3.1492 \(\int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=189 \[ -\frac {\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac {\left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {\left (15 a^2-48 a b+35 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {a b \sin ^2(c+d x)}{d}-\frac {\sec ^2(c+d x) (9 a \sin (c+d x)+11 b) (a+b \sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {b^2 \sin ^3(c+d x)}{3 d} \]

[Out]

-1/16*(15*a^2+48*a*b+35*b^2)*ln(1-sin(d*x+c))/d+1/16*(15*a^2-48*a*b+35*b^2)*ln(1+sin(d*x+c))/d-(a^2+3*b^2)*sin
(d*x+c)/d-a*b*sin(d*x+c)^2/d-1/3*b^2*sin(d*x+c)^3/d-1/8*sec(d*x+c)^2*(11*b+9*a*sin(d*x+c))*(a+b*sin(d*x+c))/d+
1/4*sec(d*x+c)^3*(a+b*sin(d*x+c))^2*tan(d*x+c)/d

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Rubi [A]  time = 0.36, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2837, 12, 1645, 1810, 633, 31} \[ -\frac {\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac {\left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {\left (15 a^2-48 a b+35 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {a b \sin ^2(c+d x)}{d}-\frac {\sec ^2(c+d x) (9 a \sin (c+d x)+11 b) (a+b \sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {b^2 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*a^2 + 48*a*b + 35*b^2)*Log[1 - Sin[c + d*x]])/(16*d) + ((15*a^2 - 48*a*b + 35*b^2)*Log[1 + Sin[c + d*x]]
)/(16*d) - ((a^2 + 3*b^2)*Sin[c + d*x])/d - (a*b*Sin[c + d*x]^2)/d - (b^2*Sin[c + d*x]^3)/(3*d) - (Sec[c + d*x
]^2*(11*b + 9*a*Sin[c + d*x])*(a + b*Sin[c + d*x]))/(8*d) + (Sec[c + d*x]^3*(a + b*Sin[c + d*x])^2*Tan[c + d*x
])/(4*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 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 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 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {x^6 (a+x)^2}{b^6 \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)^2}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{b d}\\ &=\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {(a+x) \left (-a b^6-3 b^6 x-4 a b^4 x^2-4 b^4 x^3-4 a b^2 x^4-4 b^2 x^5\right )}{\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) (11 b+9 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {b^6 \left (7 a^2+11 b^2\right )+32 a b^6 x+8 b^4 \left (a^2+2 b^2\right ) x^2+16 a b^4 x^3+8 b^4 x^4}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^5 d}\\ &=-\frac {\sec ^2(c+d x) (11 b+9 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac {\operatorname {Subst}\left (\int \left (-8 b^4 \left (a^2+3 b^2\right )-16 a b^4 x-8 b^4 x^2+\frac {5 b^6 \left (3 a^2+7 b^2\right )+48 a b^6 x}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^5 d}\\ &=-\frac {\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac {a b \sin ^2(c+d x)}{d}-\frac {b^2 \sin ^3(c+d x)}{3 d}-\frac {\sec ^2(c+d x) (11 b+9 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {5 b^6 \left (3 a^2+7 b^2\right )+48 a b^6 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^5 d}\\ &=-\frac {\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac {a b \sin ^2(c+d x)}{d}-\frac {b^2 \sin ^3(c+d x)}{3 d}-\frac {\sec ^2(c+d x) (11 b+9 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}-\frac {\left (15 a^2-48 a b+35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {\left (15 a^2+48 a b+35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=-\frac {\left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {\left (15 a^2-48 a b+35 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac {a b \sin ^2(c+d x)}{d}-\frac {b^2 \sin ^3(c+d x)}{3 d}-\frac {\sec ^2(c+d x) (11 b+9 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 1.56, size = 186, normalized size = 0.98 \[ \frac {-48 \left (a^2+3 b^2\right ) \sin (c+d x)-3 \left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))+3 \left (15 a^2-48 a b+35 b^2\right ) \log (\sin (c+d x)+1)-48 a b \sin ^2(c+d x)+\frac {3 (a+b) (9 a+13 b)}{\sin (c+d x)-1}+\frac {3 (9 a-13 b) (a-b)}{\sin (c+d x)+1}+\frac {3 (a+b)^2}{(\sin (c+d x)-1)^2}-\frac {3 (a-b)^2}{(\sin (c+d x)+1)^2}-16 b^2 \sin ^3(c+d x)}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(15*a^2 + 48*a*b + 35*b^2)*Log[1 - Sin[c + d*x]] + 3*(15*a^2 - 48*a*b + 35*b^2)*Log[1 + Sin[c + d*x]] + (3
*(a + b)^2)/(-1 + Sin[c + d*x])^2 + (3*(a + b)*(9*a + 13*b))/(-1 + Sin[c + d*x]) - 48*(a^2 + 3*b^2)*Sin[c + d*
x] - 48*a*b*Sin[c + d*x]^2 - 16*b^2*Sin[c + d*x]^3 - (3*(a - b)^2)/(1 + Sin[c + d*x])^2 + (3*(9*a - 13*b)*(a -
b))/(1 + Sin[c + d*x]))/(48*d)

