3.1502 \(\int \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=177 \[ -\frac {3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {29 a b^2 \sin (c+d x)}{8 d}-\frac {\sec ^2(c+d x) (5 a \sin (c+d x)+8 b) (a+b \sin (c+d x))^2}{8 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {b^3 \sin ^2(c+d x)}{2 d} \]

[Out]

-3/16*(a+b)*(a^2+7*a*b+8*b^2)*ln(1-sin(d*x+c))/d+3/16*(a-b)*(a^2-7*a*b+8*b^2)*ln(1+sin(d*x+c))/d-29/8*a*b^2*si
n(d*x+c)/d-1/2*b^3*sin(d*x+c)^2/d-1/8*sec(d*x+c)^2*(8*b+5*a*sin(d*x+c))*(a+b*sin(d*x+c))^2/d+1/4*sec(d*x+c)^3*
(a+b*sin(d*x+c))^3*tan(d*x+c)/d

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Rubi [A]  time = 0.37, antiderivative size = 177, 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, 1629, 633, 31} \[ -\frac {3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {29 a b^2 \sin (c+d x)}{8 d}-\frac {\sec ^2(c+d x) (5 a \sin (c+d x)+8 b) (a+b \sin (c+d x))^2}{8 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {b^3 \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + b*Sin[c + d*x])^3*Tan[c + d*x]^4,x]

[Out]

(-3*(a + b)*(a^2 + 7*a*b + 8*b^2)*Log[1 - Sin[c + d*x]])/(16*d) + (3*(a - b)*(a^2 - 7*a*b + 8*b^2)*Log[1 + Sin
[c + d*x]])/(16*d) - (29*a*b^2*Sin[c + d*x])/(8*d) - (b^3*Sin[c + d*x]^2)/(2*d) - (Sec[c + d*x]^2*(8*b + 5*a*S
in[c + d*x])*(a + b*Sin[c + d*x])^2)/(8*d) + (Sec[c + d*x]^3*(a + b*Sin[c + d*x])^3*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 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 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 \sec (c+d x) (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {x^4 (a+x)^3}{b^4 \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)^3}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {(a+x)^2 \left (-a b^4-4 b^4 x-4 a b^2 x^2-4 b^2 x^3\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b d}\\ &=-\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {(a+x) \left (b^4 \left (3 a^2+16 b^2\right )+21 a b^4 x+8 b^4 x^2\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d}\\ &=-\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\operatorname {Subst}\left (\int \left (-29 a b^4-8 b^4 x+\frac {3 \left (a b^4 \left (a^2+15 b^2\right )+8 b^4 \left (a^2+b^2\right ) x\right )}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^3 d}\\ &=-\frac {29 a b^2 \sin (c+d x)}{8 d}-\frac {b^3 \sin ^2(c+d x)}{2 d}-\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac {3 \operatorname {Subst}\left (\int \frac {a b^4 \left (a^2+15 b^2\right )+8 b^4 \left (a^2+b^2\right ) x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^3 d}\\ &=-\frac {29 a b^2 \sin (c+d x)}{8 d}-\frac {b^3 \sin ^2(c+d x)}{2 d}-\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\left (3 (a-b) \left (a^2-7 a b+8 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {\left (3 (a+b) \left (a^2+7 a b+8 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=-\frac {3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {3 (a-b) \left (a^2-7 a b+8 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {29 a b^2 \sin (c+d x)}{8 d}-\frac {b^3 \sin ^2(c+d x)}{2 d}-\frac {\sec ^2(c+d x) (8 b+5 a \sin (c+d x)) (a+b \sin (c+d x))^2}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 174, normalized size = 0.98 \[ \frac {3 \left (a^2-7 a b+8 b^2\right ) (a-b) \log (\sin (c+d x)+1)-3 (a+b) \left (a^2+7 a b+8 b^2\right ) \log (1-\sin (c+d x))-48 a b^2 \sin (c+d x)-\frac {(a-b)^3}{(\sin (c+d x)+1)^2}+\frac {(5 a-11 b) (a-b)^2}{\sin (c+d x)+1}+\frac {(a+b)^2 (5 a+11 b)}{\sin (c+d x)-1}+\frac {(a+b)^3}{(\sin (c+d x)-1)^2}-8 b^3 \sin ^2(c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + b*Sin[c + d*x])^3*Tan[c + d*x]^4,x]

