3.204 \(\int \frac {\sin ^4(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=161 \[ -\frac {2 b (a-b)^{3/2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 d}+\frac {\sin ^3(c+d x) (4 b-3 a \cos (c+d x))}{12 a^2 d}+\frac {\sin (c+d x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{8 a^4 d}+\frac {x \left (3 a^4-12 a^2 b^2+8 b^4\right )}{8 a^5} \]

[Out]

1/8*(3*a^4-12*a^2*b^2+8*b^4)*x/a^5-2*(a-b)^(3/2)*b*(a+b)^(3/2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1
/2))/a^5/d+1/8*(8*b*(a^2-b^2)-a*(3*a^2-4*b^2)*cos(d*x+c))*sin(d*x+c)/a^4/d+1/12*(4*b-3*a*cos(d*x+c))*sin(d*x+c
)^3/a^2/d

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Rubi [A]  time = 0.38, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3872, 2865, 2735, 2659, 208} \[ \frac {\sin (c+d x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{8 a^4 d}+\frac {x \left (-12 a^2 b^2+3 a^4+8 b^4\right )}{8 a^5}+\frac {\sin ^3(c+d x) (4 b-3 a \cos (c+d x))}{12 a^2 d}-\frac {2 b (a-b)^{3/2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^4 - 12*a^2*b^2 + 8*b^4)*x)/(8*a^5) - (2*(a - b)^(3/2)*b*(a + b)^(3/2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)
/2])/Sqrt[a + b]])/(a^5*d) + ((8*b*(a^2 - b^2) - a*(3*a^2 - 4*b^2)*Cos[c + d*x])*Sin[c + d*x])/(8*a^4*d) + ((4
*b - 3*a*Cos[c + d*x])*Sin[c + d*x]^3)/(12*a^2*d)

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sin ^4(c+d x)}{a+b \sec (c+d x)} \, dx &=-\int \frac {\cos (c+d x) \sin ^4(c+d x)}{-b-a \cos (c+d x)} \, dx\\ &=\frac {(4 b-3 a \cos (c+d x)) \sin ^3(c+d x)}{12 a^2 d}-\frac {\int \frac {\left (-a b+\left (3 a^2-4 b^2\right ) \cos (c+d x)\right ) \sin ^2(c+d x)}{-b-a \cos (c+d x)} \, dx}{4 a^2}\\ &=\frac {\left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{8 a^4 d}+\frac {(4 b-3 a \cos (c+d x)) \sin ^3(c+d x)}{12 a^2 d}-\frac {\int \frac {-a b \left (5 a^2-4 b^2\right )+\left (3 a^4-12 a^2 b^2+8 b^4\right ) \cos (c+d x)}{-b-a \cos (c+d x)} \, dx}{8 a^4}\\ &=\frac {\left (3 a^4-12 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {\left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{8 a^4 d}+\frac {(4 b-3 a \cos (c+d x)) \sin ^3(c+d x)}{12 a^2 d}+\frac {\left (b \left (a^2-b^2\right )^2\right ) \int \frac {1}{-b-a \cos (c+d x)} \, dx}{a^5}\\ &=\frac {\left (3 a^4-12 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {\left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{8 a^4 d}+\frac {(4 b-3 a \cos (c+d x)) \sin ^3(c+d x)}{12 a^2 d}+\frac {\left (2 b \left (a^2-b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}\\ &=\frac {\left (3 a^4-12 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac {2 (a-b)^{3/2} b (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 d}+\frac {\left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{8 a^4 d}+\frac {(4 b-3 a \cos (c+d x)) \sin ^3(c+d x)}{12 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.82, size = 172, normalized size = 1.07 \[ \frac {3 a^4 \sin (4 (c+d x))+36 a^4 c+36 a^4 d x-8 a^3 b \sin (3 (c+d x))+24 a b \left (5 a^2-4 b^2\right ) \sin (c+d x)+192 b \left (a^2-b^2\right )^{3/2} \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )-144 a^2 b^2 c-144 a^2 b^2 d x-24 \left (a^4-a^2 b^2\right ) \sin (2 (c+d x))+96 b^4 c+96 b^4 d x}{96 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*a^4*c - 144*a^2*b^2*c + 96*b^4*c + 36*a^4*d*x - 144*a^2*b^2*d*x + 96*b^4*d*x + 192*b*(a^2 - b^2)^(3/2)*Arc
Tanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 24*a*b*(5*a^2 - 4*b^2)*Sin[c + d*x] - 24*(a^4 - a^2*b^2)*S
in[2*(c + d*x)] - 8*a^3*b*Sin[3*(c + d*x)] + 3*a^4*Sin[4*(c + d*x)])/(96*a^5*d)

