3.436 \(\int \frac {\sec ^6(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=197 \[ -\frac {\sec ^5(c+d x) (b-a \sin (c+d x))}{5 d \left (a^2-b^2\right )}-\frac {2 b^6 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac {\sec ^3(c+d x) \left (a \left (4 a^2-9 b^2\right ) \sin (c+d x)+5 b^3\right )}{15 d \left (a^2-b^2\right )^2}-\frac {\sec (c+d x) \left (15 b^5-a \left (8 a^4-26 a^2 b^2+33 b^4\right ) \sin (c+d x)\right )}{15 d \left (a^2-b^2\right )^3} \]

[Out]

-2*b^6*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(7/2)/d-1/5*sec(d*x+c)^5*(b-a*sin(d*x+c))/d/
(a^2-b^2)+1/15*sec(d*x+c)^3*(5*b^3+a*(4*a^2-9*b^2)*sin(d*x+c))/(a^2-b^2)^2/d-1/15*sec(d*x+c)*(15*b^5-a*(8*a^4-
26*a^2*b^2+33*b^4)*sin(d*x+c))/(a^2-b^2)^3/d

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Rubi [A]  time = 0.50, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2696, 2866, 12, 2660, 618, 204} \[ -\frac {2 b^6 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}-\frac {\sec ^5(c+d x) (b-a \sin (c+d x))}{5 d \left (a^2-b^2\right )}+\frac {\sec ^3(c+d x) \left (a \left (4 a^2-9 b^2\right ) \sin (c+d x)+5 b^3\right )}{15 d \left (a^2-b^2\right )^2}-\frac {\sec (c+d x) \left (15 b^5-a \left (-26 a^2 b^2+8 a^4+33 b^4\right ) \sin (c+d x)\right )}{15 d \left (a^2-b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/(a + b*Sin[c + d*x]),x]

[Out]

(-2*b^6*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) - (Sec[c + d*x]^5*(b - a*Sin[c
+ d*x]))/(5*(a^2 - b^2)*d) + (Sec[c + d*x]^3*(5*b^3 + a*(4*a^2 - 9*b^2)*Sin[c + d*x]))/(15*(a^2 - b^2)^2*d) -
(Sec[c + d*x]*(15*b^5 - a*(8*a^4 - 26*a^2*b^2 + 33*b^4)*Sin[c + d*x]))/(15*(a^2 - b^2)^3*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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\sec ^6(c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac {\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}-\frac {\int \frac {\sec ^4(c+d x) \left (-4 a^2+5 b^2-4 a b \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{5 \left (a^2-b^2\right )}\\ &=-\frac {\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac {\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}+\frac {\int \frac {\sec ^2(c+d x) \left (8 a^4-18 a^2 b^2+15 b^4+2 a b \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{15 \left (a^2-b^2\right )^2}\\ &=-\frac {\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac {\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}-\frac {\sec (c+d x) \left (15 b^5-a \left (8 a^4-26 a^2 b^2+33 b^4\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^3 d}-\frac {\int \frac {15 b^6}{a+b \sin (c+d x)} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=-\frac {\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac {\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}-\frac {\sec (c+d x) \left (15 b^5-a \left (8 a^4-26 a^2 b^2+33 b^4\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^3 d}-\frac {b^6 \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=-\frac {\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac {\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}-\frac {\sec (c+d x) \left (15 b^5-a \left (8 a^4-26 a^2 b^2+33 b^4\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^3 d}-\frac {\left (2 b^6\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac {\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac {\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}-\frac {\sec (c+d x) \left (15 b^5-a \left (8 a^4-26 a^2 b^2+33 b^4\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^3 d}+\frac {\left (4 b^6\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac {2 b^6 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}-\frac {\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac {\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}-\frac {\sec (c+d x) \left (15 b^5-a \left (8 a^4-26 a^2 b^2+33 b^4\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^3 d}\\ \end {align*}