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fricas [A]  time = 0.50, size = 198, normalized size = 1.05 \[ \frac {48 \, a b \cos \left (d x + c\right )^{6} - 24 \, a b \cos \left (d x + c\right )^{4} + 3 \, {\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 144 \, a b \cos \left (d x + c\right )^{2} + 24 \, a b + 2 \, {\left (8 \, b^{2} \cos \left (d x + c\right )^{6} - 8 \, {\left (3 \, a^{2} + 10 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 3 \, {\left (9 \, a^{2} + 13 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, a^{2} + 6 \, b^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/48*(48*a*b*cos(d*x + c)^6 - 24*a*b*cos(d*x + c)^4 + 3*(15*a^2 - 48*a*b + 35*b^2)*cos(d*x + c)^4*log(sin(d*x
+ c) + 1) - 3*(15*a^2 + 48*a*b + 35*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 144*a*b*cos(d*x + c)^2 + 24*a
*b + 2*(8*b^2*cos(d*x + c)^6 - 8*(3*a^2 + 10*b^2)*cos(d*x + c)^4 - 3*(9*a^2 + 13*b^2)*cos(d*x + c)^2 + 6*a^2 +
6*b^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [A]  time = 0.31, size = 198, normalized size = 1.05 \[ -\frac {16 \, b^{2} \sin \left (d x + c\right )^{3} + 48 \, a b \sin \left (d x + c\right )^{2} + 48 \, a^{2} \sin \left (d x + c\right ) + 144 \, b^{2} \sin \left (d x + c\right ) - 3 \, {\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (36 \, a b \sin \left (d x + c\right )^{4} + 9 \, a^{2} \sin \left (d x + c\right )^{3} + 13 \, b^{2} \sin \left (d x + c\right )^{3} - 48 \, a b \sin \left (d x + c\right )^{2} - 7 \, a^{2} \sin \left (d x + c\right ) - 11 \, b^{2} \sin \left (d x + c\right ) + 16 \, a b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/48*(16*b^2*sin(d*x + c)^3 + 48*a*b*sin(d*x + c)^2 + 48*a^2*sin(d*x + c) + 144*b^2*sin(d*x + c) - 3*(15*a^2
- 48*a*b + 35*b^2)*log(abs(sin(d*x + c) + 1)) + 3*(15*a^2 + 48*a*b + 35*b^2)*log(abs(sin(d*x + c) - 1)) - 6*(3
6*a*b*sin(d*x + c)^4 + 9*a^2*sin(d*x + c)^3 + 13*b^2*sin(d*x + c)^3 - 48*a*b*sin(d*x + c)^2 - 7*a^2*sin(d*x +
c) - 11*b^2*sin(d*x + c) + 16*a*b)/(sin(d*x + c)^2 - 1)^2)/d

