3.460 \(\int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=264 \[ \frac {7 a b \sec ^3(c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {b \sec ^3(c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac {\sec ^3(c+d x) \left (5 b \left (6 a^2+b^2\right )-a \left (2 a^2+33 b^2\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )^3}+\frac {5 b^4 \left (6 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {\sec (c+d x) \left (15 b^3 \left (6 a^2+b^2\right )+a \left (4 a^4-28 a^2 b^2-81 b^4\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )^4} \]

[Out]

5*b^4*(6*a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(9/2)/d+1/2*b*sec(d*x+c)^3/(a^2-b
^2)/d/(a+b*sin(d*x+c))^2+7/2*a*b*sec(d*x+c)^3/(a^2-b^2)^2/d/(a+b*sin(d*x+c))-1/6*sec(d*x+c)^3*(5*b*(6*a^2+b^2)
-a*(2*a^2+33*b^2)*sin(d*x+c))/(a^2-b^2)^3/d+1/6*sec(d*x+c)*(15*b^3*(6*a^2+b^2)+a*(4*a^4-28*a^2*b^2-81*b^4)*sin
(d*x+c))/(a^2-b^2)^4/d

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Rubi [A]  time = 0.64, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2694, 2864, 2866, 12, 2660, 618, 204} \[ \frac {5 b^4 \left (6 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{9/2}}+\frac {7 a b \sec ^3(c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {b \sec ^3(c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac {\sec ^3(c+d x) \left (5 b \left (6 a^2+b^2\right )-a \left (2 a^2+33 b^2\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )^3}+\frac {\sec (c+d x) \left (a \left (-28 a^2 b^2+4 a^4-81 b^4\right ) \sin (c+d x)+15 b^3 \left (6 a^2+b^2\right )\right )}{6 d \left (a^2-b^2\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b^4*(6*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(9/2)*d) + (b*Sec[c + d*x]
^3)/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) + (7*a*b*Sec[c + d*x]^3)/(2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x]))
- (Sec[c + d*x]^3*(5*b*(6*a^2 + b^2) - a*(2*a^2 + 33*b^2)*Sin[c + d*x]))/(6*(a^2 - b^2)^3*d) + (Sec[c + d*x]*
(15*b^3*(6*a^2 + b^2) + a*(4*a^4 - 28*a^2*b^2 - 81*b^4)*Sin[c + d*x]))/(6*(a^2 - b^2)^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 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 2694

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

Rule 2864

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

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 ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\frac {b \sec ^3(c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {\int \frac {\sec ^4(c+d x) (-2 a+5 b \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac {b \sec ^3(c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {7 a b \sec ^3(c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\int \frac {\sec ^4(c+d x) \left (2 a^2+5 b^2-28 a b \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac {b \sec ^3(c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {7 a b \sec ^3(c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 b \left (6 a^2+b^2\right )-a \left (2 a^2+33 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}-\frac {\int \frac {\sec ^2(c+d x) \left (-4 a^4+24 a^2 b^2+15 b^4-2 a b \left (2 a^2+33 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{6 \left (a^2-b^2\right )^3}\\ &=\frac {b \sec ^3(c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {7 a b \sec ^3(c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 b \left (6 a^2+b^2\right )-a \left (2 a^2+33 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}+\frac {\sec (c+d x) \left (15 b^3 \left (6 a^2+b^2\right )+a \left (4 a^4-28 a^2 b^2-81 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^4 d}+\frac {\int \frac {15 b^4 \left (6 a^2+b^2\right )}{a+b \sin (c+d x)} \, dx}{6 \left (a^2-b^2\right )^4}\\ &=\frac {b \sec ^3(c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {7 a b \sec ^3(c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 b \left (6 a^2+b^2\right )-a \left (2 a^2+33 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}+\frac {\sec (c+d x) \left (15 b^3 \left (6 a^2+b^2\right )+a \left (4 a^4-28 a^2 b^2-81 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^4 d}+\frac {\left (5 b^4 \left (6 a^2+b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^4}\\ &=\frac {b \sec ^3(c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {7 a b \sec ^3(c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 b \left (6 a^2+b^2\right )-a \left (2 a^2+33 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}+\frac {\sec (c+d x) \left (15 b^3 \left (6 a^2+b^2\right )+a \left (4 a^4-28 a^2 b^2-81 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^4 d}+\frac {\left (5 b^4 \left (6 a^2+b^2\right )\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 )^4 d}\\ &=\frac {b \sec ^3(c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {7 a b \sec ^3(c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 b \left (6 a^2+b^2\right )-a \left (2 a^2+33 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}+\frac {\sec (c+d x) \left (15 b^3 \left (6 a^2+b^2\right )+a \left (4 a^4-28 a^2 b^2-81 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^4 d}-\frac {\left (10 b^4 \left (6 a^2+b^2\right )\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 )^4 d}\\ &=\frac {5 b^4 \left (6 a^2+b^2\right ) \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 )^{9/2} d}+\frac {b \sec ^3(c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {7 a b \sec ^3(c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 b \left (6 a^2+b^2\right )-a \left (2 a^2+33 b^2\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}+\frac {\sec (c+d x) \left (15 b^3 \left (6 a^2+b^2\right )+a \left (4 a^4-28 a^2 b^2-81 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^4 d}\\ \end {align*}