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Mathematica [A]  time = 2.53, size = 370, normalized size = 1.88 \[ \frac {\sec ^5(c+d x) \left (640 a^5 \sin (c+d x)+320 a^5 \sin (3 (c+d x))+64 a^5 \sin (5 (c+d x))+45 a^4 b \cos (3 (c+d x))+9 a^4 b \cos (5 (c+d x))-384 a^4 b-1600 a^3 b^2 \sin (c+d x)-1040 a^3 b^2 \sin (3 (c+d x))-208 a^3 b^2 \sin (5 (c+d x))-190 a^2 b^3 \cos (3 (c+d x))-38 a^2 b^3 \cos (5 (c+d x))+1088 a^2 b^3+320 b^3 \left (a^2-4 b^2\right ) \cos (2 (c+d x))+10 b \left (9 a^4-38 a^2 b^2+149 b^4\right ) \cos (c+d x)+1200 a b^4 \sin (c+d x)+1080 a b^4 \sin (3 (c+d x))+264 a b^4 \sin (5 (c+d x))+745 b^5 \cos (3 (c+d x))-240 b^5 \cos (4 (c+d x))+149 b^5 \cos (5 (c+d x))-1424 b^5\right )}{1920 d (a-b)^3 (a+b)^3}-\frac {2 b^6 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/(a + b*Sin[c + d*x]),x]

[Out]

(-2*b^6*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + (Sec[c + d*x]^5*(-384*a^4*b
+ 1088*a^2*b^3 - 1424*b^5 + 10*b*(9*a^4 - 38*a^2*b^2 + 149*b^4)*Cos[c + d*x] + 320*b^3*(a^2 - 4*b^2)*Cos[2*(c
+ d*x)] + 45*a^4*b*Cos[3*(c + d*x)] - 190*a^2*b^3*Cos[3*(c + d*x)] + 745*b^5*Cos[3*(c + d*x)] - 240*b^5*Cos[4*
(c + d*x)] + 9*a^4*b*Cos[5*(c + d*x)] - 38*a^2*b^3*Cos[5*(c + d*x)] + 149*b^5*Cos[5*(c + d*x)] + 640*a^5*Sin[c
+ d*x] - 1600*a^3*b^2*Sin[c + d*x] + 1200*a*b^4*Sin[c + d*x] + 320*a^5*Sin[3*(c + d*x)] - 1040*a^3*b^2*Sin[3*
(c + d*x)] + 1080*a*b^4*Sin[3*(c + d*x)] + 64*a^5*Sin[5*(c + d*x)] - 208*a^3*b^2*Sin[5*(c + d*x)] + 264*a*b^4*
Sin[5*(c + d*x)]))/(1920*(a - b)^3*(a + b)^3*d)

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fricas [A]  time = 0.52, size = 666, normalized size = 3.38 \[ \left [\frac {15 \, \sqrt {-a^{2} + b^{2}} b^{6} \cos \left (d x + c\right )^{5} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 6 \, a^{6} b + 18 \, a^{4} b^{3} - 18 \, a^{2} b^{5} + 6 \, b^{7} - 30 \, {\left (a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} + 10 \, {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{7} - 9 \, a^{5} b^{2} + 9 \, a^{3} b^{4} - 3 \, a b^{6} + {\left (8 \, a^{7} - 34 \, a^{5} b^{2} + 59 \, a^{3} b^{4} - 33 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{7} - 17 \, a^{5} b^{2} + 22 \, a^{3} b^{4} - 9 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{5}}, \frac {15 \, \sqrt {a^{2} - b^{2}} b^{6} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{5} - 3 \, a^{6} b + 9 \, a^{4} b^{3} - 9 \, a^{2} b^{5} + 3 \, b^{7} - 15 \, {\left (a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, a^{7} - 9 \, a^{5} b^{2} + 9 \, a^{3} b^{4} - 3 \, a b^{6} + {\left (8 \, a^{7} - 34 \, a^{5} b^{2} + 59 \, a^{3} b^{4} - 33 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a^{7} - 17 \, a^{5} b^{2} + 22 \, a^{3} b^{4} - 9 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*sqrt(-a^2 + b^2)*b^6*cos(d*x + c)^5*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b
^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x +
c) - a^2 - b^2)) - 6*a^6*b + 18*a^4*b^3 - 18*a^2*b^5 + 6*b^7 - 30*(a^2*b^5 - b^7)*cos(d*x + c)^4 + 10*(a^4*b^3
- 2*a^2*b^5 + b^7)*cos(d*x + c)^2 + 2*(3*a^7 - 9*a^5*b^2 + 9*a^3*b^4 - 3*a*b^6 + (8*a^7 - 34*a^5*b^2 + 59*a^3
*b^4 - 33*a*b^6)*cos(d*x + c)^4 + (4*a^7 - 17*a^5*b^2 + 22*a^3*b^4 - 9*a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((
a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*d*cos(d*x + c)^5), 1/15*(15*sqrt(a^2 - b^2)*b^6*arctan(-(a*sin(
d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c)^5 - 3*a^6*b + 9*a^4*b^3 - 9*a^2*b^5 + 3*b^7 - 15*(a
^2*b^5 - b^7)*cos(d*x + c)^4 + 5*(a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^2 + (3*a^7 - 9*a^5*b^2 + 9*a^3*b^4 -
3*a*b^6 + (8*a^7 - 34*a^5*b^2 + 59*a^3*b^4 - 33*a*b^6)*cos(d*x + c)^4 + (4*a^7 - 17*a^5*b^2 + 22*a^3*b^4 - 9*
a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*d*cos(d*x + c)^5)]