[Out]

(-3*(a + b)*(a^2 + 7*a*b + 8*b^2)*Log[1 - Sin[c + d*x]] + 3*(a - b)*(a^2 - 7*a*b + 8*b^2)*Log[1 + Sin[c + d*x]
] + (a + b)^3/(-1 + Sin[c + d*x])^2 + ((a + b)^2*(5*a + 11*b))/(-1 + Sin[c + d*x]) - 48*a*b^2*Sin[c + d*x] - 8
*b^3*Sin[c + d*x]^2 - (a - b)^3/(1 + Sin[c + d*x])^2 + ((5*a - 11*b)*(a - b)^2)/(1 + Sin[c + d*x]))/(16*d)

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fricas [A]  time = 0.49, size = 208, normalized size = 1.18 \[ \frac {8 \, b^{3} \cos \left (d x + c\right )^{6} - 4 \, b^{3} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{3} + 8 \, a^{2} b + 15 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 12 \, a^{2} b + 4 \, b^{3} - 24 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (24 \, a b^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{3} - 6 \, a b^{2} + {\left (5 \, a^{3} + 27 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

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))^3,x, algorithm="fricas")

[Out]

1/16*(8*b^3*cos(d*x + c)^6 - 4*b^3*cos(d*x + c)^4 + 3*(a^3 - 8*a^2*b + 15*a*b^2 - 8*b^3)*cos(d*x + c)^4*log(si
n(d*x + c) + 1) - 3*(a^3 + 8*a^2*b + 15*a*b^2 + 8*b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 12*a^2*b + 4*b^
3 - 24*(2*a^2*b + b^3)*cos(d*x + c)^2 - 2*(24*a*b^2*cos(d*x + c)^4 - 2*a^3 - 6*a*b^2 + (5*a^3 + 27*a*b^2)*cos(
d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [A]  time = 0.32, size = 221, normalized size = 1.25 \[ -\frac {8 \, b^{3} \sin \left (d x + c\right )^{2} + 48 \, a b^{2} \sin \left (d x + c\right ) - 3 \, {\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (a^{3} + 8 \, a^{2} b + 15 \, a b^{2} + 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, a^{2} b \sin \left (d x + c\right )^{4} + 18 \, b^{3} \sin \left (d x + c\right )^{4} + 5 \, a^{3} \sin \left (d x + c\right )^{3} + 27 \, a b^{2} \sin \left (d x + c\right )^{3} - 12 \, a^{2} b \sin \left (d x + c\right )^{2} - 24 \, b^{3} \sin \left (d x + c\right )^{2} - 3 \, a^{3} \sin \left (d x + c\right ) - 21 \, a b^{2} \sin \left (d x + c\right ) + 8 \, b^{3}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

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))^3,x, algorithm="giac")

[Out]

-1/16*(8*b^3*sin(d*x + c)^2 + 48*a*b^2*sin(d*x + c) - 3*(a^3 - 8*a^2*b + 15*a*b^2 - 8*b^3)*log(abs(sin(d*x + c
) + 1)) + 3*(a^3 + 8*a^2*b + 15*a*b^2 + 8*b^3)*log(abs(sin(d*x + c) - 1)) - 2*(18*a^2*b*sin(d*x + c)^4 + 18*b^
3*sin(d*x + c)^4 + 5*a^3*sin(d*x + c)^3 + 27*a*b^2*sin(d*x + c)^3 - 12*a^2*b*sin(d*x + c)^2 - 24*b^3*sin(d*x +
c)^2 - 3*a^3*sin(d*x + c) - 21*a*b^2*sin(d*x + c) + 8*b^3)/(sin(d*x + c)^2 - 1)^2)/d

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maple [B]  time = 0.33, size = 385, normalized size = 2.18 \[ \frac {a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {3 a^{3} \sin \left (d x +c \right )}{8 d}+\frac {3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 a^{2} b \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {3 a^{2} b \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 a^{2} b \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {3 a \,b^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {9 a \,b^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {9 a \,b^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}-\frac {15 a \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {45 a \,b^{2} \sin \left (d x +c \right )}{8 d}+\frac {45 a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b^{3} \left (\sin ^{8}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {b^{3} \left (\sin ^{8}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}-\frac {b^{3} \left (\sin ^{6}\left (d x +c \right )\right )}{2 d}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4 d}-\frac {3 b^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