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maple [A]  time = 0.32, size = 355, normalized size = 1.88 \[ \frac {a^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {3 a^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {3 a^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}-\frac {5 a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {15 a^{2} \sin \left (d x +c \right )}{8 d}+\frac {15 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {a b \left (\sin ^{8}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{4}}-\frac {a b \left (\sin ^{8}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}-\frac {a b \left (\sin ^{6}\left (d x +c \right )\right )}{d}-\frac {3 a b \left (\sin ^{4}\left (d x +c \right )\right )}{2 d}-\frac {3 a b \left (\sin ^{2}\left (d x +c \right )\right )}{d}-\frac {6 a b \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {b^{2} \left (\sin ^{9}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {5 b^{2} \left (\sin ^{9}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {5 b^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{8 d}-\frac {7 b^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}-\frac {35 b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{24 d}-\frac {35 b^{2} \sin \left (d x +c \right )}{8 d}+\frac {35 b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^6*(a+b*sin(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.30, size = 180, normalized size = 0.95 \[ -\frac {16 \, b^{2} \sin \left (d x + c\right )^{3} + 48 \, a b \sin \left (d x + c\right )^{2} - 3 \, {\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 48 \, {\left (a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right ) - \frac {6 \, {\left (24 \, a b \sin \left (d x + c\right )^{2} + {\left (9 \, a^{2} + 13 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 20 \, a b - {\left (7 \, a^{2} + 11 \, b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/48*(16*b^2*sin(d*x + c)^3 + 48*a*b*sin(d*x + c)^2 - 3*(15*a^2 - 48*a*b + 35*b^2)*log(sin(d*x + c) + 1) + 3*
(15*a^2 + 48*a*b + 35*b^2)*log(sin(d*x + c) - 1) + 48*(a^2 + 3*b^2)*sin(d*x + c) - 6*(24*a*b*sin(d*x + c)^2 +
(9*a^2 + 13*b^2)*sin(d*x + c)^3 - 20*a*b - (7*a^2 + 11*b^2)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 +
1))/d

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mupad [B]  time = 12.33, size = 433, normalized size = 2.29 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {15\,a^2}{8}-6\,a\,b+\frac {35\,b^2}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {15\,a^2}{8}+6\,a\,b+\frac {35\,b^2}{8}\right )}{d}-\frac {\left (-\frac {15\,a^2}{4}-\frac {35\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\left (\frac {5\,a^2}{2}+\frac {35\,b^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {47\,a^2}{4}+\frac {329\,b^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+32\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (11\,a^2-17\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {47\,a^2}{4}+\frac {329\,b^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {5\,a^2}{2}+\frac {35\,b^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-\frac {15\,a^2}{4}-\frac {35\,b^2}{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 {6\,a\,b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^6*(a + b*sin(c + d*x))^2)/cos(c + d*x)^5,x)

[Out]

(log(tan(c/2 + (d*x)/2) + 1)*((15*a^2)/8 - 6*a*b + (35*b^2)/8))/d - (log(tan(c/2 + (d*x)/2) - 1)*(6*a*b + (15*
a^2)/8 + (35*b^2)/8))/d - (tan(c/2 + (d*x)/2)^7*(11*a^2 - 17*b^2) + tan(c/2 + (d*x)/2)^3*((5*a^2)/2 + (35*b^2)
/6) + tan(c/2 + (d*x)/2)^11*((5*a^2)/2 + (35*b^2)/6) - tan(c/2 + (d*x)/2)^13*((15*a^2)/4 + (35*b^2)/4) + tan(c
/2 + (d*x)/2)^5*((47*a^2)/4 + (329*b^2)/12) + tan(c/2 + (d*x)/2)^9*((47*a^2)/4 + (329*b^2)/12) - tan(c/2 + (d*
x)/2)*((15*a^2)/4 + (35*b^2)/4) - 12*a*b*tan(c/2 + (d*x)/2)^2 + 12*a*b*tan(c/2 + (d*x)/2)^4 + 32*a*b*tan(c/2 +
(d*x)/2)^6 + 32*a*b*tan(c/2 + (d*x)/2)^8 + 12*a*b*tan(c/2 + (d*x)/2)^10 - 12*a*b*tan(c/2 + (d*x)/2)^12)/(d*(t
an(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)) + (6*a*b*log(tan(c/2 + (d*x)/2)^2 + 1))/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)**6*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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