3.448 \(\int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=193 \[ \frac {b \sec ^3(c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^2}+\frac {10 a b^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac {\sec (c+d x) \left (\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)+15 a b^3\right )}{3 d \left (a^2-b^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.37, antiderivative size = 193, 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 = {2694, 2866, 12, 2660, 618, 204} \[ \frac {10 a b^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac {b \sec ^3(c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^2}+\frac {\sec (c+d x) \left (\left (-9 a^2 b^2+2 a^4-8 b^4\right ) \sin (c+d x)+15 a b^3\right )}{3 d \left (a^2-b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + b*Sin[c + d*x])^2,x]

[Out]

(10*a*b^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + (b*Sec[c + d*x]^3)/((a^2 -
b^2)*d*(a + b*Sin[c + d*x])) - (Sec[c + d*x]^3*(5*a*b - (a^2 + 4*b^2)*Sin[c + d*x]))/(3*(a^2 - b^2)^2*d) + (S
ec[c + d*x]*(15*a*b^3 + (2*a^4 - 9*a^2*b^2 - 8*b^4)*Sin[c + d*x]))/(3*(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 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 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))^2} \, dx &=\frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\int \frac {\sec ^4(c+d x) (-a+4 b \sin (c+d x))}{a+b \sin (c+d x)} \, dx}{-a^2+b^2}\\ &=\frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\int \frac {\sec ^2(c+d x) \left (a \left (2 a^2-7 b^2\right )+2 b \left (a^2+4 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=\frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac {\int -\frac {15 a b^4}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^3}\\ &=\frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}+\frac {\left (5 a b^4\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=\frac {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}+\frac {\left (10 a b^4\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 {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac {\left (20 a b^4\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 {10 a b^4 \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 {b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.86, size = 336, normalized size = 1.74 \[ \frac {\frac {120 a b^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {12 b^5 \cos (c+d x)}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))}+\frac {4 (2 a+5 b) \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {4 (2 a-5 b) \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {1}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {1}{(a-b)^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + b*Sin[c + d*x])^2,x]

[Out]

((120*a*b^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) + 1/((a + b)^2*(Cos[(c + d*x)/
2] - Sin[(c + d*x)/2])^2) + (2*Sin[(c + d*x)/2])/((a + b)^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + (4*(2*a
+ 5*b)*Sin[(c + d*x)/2])/((a + b)^3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*Sin[(c + d*x)/2])/((a - b)^2*
(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - 1/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(2*a - 5
*b)*Sin[(c + d*x)/2])/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (12*b^5*Cos[c + d*x])/((a - b)^3*(a
+ b)^3*(a + b*Sin[c + d*x])))/(12*d)

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fricas [A]  time = 0.53, size = 782, normalized size = 4.05 \[ \left [-\frac {2 \, a^{6} b - 6 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - 2 \, b^{7} + 2 \, {\left (2 \, a^{6} b - 11 \, a^{4} b^{3} + a^{2} b^{5} + 8 \, b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b + 2 \, a^{4} b^{3} - 7 \, a^{2} b^{5} + 4 \, b^{7}\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left (a b^{5} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a^{2} b^{4} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a^{2} + b^{2}} \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 ) - 2 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6} + {\left (2 \, a^{7} - 11 \, a^{5} b^{2} + 16 \, a^{3} b^{4} - 7 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right )^{3}\right )}}, -\frac {a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7} + {\left (2 \, a^{6} b - 11 \, a^{4} b^{3} + a^{2} b^{5} + 8 \, b^{7}\right )} \cos \left (d x + c\right )^{4} - {\left (a^{6} b + 2 \, a^{4} b^{3} - 7 \, a^{2} b^{5} + 4 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (a b^{5} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a^{2} b^{4} \cos \left (d x + c\right )^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6} + {\left (2 \, a^{7} - 11 \, a^{5} b^{2} + 16 \, a^{3} b^{4} - 7 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left ({\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right )^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/6*(2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 + 2*(2*a^6*b - 11*a^4*b^3 + a^2*b^5 + 8*b^7)*cos(d*x + c)^4 - 2
*(a^6*b + 2*a^4*b^3 - 7*a^2*b^5 + 4*b^7)*cos(d*x + c)^2 - 15*(a*b^5*cos(d*x + c)^3*sin(d*x + c) + a^2*b^4*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*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 + (2*a^7 - 11*a^5*b^2 + 16*a^3*b^4 - 7*a*b^6)*cos(d*x + c)^2)*sin(d*
x + c))/((a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^3*sin(d*x + c) + (a^9 - 4*a^7*b^2 +
6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3), -1/3*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 + (2*a^6*b - 11*a^
4*b^3 + a^2*b^5 + 8*b^7)*cos(d*x + c)^4 - (a^6*b + 2*a^4*b^3 - 7*a^2*b^5 + 4*b^7)*cos(d*x + c)^2 + 15*(a*b^5*c
os(d*x + c)^3*sin(d*x + c) + a^2*b^4*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))) - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 + (2*a^7 - 11*a^5*b^2 + 16*a^3*b^4 - 7*a*b^6)*cos(d
*x + c)^2)*sin(d*x + c))/((a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^3*sin(d*x + c) + (a
^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3)]