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fricas [A]  time = 0.66, size = 393, normalized size = 2.44 \[ \left [\frac {3 \, {\left (3 \, a^{4} - 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} d x - 12 \, {\left (a^{2} b - b^{3}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (6 \, a^{4} \cos \left (d x + c\right )^{3} - 8 \, a^{3} b \cos \left (d x + c\right )^{2} + 32 \, a^{3} b - 24 \, a b^{3} - 3 \, {\left (5 \, a^{4} - 4 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, a^{5} d}, \frac {3 \, {\left (3 \, a^{4} - 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} d x - 24 \, {\left (a^{2} b - b^{3}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (6 \, a^{4} \cos \left (d x + c\right )^{3} - 8 \, a^{3} b \cos \left (d x + c\right )^{2} + 32 \, a^{3} b - 24 \, a b^{3} - 3 \, {\left (5 \, a^{4} - 4 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, a^{5} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/24*(3*(3*a^4 - 12*a^2*b^2 + 8*b^4)*d*x - 12*(a^2*b - b^3)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 -
2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2
+ 2*a*b*cos(d*x + c) + b^2)) + (6*a^4*cos(d*x + c)^3 - 8*a^3*b*cos(d*x + c)^2 + 32*a^3*b - 24*a*b^3 - 3*(5*a^
4 - 4*a^2*b^2)*cos(d*x + c))*sin(d*x + c))/(a^5*d), 1/24*(3*(3*a^4 - 12*a^2*b^2 + 8*b^4)*d*x - 24*(a^2*b - b^3
)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (6*a^4*cos(d*x
+ c)^3 - 8*a^3*b*cos(d*x + c)^2 + 32*a^3*b - 24*a*b^3 - 3*(5*a^4 - 4*a^2*b^2)*cos(d*x + c))*sin(d*x + c))/(a^5
*d)]

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giac [B]  time = 0.30, size = 407, normalized size = 2.53 \[ \frac {\frac {3 \, {\left (3 \, a^{4} - 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} {\left (d x + c\right )}}{a^{5}} - \frac {48 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{5}} + \frac {2 \, {\left (9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 33 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 104 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 33 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 104 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*(3*a^4 - 12*a^2*b^2 + 8*b^4)*(d*x + c)/a^5 - 48*(a^4*b - 2*a^2*b^3 + b^5)*(pi*floor(1/2*(d*x + c)/pi +
1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/(sqrt(-a^
2 + b^2)*a^5) + 2*(9*a^3*tan(1/2*d*x + 1/2*c)^7 + 24*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 12*a*b^2*tan(1/2*d*x + 1/2
*c)^7 - 24*b^3*tan(1/2*d*x + 1/2*c)^7 + 33*a^3*tan(1/2*d*x + 1/2*c)^5 + 104*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 12*
a*b^2*tan(1/2*d*x + 1/2*c)^5 - 72*b^3*tan(1/2*d*x + 1/2*c)^5 - 33*a^3*tan(1/2*d*x + 1/2*c)^3 + 104*a^2*b*tan(1
/2*d*x + 1/2*c)^3 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 72*b^3*tan(1/2*d*x + 1/2*c)^3 - 9*a^3*tan(1/2*d*x + 1/2*
c) + 24*a^2*b*tan(1/2*d*x + 1/2*c) + 12*a*b^2*tan(1/2*d*x + 1/2*c) - 24*b^3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*
x + 1/2*c)^2 + 1)^4*a^4))/d

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maple [B]  time = 0.45, size = 769, normalized size = 4.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^4/(a+b*sec(d*x+c)),x)

[Out]

-2/d*b/a/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))+4/d*b^3/a^3/((a-b)*(a+b))^(
1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-2/d*b^5/a^5/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x
+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))+3/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7+2/d/a^2/(1+tan(1/2*
d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*b-1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*b^2-2/d/a^4/(1+
tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*b^3+26/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*b-
1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*b^2-6/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2
*c)^5*b^3+11/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5-11/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2
*d*x+1/2*c)^3+1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*b^2+26/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^
4*tan(1/2*d*x+1/2*c)^3*b-6/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*b^3+2/d/a^2/(1+tan(1/2*d*x+1/
2*c)^2)^4*tan(1/2*d*x+1/2*c)*b-2/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*b^3-3/4/d/a/(1+tan(1/2*d*
x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)+1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*b^2-3/d/a^3*arctan(tan(
1/2*d*x+1/2*c))*b^2+2/d/a^5*arctan(tan(1/2*d*x+1/2*c))*b^4+3/4/a/d*arctan(tan(1/2*d*x+1/2*c))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 2.29, size = 317, normalized size = 1.97 \[ \frac {\frac {5\,b\,\sin \left (c+d\,x\right )}{4}-\frac {b\,\sin \left (3\,c+3\,d\,x\right )}{12}}{a^2\,d}-\frac {3\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {b^2\,\sin \left (2\,c+2\,d\,x\right )}{4}}{a^3\,d}+\frac {\frac {3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}-\frac {\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {\sin \left (4\,c+4\,d\,x\right )}{32}}{a\,d}-\frac {b^3\,\sin \left (c+d\,x\right )}{a^4\,d}+\frac {2\,b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^5\,d}-\frac {2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b-\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )\,\sqrt {a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{a^5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5*b*sin(c + d*x))/4 - (b*sin(3*c + 3*d*x))/12)/(a^2*d) - (3*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
- (b^2*sin(2*c + 2*d*x))/4)/(a^3*d) + ((3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 - sin(2*c + 2*d*x)/4
+ sin(4*c + 4*d*x)/32)/(a*d) - (b^3*sin(c + d*x))/(a^4*d) + (2*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
)/(a^5*d) - (2*b*atanh((sin(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/(a^3*cos(c/2 + (d*x)/2)
- b^3*cos(c/2 + (d*x)/2) - a*b^2*cos(c/2 + (d*x)/2) + a^2*b*cos(c/2 + (d*x)/2)))*(a^6 - b^6 + 3*a^2*b^4 - 3*a^
4*b^2)^(1/2))/(a^5*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{4}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**4/(a+b*sec(d*x+c)),x)

[Out]

Integral(sin(c + d*x)**4/(a + b*sec(c + d*x)), x)

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