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Mathematica [A]  time = 2.81, size = 380, normalized size = 1.44 \[ \frac {\frac {60 b^4 \left (6 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{9/2}}+\frac {66 a b^5 \cos (c+d x)}{(a-b)^4 (a+b)^4 (a+b \sin (c+d x))}+\frac {6 b^5 \cos (c+d x)}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^2}+\frac {2 (4 a+13 b) \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^4 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 (4 a-13 b) \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^4 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {1}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {1}{(a-b)^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((60*b^4*(6*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(9/2) + 1/((a + b)^3*(Cos
[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (2*Sin[(c + d*x)/2])/((a + b)^3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3
) + (2*(4*a + 13*b)*Sin[(c + d*x)/2])/((a + b)^4*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*Sin[(c + d*x)/2])
/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - 1/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) +
(2*(4*a - 13*b)*Sin[(c + d*x)/2])/((a - b)^4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (6*b^5*Cos[c + d*x])/((
a - b)^3*(a + b)^3*(a + b*Sin[c + d*x])^2) + (66*a*b^5*Cos[c + d*x])/((a - b)^4*(a + b)^4*(a + b*Sin[c + d*x])
))/(12*d)

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fricas [B]  time = 0.88, size = 1200, normalized size = 4.55 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/12*(4*a^8*b - 16*a^6*b^3 + 24*a^4*b^5 - 16*a^2*b^7 + 4*b^9 + 2*(8*a^8*b - 64*a^6*b^3 - 16*a^4*b^5 + 87*a^2*
b^7 - 15*b^9)*cos(d*x + c)^4 - 4*(2*a^8*b - a^6*b^3 - 9*a^4*b^5 + 13*a^2*b^7 - 5*b^9)*cos(d*x + c)^2 - 15*((6*
a^2*b^6 + b^8)*cos(d*x + c)^5 - 2*(6*a^3*b^5 + a*b^7)*cos(d*x + c)^3*sin(d*x + c) - (6*a^4*b^4 + 7*a^2*b^6 + b
^8)*cos(d*x + c)^3)*sqrt(-a^2 + b^2)*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)) - 2*(2*a^9 - 8*a^7*b^2 + 12*a^5*b^4 - 8*a^3*b^6 + 2*a*b^8 - (4*a^7*b^2 - 32*a^5*b^4 - 53*a^3*b^6 + 81*
a*b^8)*cos(d*x + c)^4 + 2*(2*a^9 - 15*a^7*b^2 + 33*a^5*b^4 - 29*a^3*b^6 + 9*a*b^8)*cos(d*x + c)^2)*sin(d*x + c
))/((a^10*b^2 - 5*a^8*b^4 + 10*a^6*b^6 - 10*a^4*b^8 + 5*a^2*b^10 - b^12)*d*cos(d*x + c)^5 - 2*(a^11*b - 5*a^9*
b^3 + 10*a^7*b^5 - 10*a^5*b^7 + 5*a^3*b^9 - a*b^11)*d*cos(d*x + c)^3*sin(d*x + c) - (a^12 - 4*a^10*b^2 + 5*a^8
*b^4 - 5*a^4*b^8 + 4*a^2*b^10 - b^12)*d*cos(d*x + c)^3), 1/6*(2*a^8*b - 8*a^6*b^3 + 12*a^4*b^5 - 8*a^2*b^7 + 2
*b^9 + (8*a^8*b - 64*a^6*b^3 - 16*a^4*b^5 + 87*a^2*b^7 - 15*b^9)*cos(d*x + c)^4 - 2*(2*a^8*b - a^6*b^3 - 9*a^4
*b^5 + 13*a^2*b^7 - 5*b^9)*cos(d*x + c)^2 - 15*((6*a^2*b^6 + b^8)*cos(d*x + c)^5 - 2*(6*a^3*b^5 + a*b^7)*cos(d
*x + c)^3*sin(d*x + c) - (6*a^4*b^4 + 7*a^2*b^6 + b^8)*cos(d*x + c)^3)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c)
+ b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - (2*a^9 - 8*a^7*b^2 + 12*a^5*b^4 - 8*a^3*b^6 + 2*a*b^8 - (4*a^7*b^2 - 3
2*a^5*b^4 - 53*a^3*b^6 + 81*a*b^8)*cos(d*x + c)^4 + 2*(2*a^9 - 15*a^7*b^2 + 33*a^5*b^4 - 29*a^3*b^6 + 9*a*b^8)
*cos(d*x + c)^2)*sin(d*x + c))/((a^10*b^2 - 5*a^8*b^4 + 10*a^6*b^6 - 10*a^4*b^8 + 5*a^2*b^10 - b^12)*d*cos(d*x
+ c)^5 - 2*(a^11*b - 5*a^9*b^3 + 10*a^7*b^5 - 10*a^5*b^7 + 5*a^3*b^9 - a*b^11)*d*cos(d*x + c)^3*sin(d*x + c)
- (a^12 - 4*a^10*b^2 + 5*a^8*b^4 - 5*a^4*b^8 + 4*a^2*b^10 - b^12)*d*cos(d*x + c)^3)]