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giac [B]  time = 1.23, size = 584, normalized size = 2.96 \[ -\frac {2 \, {\left (\frac {15 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{6}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {15 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 45 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 15 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 45 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 45 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 20 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 80 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 90 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 58 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 166 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 198 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 80 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 140 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 20 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 10 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 70 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 45 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 45 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{4} b + 11 \, a^{2} b^{3} - 23 \, b^{5}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}\right )}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*b^6
/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + (15*a^5*tan(1/2*d*x + 1/2*c)^9 - 45*a^3*b^2*tan(1/2*d
*x + 1/2*c)^9 + 45*a*b^4*tan(1/2*d*x + 1/2*c)^9 - 15*a^4*b*tan(1/2*d*x + 1/2*c)^8 + 45*a^2*b^3*tan(1/2*d*x + 1
/2*c)^8 - 45*b^5*tan(1/2*d*x + 1/2*c)^8 - 20*a^5*tan(1/2*d*x + 1/2*c)^7 + 80*a^3*b^2*tan(1/2*d*x + 1/2*c)^7 -
120*a*b^4*tan(1/2*d*x + 1/2*c)^7 - 30*a^2*b^3*tan(1/2*d*x + 1/2*c)^6 + 90*b^5*tan(1/2*d*x + 1/2*c)^6 + 58*a^5*
tan(1/2*d*x + 1/2*c)^5 - 166*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 198*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 30*a^4*b*tan(
1/2*d*x + 1/2*c)^4 + 80*a^2*b^3*tan(1/2*d*x + 1/2*c)^4 - 140*b^5*tan(1/2*d*x + 1/2*c)^4 - 20*a^5*tan(1/2*d*x +
1/2*c)^3 + 80*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 120*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 10*a^2*b^3*tan(1/2*d*x + 1/
2*c)^2 + 70*b^5*tan(1/2*d*x + 1/2*c)^2 + 15*a^5*tan(1/2*d*x + 1/2*c) - 45*a^3*b^2*tan(1/2*d*x + 1/2*c) + 45*a*
b^4*tan(1/2*d*x + 1/2*c) - 3*a^4*b + 11*a^2*b^3 - 23*b^5)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/2*d*x +
1/2*c)^2 - 1)^5))/d