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giac [B]  time = 0.67, size = 427, normalized size = 2.21 \[ \frac {2 \, {\left (\frac {15 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a b^{4}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, {\left (b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a b^{5}\right )}}{{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}} - \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{3} b + 14 \, a b^{3}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

2/3*(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)))*a*b^4
/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + 3*(b^6*tan(1/2*d*x + 1/2*c) + a*b^5)/((a^7 - 3*a^5*b^
2 + 3*a^3*b^4 - a*b^6)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)) - (3*a^4*tan(1/2*d*x + 1/2*c
)^5 - 9*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*b^4*tan(1/2*d*x + 1/2*c)^5 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^4 + 18*a*
b^3*tan(1/2*d*x + 1/2*c)^4 - 2*a^4*tan(1/2*d*x + 1/2*c)^3 + 18*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 8*b^4*tan(1/2*
d*x + 1/2*c)^3 - 24*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^4*tan(1/2*d*x + 1/2*c) - 9*a^2*b^2*tan(1/2*d*x + 1/2*c)
- 6*b^4*tan(1/2*d*x + 1/2*c) - 2*a^3*b + 14*a*b^3)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/2*d*x + 1/2*c)
^2 - 1)^3))/d

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maple [B]  time = 0.32, size = 370, normalized size = 1.92 \[ -\frac {1}{3 d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a}{d \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b}{d \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 b^{6} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a}+\frac {2 b^{5}}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {10 b^{4} a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {a^{2}-b^{2}}}-\frac {1}{3 d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a}{d \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 b}{d \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+b*sin(d*x+c))^2,x)

[Out]

-1/3/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)^3-1/2/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)^2-1/d/(a+b)^3/(tan(1/2*d*x+1/2*c)
-1)*a-2/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)*b+2/d*b^6/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c
)*b+a)/a*tan(1/2*d*x+1/2*c)+2/d*b^5/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)+10/d*b^4
/(a-b)^3/(a+b)^3*a/(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)^2/(tan
(1/2*d*x+1/2*c)+1)^3+1/2/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)^2-1/d/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)*a+2/d/(a-b)^3/(
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))^2,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 = 8.51, size = 727, normalized size = 3.77 \[ \frac {\frac {2\,\left (-2\,a^4\,b+14\,a^2\,b^3+3\,b^5\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {10\,b^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-4\,a^4\,b+28\,a^2\,b^3+21\,b^5\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 )\,\left (3\,a^6-13\,a^4\,b^2+22\,a^2\,b^4+3\,b^6\right )}{3\,a\,\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 )}^7\,\left (-a^6+3\,a^4\,b^2+2\,a^2\,b^4+b^6\right )}{a\,\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 )}^5\,\left (a^6-3\,a^4\,b^2+38\,a^2\,b^4+9\,b^6\right )}{3\,a\,\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 )}^3\,\left (a^6+9\,a^4\,b^2-46\,a^2\,b^4-9\,b^6\right )}{3\,a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {10\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-2\,a^4+6\,a^2\,b^2+5\,b^4\right )}{3\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}+\frac {10\,a\,b^4\,\mathrm {atan}\left (\frac {\frac {5\,a\,b^4\,\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 {10\,a^2\,b^4\,\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}}}{10\,a\,b^4}\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)^4*(a + b*sin(c + d*x))^2),x)

[Out]

((2*(3*b^5 - 2*a^4*b + 14*a^2*b^3))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (10*b^5*tan(c/2 + (d*x)/2)^6)/(a
^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (2*tan(c/2 + (d*x)/2)^2*(21*b^5 - 4*a^4*b + 28*a^2*b^3))/(3*(a^6 - b^6 + 3
*a^2*b^4 - 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)*(3*a^6 + 3*b^6 + 22*a^2*b^4 - 13*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)^7*(b^6 - a^6 + 2*a^2*b^4 + 3*a^4*b^2))/(a*(a^6 - b^6 + 3*a^2*b^4
- 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^5*(a^6 + 9*b^6 + 38*a^2*b^4 - 3*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)^3*(a^6 - 9*b^6 - 46*a^2*b^4 + 9*a^4*b^2))/(3*a*(a^6 - b^6 + 3*a^2*b^4 - 3
*a^4*b^2)) + (10*b*tan(c/2 + (d*x)/2)^4*(5*b^4 - 2*a^4 + 6*a^2*b^2))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)))/
(d*(a + 2*b*tan(c/2 + (d*x)/2) - 2*a*tan(c/2 + (d*x)/2)^2 + 2*a*tan(c/2 + (d*x)/2)^6 - a*tan(c/2 + (d*x)/2)^8
- 6*b*tan(c/2 + (d*x)/2)^3 + 6*b*tan(c/2 + (d*x)/2)^5 - 2*b*tan(c/2 + (d*x)/2)^7)) + (10*a*b^4*atan(((5*a*b^4*
(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)) + (10*a^2*b^4*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)))/(10*a*b^4)))/(d*(a + b)^(7/2)*(a - b)^(7/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+b*sin(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)**4/(a + b*sin(c + d*x))**2, x)

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