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))^3,x)

[Out]

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

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maxima [A]  time = 0.32, size = 190, normalized size = 1.07 \[ -\frac {8 \, b^{3} \sin \left (d x + c\right )^{2} + 48 \, a b^{2} \sin \left (d x + c\right ) - 3 \, {\left (a^{3} - 8 \, a^{2} b + 15 \, a b^{2} - 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (a^{3} + 8 \, a^{2} b + 15 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (5 \, a^{3} + 27 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} - 18 \, a^{2} b - 10 \, b^{3} + 12 \, {\left (2 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )^{2} - 3 \, {\left (a^{3} + 7 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

-1/16*(8*b^3*sin(d*x + c)^2 + 48*a*b^2*sin(d*x + c) - 3*(a^3 - 8*a^2*b + 15*a*b^2 - 8*b^3)*log(sin(d*x + c) +
1) + 3*(a^3 + 8*a^2*b + 15*a*b^2 + 8*b^3)*log(sin(d*x + c) - 1) - 2*((5*a^3 + 27*a*b^2)*sin(d*x + c)^3 - 18*a^
2*b - 10*b^3 + 12*(2*a^2*b + b^3)*sin(d*x + c)^2 - 3*(a^3 + 7*a*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 = 12.11, size = 449, normalized size = 2.54 \[ \frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (3\,a^2\,b+3\,b^3\right )}{d}-\frac {\left (-\frac {3\,a^3}{4}-\frac {45\,a\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (-6\,a^2\,b-6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {5\,a^3}{4}+\frac {75\,a\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (12\,a^2\,b+12\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (\frac {15\,a^3}{2}+\frac {33\,a\,b^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (36\,a^2\,b+4\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {15\,a^3}{2}+\frac {33\,a\,b^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (12\,a^2\,b+12\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {5\,a^3}{4}+\frac {75\,a\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-6\,a^2\,b-6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-\frac {3\,a^3}{4}-\frac {45\,a\,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 )}^{12}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\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+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a+b\right )\,\left (a^2+7\,a\,b+8\,b^2\right )}{8\,d}+\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-b\right )\,\left (a^2-7\,a\,b+8\,b^2\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^4*(a + b*sin(c + d*x))^3)/cos(c + d*x)^5,x)

[Out]

(log(tan(c/2 + (d*x)/2)^2 + 1)*(3*a^2*b + 3*b^3))/d - (tan(c/2 + (d*x)/2)^5*((33*a*b^2)/2 + (15*a^3)/2) - tan(
c/2 + (d*x)/2)*((45*a*b^2)/4 + (3*a^3)/4) + tan(c/2 + (d*x)/2)^7*((33*a*b^2)/2 + (15*a^3)/2) - tan(c/2 + (d*x)
/2)^11*((45*a*b^2)/4 + (3*a^3)/4) + tan(c/2 + (d*x)/2)^3*((75*a*b^2)/4 + (5*a^3)/4) + tan(c/2 + (d*x)/2)^9*((7
5*a*b^2)/4 + (5*a^3)/4) - tan(c/2 + (d*x)/2)^2*(6*a^2*b + 6*b^3) - tan(c/2 + (d*x)/2)^10*(6*a^2*b + 6*b^3) + t
an(c/2 + (d*x)/2)^4*(12*a^2*b + 12*b^3) + tan(c/2 + (d*x)/2)^8*(12*a^2*b + 12*b^3) + tan(c/2 + (d*x)/2)^6*(36*
a^2*b + 4*b^3))/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2
)^8 + 2*tan(c/2 + (d*x)/2)^10 - tan(c/2 + (d*x)/2)^12 - 1)) - (3*log(tan(c/2 + (d*x)/2) - 1)*(a + b)*(7*a*b +
a^2 + 8*b^2))/(8*d) + (3*log(tan(c/2 + (d*x)/2) + 1)*(a - b)*(a^2 - 7*a*b + 8*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)**4*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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