sec(d*x+c)^2*(a+b*sin(d*x+c))^2*(8*a+11*b*sin(d*x+c))/d

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Rubi [A]  time = 0.34, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2721, 1645, 1629, 633, 31} \[ -\frac {b \left (24 a^2+35 b^2\right ) \sin (c+d x)}{8 d}-\frac {(a+b) \left (8 a^2+37 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}-\frac {(a-b) \left (8 a^2-37 a b+35 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}-\frac {b^3 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^3*Tan[c + d*x]^5,x]

[Out]

-((a + b)*(8*a^2 + 37*a*b + 35*b^2)*Log[1 - Sin[c + d*x]])/(16*d) - ((a - b)*(8*a^2 - 37*a*b + 35*b^2)*Log[1 +
Sin[c + d*x]])/(16*d) - (b*(24*a^2 + 35*b^2)*Sin[c + d*x])/(8*d) - (3*a*b^2*Sin[c + d*x]^2)/(2*d) - (b^3*Sin[
c + d*x]^3)/(3*d) + (Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3)/(4*d) - (Sec[c + d*x]^2*(a + b*Sin[c + d*x])^2*(8*
a + 11*b*Sin[c + d*x]))/(8*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

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 1645

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, Simp[((d + e*x)^m*(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] + Dist[1/(2*a*c*(p
+ 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*c*(p + 1)*(d + e*x)*Q - a*e*g*m + c*d*f*(2*p
+ 3) + c*e*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 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 (a+b \sin (c+d x))^3 \tan ^5(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5 (a+x)^3}{\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))^3}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {(a+x)^2 \left (-3 b^6-4 a b^4 x-4 b^4 x^2-4 a b^2 x^3-4 b^2 x^4\right )}{\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))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}+\frac {\operatorname {Subst}\left (\int \frac {(a+x) \left (21 a b^6+b^4 \left (8 a^2+27 b^2\right ) x+16 a b^4 x^2+8 b^4 x^3\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}+\frac {\operatorname {Subst}\left (\int \left (-24 a^2 b^4-35 b^6-24 a b^4 x-8 b^4 x^2+\frac {5 b^6 \left (9 a^2+7 b^2\right )+8 a b^4 \left (a^2+9 b^2\right ) x}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=-\frac {b \left (24 a^2+35 b^2\right ) \sin (c+d x)}{8 d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}+\frac {\operatorname {Subst}\left (\int \frac {5 b^6 \left (9 a^2+7 b^2\right )+8 a b^4 \left (a^2+9 b^2\right ) x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^4 d}\\ &=-\frac {b \left (24 a^2+35 b^2\right ) \sin (c+d x)}{8 d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}+\frac {\left ((a-b) \left (8 a^2-37 a b+35 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {\left ((a+b) \left (8 a^2+37 a b+35 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=-\frac {(a+b) \left (8 a^2+37 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}-\frac {(a-b) \left (8 a^2-37 a b+35 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {b \left (24 a^2+35 b^2\right ) \sin (c+d x)}{8 d}-\frac {3 a b^2 \sin ^2(c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (8 a+11 b \sin (c+d x))}{8 d}\\ \end {align*}

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Mathematica [A]  time = 1.02, size = 199, normalized size = 0.99 \[ -\frac {144 b \left (a^2+b^2\right ) \sin (c+d x)+3 \left (8 a^2-37 a b+35 b^2\right ) (a-b) \log (\sin (c+d x)+1)+3 (a+b) \left (8 a^2+37 a b+35 b^2\right ) \log (1-\sin (c+d x))+72 a b^2 \sin ^2(c+d x)-\frac {3 (a-b)^3}{(\sin (c+d x)+1)^2}+\frac {3 (7 a-13 b) (a-b)^2}{\sin (c+d x)+1}-\frac {3 (a+b)^2 (7 a+13 b)}{\sin (c+d x)-1}-\frac {3 (a+b)^3}{(\sin (c+d x)-1)^2}+16 b^3 \sin ^3(c+d x)}{48 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^3*Tan[c + d*x]^5,x]

[Out]

-1/48*(3*(a + b)*(8*a^2 + 37*a*b + 35*b^2)*Log[1 - Sin[c + d*x]] + 3*(a - b)*(8*a^2 - 37*a*b + 35*b^2)*Log[1 +
Sin[c + d*x]] - (3*(a + b)^3)/(-1 + Sin[c + d*x])^2 - (3*(a + b)^2*(7*a + 13*b))/(-1 + Sin[c + d*x]) + 144*b*
(a^2 + b^2)*Sin[c + d*x] + 72*a*b^2*Sin[c + d*x]^2 + 16*b^3*Sin[c + d*x]^3 - (3*(a - b)^3)/(1 + Sin[c + d*x])^
2 + (3*(7*a - 13*b)*(a - b)^2)/(1 + Sin[c + d*x]))/d