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maple [B]  time = 0.18, size = 525, normalized size = 2.66 \[ -\frac {2}{5 d \left (2 a +2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {1}{2 d \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {7 a}{8 d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {9 b}{8 d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {11 a}{12 d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {13 b}{12 d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a^{2}}{d \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {21 a b}{8 d \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {15 b^{2}}{8 d \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b^{6} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {a^{2}-b^{2}}}-\frac {2}{5 d \left (2 a -2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{2 d \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {7 a}{8 d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {9 b}{8 d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {11 a}{12 d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {13 b}{12 d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {a^{2}}{d \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {21 a b}{8 d \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {15 b^{2}}{8 d \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/d/(2*a+2*b)/(tan(1/2*d*x+1/2*c)-1)^5-1/2/d/(a+b)/(tan(1/2*d*x+1/2*c)-1)^4-7/8/d/(a+b)^2/(tan(1/2*d*x+1/2*
c)-1)^2*a-9/8/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)^2*b-11/12/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)^3*a-13/12/d/(a+b)^2/
(tan(1/2*d*x+1/2*c)-1)^3*b-1/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)*a^2-21/8/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)*a*b-15
/8/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)*b^2-2/d*b^6/(a-b)^3/(a+b)^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/
2*c)+2*b)/(a^2-b^2)^(1/2))-2/5/d/(2*a-2*b)/(tan(1/2*d*x+1/2*c)+1)^5+1/2/d/(a-b)/(tan(1/2*d*x+1/2*c)+1)^4+7/8/d
/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)^2*a-9/8/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)^2*b-11/12/d/(a-b)^2/(tan(1/2*d*x+1/2*
c)+1)^3*a+13/12/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)^3*b-1/d/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)*a^2+21/8/d/(a-b)^3/(ta
n(1/2*d*x+1/2*c)+1)*a*b-15/8/d/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)*b^2

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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(sec(d*x+c)^6/(a+b*sin(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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 8.06, size = 774, normalized size = 3.93 \[ \frac {\frac {2\,\left (3\,a^4\,b-11\,a^2\,b^3+23\,b^5\right )}{15\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^5-3\,a^3\,b^2+3\,a\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (7\,b^5-a^2\,b^3\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (3\,b^5-a^2\,b^3\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^5-4\,a^3\,b^2+6\,a\,b^4\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (a^5-3\,a^3\,b^2+3\,a\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (a^5-4\,a^3\,b^2+6\,a\,b^4\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (29\,a^5-83\,a^3\,b^2+99\,a\,b^4\right )}{15\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (a^4\,b-3\,a^2\,b^3+3\,b^5\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^4\,b-8\,a^2\,b^3+14\,b^5\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,b^6\,\mathrm {atan}\left (\frac {\frac {b^6\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {2\,a\,b^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}}{2\,b^6}\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^6*(a + b*sin(c + d*x))),x)

[Out]

((2*(3*a^4*b + 23*b^5 - 11*a^2*b^3))/(15*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (2*tan(c/2 + (d*x)/2)*(3*a*b^4
+ a^5 - 3*a^3*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (4*tan(c/2 + (d*x)/2)^2*(7*b^5 - a^2*b^3))/(3*(a^6
- b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (4*tan(c/2 + (d*x)/2)^6*(3*b^5 - a^2*b^3))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^
2) + (8*tan(c/2 + (d*x)/2)^3*(6*a*b^4 + a^5 - 4*a^3*b^2))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (2*tan(c/2
+ (d*x)/2)^9*(3*a*b^4 + a^5 - 3*a^3*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (8*tan(c/2 + (d*x)/2)^7*(6*a*
b^4 + a^5 - 4*a^3*b^2))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (4*tan(c/2 + (d*x)/2)^5*(99*a*b^4 + 29*a^5 -
83*a^3*b^2))/(15*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^8*(a^4*b + 3*b^5 - 3*a^2*b^3))/
(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (4*tan(c/2 + (d*x)/2)^4*(3*a^4*b + 14*b^5 - 8*a^2*b^3))/(3*(a^6 - b^6 +
3*a^2*b^4 - 3*a^4*b^2)))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*ta
n(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1)) - (2*b^6*atan(((b^6*(2*a^6*b - 2*b^7 + 6*a^2*b^5 - 6*a^4*b^3)
)/((a + b)^(7/2)*(a - b)^(7/2)) + (2*a*b^6*tan(c/2 + (d*x)/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/((a + b)^(7
/2)*(a - b)^(7/2)))/(2*b^6)))/(d*(a + b)^(7/2)*(a - b)^(7/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{6}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6/(a+b*sin(d*x+c)),x)

[Out]

Integral(sec(c + d*x)**6/(a + b*sin(c + d*x)), x)

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