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giac [B]  time = 9.09, size = 622, normalized size = 2.36 \[ \frac {\frac {15 \, {\left (6 \, a^{2} b^{4} + b^{6}\right )} {\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 )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, {\left (13 \, a^{3} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a^{4} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 23 \, a^{2} b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, a^{3} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{4} b^{5} - a^{2} b^{7}\right )}}{{\left (a^{10} - 4 \, a^{8} b^{2} + 6 \, a^{6} b^{4} - 4 \, a^{4} b^{6} + a^{2} b^{8}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}} - \frac {2 \, {\left (3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 36 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 9 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 32 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 42 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{4} b + 32 \, a^{2} b^{3} + 7 \, b^{5}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/3*(15*(6*a^2*b^4 + b^6)*(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)))/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*sqrt(a^2 - b^2)) + 3*(13*a^3*b^6*tan(1/2*d*x + 1
/2*c)^3 - 2*a*b^8*tan(1/2*d*x + 1/2*c)^3 + 12*a^4*b^5*tan(1/2*d*x + 1/2*c)^2 + 23*a^2*b^7*tan(1/2*d*x + 1/2*c)
^2 - 2*b^9*tan(1/2*d*x + 1/2*c)^2 + 35*a^3*b^6*tan(1/2*d*x + 1/2*c) - 2*a*b^8*tan(1/2*d*x + 1/2*c) + 12*a^4*b^
5 - a^2*b^7)/((a^10 - 4*a^8*b^2 + 6*a^6*b^4 - 4*a^4*b^6 + a^2*b^8)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x
+ 1/2*c) + a)^2) - 2*(3*a^5*tan(1/2*d*x + 1/2*c)^5 - 12*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 - 27*a*b^4*tan(1/2*d*x
+ 1/2*c)^5 - 9*a^4*b*tan(1/2*d*x + 1/2*c)^4 + 36*a^2*b^3*tan(1/2*d*x + 1/2*c)^4 + 9*b^5*tan(1/2*d*x + 1/2*c)^
4 - 2*a^5*tan(1/2*d*x + 1/2*c)^3 + 32*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 42*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 60*a^
2*b^3*tan(1/2*d*x + 1/2*c)^2 - 12*b^5*tan(1/2*d*x + 1/2*c)^2 + 3*a^5*tan(1/2*d*x + 1/2*c) - 12*a^3*b^2*tan(1/2
*d*x + 1/2*c) - 27*a*b^4*tan(1/2*d*x + 1/2*c) - 3*a^4*b + 32*a^2*b^3 + 7*b^5)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 -
4*a^2*b^6 + b^8)*(tan(1/2*d*x + 1/2*c)^2 - 1)^3))/d