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fricas [A]  time = 0.49, size = 238, normalized size = 1.18 \[ \frac {72 \, a b^{2} \cos \left (d x + c\right )^{6} - 36 \, a b^{2} \cos \left (d x + c\right )^{4} - 3 \, {\left (8 \, a^{3} - 45 \, a^{2} b + 72 \, a b^{2} - 35 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, a^{3} + 45 \, a^{2} b + 72 \, a b^{2} + 35 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 12 \, a^{3} + 36 \, a b^{2} - 24 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, b^{3} \cos \left (d x + c\right )^{6} - 8 \, {\left (9 \, a^{2} b + 10 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 18 \, a^{2} b + 6 \, b^{3} - 3 \, {\left (27 \, a^{2} b + 13 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

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))^3,x, algorithm="fricas")

[Out]

1/48*(72*a*b^2*cos(d*x + c)^6 - 36*a*b^2*cos(d*x + c)^4 - 3*(8*a^3 - 45*a^2*b + 72*a*b^2 - 35*b^3)*cos(d*x + c
)^4*log(sin(d*x + c) + 1) - 3*(8*a^3 + 45*a^2*b + 72*a*b^2 + 35*b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 1
2*a^3 + 36*a*b^2 - 24*(2*a^3 + 9*a*b^2)*cos(d*x + c)^2 + 2*(8*b^3*cos(d*x + c)^6 - 8*(9*a^2*b + 10*b^3)*cos(d*
x + c)^4 + 18*a^2*b + 6*b^3 - 3*(27*a^2*b + 13*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [A]  time = 0.34, size = 251, normalized size = 1.24 \[ -\frac {16 \, b^{3} \sin \left (d x + c\right )^{3} + 72 \, a b^{2} \sin \left (d x + c\right )^{2} + 144 \, a^{2} b \sin \left (d x + c\right ) + 144 \, b^{3} \sin \left (d x + c\right ) + 3 \, {\left (8 \, a^{3} - 45 \, a^{2} b + 72 \, a b^{2} - 35 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (8 \, a^{3} + 45 \, a^{2} b + 72 \, a b^{2} + 35 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (6 \, a^{3} \sin \left (d x + c\right )^{4} + 54 \, a b^{2} \sin \left (d x + c\right )^{4} + 27 \, a^{2} b \sin \left (d x + c\right )^{3} + 13 \, b^{3} \sin \left (d x + c\right )^{3} - 4 \, a^{3} \sin \left (d x + c\right )^{2} - 72 \, a b^{2} \sin \left (d x + c\right )^{2} - 21 \, a^{2} b \sin \left (d x + c\right ) - 11 \, b^{3} \sin \left (d x + c\right ) + 24 \, a b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{48 \, d} \]

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))^3,x, algorithm="giac")

[Out]

-1/48*(16*b^3*sin(d*x + c)^3 + 72*a*b^2*sin(d*x + c)^2 + 144*a^2*b*sin(d*x + c) + 144*b^3*sin(d*x + c) + 3*(8*
a^3 - 45*a^2*b + 72*a*b^2 - 35*b^3)*log(abs(sin(d*x + c) + 1)) + 3*(8*a^3 + 45*a^2*b + 72*a*b^2 + 35*b^3)*log(
abs(sin(d*x + c) - 1)) - 6*(6*a^3*sin(d*x + c)^4 + 54*a*b^2*sin(d*x + c)^4 + 27*a^2*b*sin(d*x + c)^3 + 13*b^3*
sin(d*x + c)^3 - 4*a^3*sin(d*x + c)^2 - 72*a*b^2*sin(d*x + c)^2 - 21*a^2*b*sin(d*x + c) - 11*b^3*sin(d*x + c)
+ 24*a*b^2)/(sin(d*x + c)^2 - 1)^2)/d

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maple [B]  time = 0.34, size = 420, normalized size = 2.08 \[ \frac {a^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {3 a^{2} b \left (\sin ^{7}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {9 a^{2} b \left (\sin ^{7}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {9 a^{2} b \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}-\frac {15 a^{2} b \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {45 a^{2} b \sin \left (d x +c \right )}{8 d}+\frac {45 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 a \,b^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {3 a \,b^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}-\frac {3 a \,b^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{2 d}-\frac {9 a \,b^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d}-\frac {9 a \,b^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {9 a \,b^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {b^{3} \left (\sin ^{9}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {5 b^{3} \left (\sin ^{9}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {5 b^{3} \left (\sin ^{7}\left (d x +c \right )\right )}{8 d}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right ) b^{3}}{8 d}-\frac {35 b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{24 d}-\frac {35 b^{3} \sin \left (d x +c \right )}{8 d}+\frac {35 b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

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))^3,x)

[Out]

1/4/d*a^3*tan(d*x+c)^4-1/2/d*a^3*tan(d*x+c)^2-1/d*a^3*ln(cos(d*x+c))+3/4/d*a^2*b*sin(d*x+c)^7/cos(d*x+c)^4-9/8
/d*a^2*b*sin(d*x+c)^7/cos(d*x+c)^2-9/8/d*a^2*b*sin(d*x+c)^5-15/8/d*a^2*b*sin(d*x+c)^3-45/8*a^2*b*sin(d*x+c)/d+
45/8/d*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+3/4/d*a*b^2*sin(d*x+c)^8/cos(d*x+c)^4-3/2/d*a*b^2*sin(d*x+c)^8/cos(d*x+
c)^2-3/2/d*a*b^2*sin(d*x+c)^6-9/4/d*a*b^2*sin(d*x+c)^4-9/2*a*b^2*sin(d*x+c)^2/d-9/d*a*b^2*ln(cos(d*x+c))+1/4/d
*b^3*sin(d*x+c)^9/cos(d*x+c)^4-5/8/d*b^3*sin(d*x+c)^9/cos(d*x+c)^2-5/8/d*b^3*sin(d*x+c)^7-7/8/d*sin(d*x+c)^5*b
^3-35/24*b^3*sin(d*x+c)^3/d-35/8*b^3*sin(d*x+c)/d+35/8/d*b^3*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.35, size = 217, normalized size = 1.07 \[ -\frac {16 \, b^{3} \sin \left (d x + c\right )^{3} + 72 \, a b^{2} \sin \left (d x + c\right )^{2} + 3 \, {\left (8 \, a^{3} - 45 \, a^{2} b + 72 \, a b^{2} - 35 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (8 \, a^{3} + 45 \, a^{2} b + 72 \, a b^{2} + 35 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, {\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right ) - \frac {6 \, {\left ({\left (27 \, a^{2} b + 13 \, b^{3}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{3} - 30 \, a b^{2} + 4 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (21 \, a^{2} b + 11 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{48 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

-1/48*(16*b^3*sin(d*x + c)^3 + 72*a*b^2*sin(d*x + c)^2 + 3*(8*a^3 - 45*a^2*b + 72*a*b^2 - 35*b^3)*log(sin(d*x
+ c) + 1) + 3*(8*a^3 + 45*a^2*b + 72*a*b^2 + 35*b^3)*log(sin(d*x + c) - 1) + 144*(a^2*b + b^3)*sin(d*x + c) -
6*((27*a^2*b + 13*b^3)*sin(d*x + c)^3 - 6*a^3 - 30*a*b^2 + 4*(2*a^3 + 9*a*b^2)*sin(d*x + c)^2 - (21*a^2*b + 11
*b^3)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

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mupad [B]  time = 12.12, size = 512, normalized size = 2.53 \[ \frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^3+9\,a\,b^2\right )}{d}-\frac {\left (-\frac {45\,a^2\,b}{4}-\frac {35\,b^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+\left (-2\,a^3-18\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\left (\frac {15\,a^2\,b}{2}+\frac {35\,b^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (2\,a^3+18\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {141\,a^2\,b}{4}+\frac {329\,b^3}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (16\,a^3+48\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (33\,a^2\,b-17\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (16\,a^3+48\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {141\,a^2\,b}{4}+\frac {329\,b^3}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (2\,a^3+18\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {15\,a^2\,b}{2}+\frac {35\,b^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,a^3-18\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-\frac {45\,a^2\,b}{4}-\frac {35\,b^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^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 (a-b\right )\,\left (8\,a^2-37\,a\,b+35\,b^2\right )}{8\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a+b\right )\,\left (8\,a^2+37\,a\,b+35\,b^2\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^5*(a + b*sin(c + d*x))^3)/cos(c + d*x)^5,x)

[Out]

(log(tan(c/2 + (d*x)/2)^2 + 1)*(9*a*b^2 + a^3))/d - (tan(c/2 + (d*x)/2)^4*(18*a*b^2 + 2*a^3) - tan(c/2 + (d*x)
/2)^2*(18*a*b^2 + 2*a^3) - tan(c/2 + (d*x)/2)*((45*a^2*b)/4 + (35*b^3)/4) + tan(c/2 + (d*x)/2)^10*(18*a*b^2 +
2*a^3) - tan(c/2 + (d*x)/2)^12*(18*a*b^2 + 2*a^3) + tan(c/2 + (d*x)/2)^6*(48*a*b^2 + 16*a^3) + tan(c/2 + (d*x)
/2)^8*(48*a*b^2 + 16*a^3) + tan(c/2 + (d*x)/2)^7*(33*a^2*b - 17*b^3) + tan(c/2 + (d*x)/2)^3*((15*a^2*b)/2 + (3
5*b^3)/6) + tan(c/2 + (d*x)/2)^11*((15*a^2*b)/2 + (35*b^3)/6) - tan(c/2 + (d*x)/2)^13*((45*a^2*b)/4 + (35*b^3)
/4) + tan(c/2 + (d*x)/2)^5*((141*a^2*b)/4 + (329*b^3)/12) + tan(c/2 + (d*x)/2)^9*((141*a^2*b)/4 + (329*b^3)/12
))/(d*(tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 + 3*tan
(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 - tan(c/2 + (d*x)/2)^14 - 1)) - (log(tan(c/2 + (d*x)/2) + 1)*(a - b
)*(8*a^2 - 37*a*b + 35*b^2))/(8*d) - (log(tan(c/2 + (d*x)/2) - 1)*(a + b)*(37*a*b + 8*a^2 + 35*b^2))/(8*d)

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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))**3,x)

[Out]

Timed out

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