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maple [B]  time = 0.34, size = 854, normalized size = 3.23 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)^3-1/2/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)^2-1/d/(a+b)^4/(tan(1/2*d*x+1/2*c)
-1)*a-5/2/d/(a+b)^4/(tan(1/2*d*x+1/2*c)-1)*b+13/d*b^6/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/
2*c)*b+a)^2*a*tan(1/2*d*x+1/2*c)^3-2/d*b^8/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2
/a*tan(1/2*d*x+1/2*c)^3+12/d*b^5/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a^2*tan(1
/2*d*x+1/2*c)^2+23/d*b^7/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c
)^2-2/d*b^9/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a^2*tan(1/2*d*x+1/2*c)^2+35/d*
b^6/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a*tan(1/2*d*x+1/2*c)-2/d*b^8/(a-b)^4/(
a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a*tan(1/2*d*x+1/2*c)+12/d*b^5/(a-b)^4/(a+b)^4/(tan(
1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a^2-1/d*b^7/(a-b)^4/(a+b)^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d
*x+1/2*c)*b+a)^2+30/d*b^4/(a-b)^4/(a+b)^4/(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))*a^2+5/d*b^6/(a-b)^4/(a+b)^4/(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))-1/3/
d/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)^3+1/2/d/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)^2-1/d/(a-b)^4/(tan(1/2*d*x+1/2*c)+1)*a
+5/2/d/(a-b)^4/(tan(1/2*d*x+1/2*c)+1)*b

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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)^4/(a+b*sin(d*x+c))^3,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 = 9.37, size = 1167, normalized size = 4.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + b*sin(c + d*x))^3),x)

[Out]

(5*b^4*atan(((5*b^4*(6*a^2 + b^2)*(2*a^8*b + 2*b^9 - 8*a^2*b^7 + 12*a^4*b^5 - 8*a^6*b^3))/(2*(a + b)^(9/2)*(a
- b)^(9/2)) + (5*a*b^4*tan(c/2 + (d*x)/2)*(6*a^2 + b^2)*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2))/((a +
b)^(9/2)*(a - b)^(9/2)))/(5*b^6 + 30*a^2*b^4))*(6*a^2 + b^2))/(d*(a + b)^(9/2)*(a - b)^(9/2)) - ((2*tan(c/2 +
(d*x)/2)^5*(255*a*b^6 + 2*a^7 + 62*a^3*b^4 - 4*a^5*b^2))/(3*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2))
- (6*a^6*b + 3*b^7 - 50*a^2*b^5 - 64*a^4*b^3)/(3*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (4*tan(c/2
+ (d*x)/2)^7*(2*a^6 + 3*b^6 + 36*a^2*b^4 - 6*a^4*b^2))/(3*a*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (2*tan(c/2
+ (d*x)/2)^2*(6*a^6*b + 3*b^7 - 64*a^2*b^5 - 50*a^4*b^3))/(3*a^2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (tan(
c/2 + (d*x)/2)^9*(13*a^2*b^6 - 2*b^8 - 2*a^8 + 18*a^4*b^4 + 8*a^6*b^2))/(a*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4
- 4*a^6*b^2)) - (tan(c/2 + (d*x)/2)^8*(23*a^2*b^7 - 2*b^9 - 2*a^8*b + 78*a^4*b^5 + 8*a^6*b^3))/(a^2*(a^8 + b^8
- 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2)) + (tan(c/2 + (d*x)/2)*(6*a^8 - 6*b^8 + 161*a^2*b^6 + 202*a^4*b^4 - 48*a
^6*b^2))/(3*a*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (4*tan(c/2 + (d*x)/2)^3*(2*a^8 + 3*b^8 - 133*
a^2*b^6 - 86*a^4*b^4 + 4*a^6*b^2))/(3*a*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (2*tan(c/2 + (d*x)/
2)^4*(8*a^8*b - 9*b^9 + 156*a^2*b^7 + 188*a^4*b^5 - 28*a^6*b^3))/(3*a^2*(a^2 - b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3
*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^6*(141*a^2*b^7 - 9*b^9 - 14*a^8*b + 246*a^4*b^5 + 56*a^6*b^3))/(3*a^2*(a^2
- b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)))/(d*(tan(c/2 + (d*x)/2)^4*(2*a^2 + 12*b^2) - tan(c/2 + (d*x)/2)^6*
(2*a^2 + 12*b^2) + a^2*tan(c/2 + (d*x)/2)^10 - a^2 + tan(c/2 + (d*x)/2)^2*(a^2 - 4*b^2) - tan(c/2 + (d*x)/2)^8
*(a^2 - 4*b^2) + 8*a*b*tan(c/2 + (d*x)/2)^3 - 8*a*b*tan(c/2 + (d*x)/2)^7 + 4*a*b*tan(c/2 + (d*x)/2)^9 - 4*a*b*
tan(c/2 + (d*x)/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+b*sin(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**4/(a + b*sin(c + d*x))**3, x)

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