3.1474 \(\int \frac {\sec (c+d x) \tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=204 \[ \frac {3 a b \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2} \]

[Out]

3*a*b*(2*a^2+3*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(7/2)/d-1/2*a*sec(d*x+c)/(a^2-b
^2)/d/(a+b*sin(d*x+c))^2-1/2*(3*a^2+2*b^2)*sec(d*x+c)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))+1/2*sec(d*x+c)*(3*a*(2*a^
2+3*b^2)-b*(11*a^2+4*b^2)*sin(d*x+c))/(a^2-b^2)^3/d

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Rubi [A]  time = 0.36, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2864, 2866, 12, 2660, 618, 204} \[ \frac {3 a b \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 d \left (a^2-b^2\right )^3}-\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*b*(2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) - (a*Sec[c + d*
x])/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) - ((3*a^2 + 2*b^2)*Sec[c + d*x])/(2*(a^2 - b^2)^2*d*(a + b*Sin[c
+ d*x])) + (Sec[c + d*x]*(3*a*(2*a^2 + 3*b^2) - b*(11*a^2 + 4*b^2)*Sin[c + d*x]))/(2*(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 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 (c+d x) \tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx &=-\frac {a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {\int \frac {\sec ^2(c+d x) (2 b-3 a \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\int \frac {\sec ^2(c+d x) \left (-5 a b+2 \left (3 a^2+2 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac {a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}-\frac {\int -\frac {3 a b \left (2 a^2+3 b^2\right )}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac {a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}+\frac {\left (3 a b \left (2 a^2+3 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac {a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}+\frac {\left (3 a b \left (2 a^2+3 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 )^3 d}\\ &=-\frac {a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}-\frac {\left (6 a b \left (2 a^2+3 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 )^3 d}\\ &=\frac {3 a b \left (2 a^2+3 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 )^{7/2} d}-\frac {a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d}\\ \end {align*}

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Mathematica [A]  time = 3.03, size = 206, normalized size = 1.01 \[ \frac {\frac {6 a b \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {b^2 \cos (c+d x) \left (b \left (5 a^2+2 b^2\right ) \sin (c+d x)+a \left (6 a^2+b^2\right )\right )}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^2}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {2}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {2}{(a-b)^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a*b*(2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) + Sin[(c + d*x)/2]
*(2/((a + b)^3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - 2/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) +
(b^2*Cos[c + d*x]*(a*(6*a^2 + b^2) + b*(5*a^2 + 2*b^2)*Sin[c + d*x]))/((a - b)^3*(a + b)^3*(a + b*Sin[c + d*x
])^2))/(2*d)

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fricas [B]  time = 0.51, size = 895, normalized size = 4.39 \[ \left [-\frac {4 \, a^{7} - 12 \, a^{5} b^{2} + 12 \, a^{3} b^{4} - 4 \, a b^{6} + 2 \, {\left (16 \, a^{5} b^{2} - 17 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (2 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (2 \, a^{5} b + 5 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )\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 (2 \, a^{6} b - 6 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - 2 \, b^{7} - {\left (11 \, a^{4} b^{3} - 7 \, a^{2} b^{5} - 4 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{10} - 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} + 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )\right )}}, -\frac {2 \, a^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} + {\left (16 \, a^{5} b^{2} - 17 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (2 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (2 \, a^{5} b + 5 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )\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 (2 \, a^{6} b - 6 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - 2 \, b^{7} - {\left (11 \, a^{4} b^{3} - 7 \, a^{2} b^{5} - 4 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{10} - 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} + 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(4*a^7 - 12*a^5*b^2 + 12*a^3*b^4 - 4*a*b^6 + 2*(16*a^5*b^2 - 17*a^3*b^4 + a*b^6)*cos(d*x + c)^2 - 3*((2*
a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 - 2*(2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)*sin(d*x + c) - (2*a^5*b + 5*a^3*b^3
+ 3*a*b^5)*cos(d*x + c))*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^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 - (11*a^4*b^3 - 7*a^2*b^5 - 4*b^7)*cos(d*x + c)^2)
*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3 +
6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^
2*b^8 + b^10)*d*cos(d*x + c)), -1/2*(2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 + (16*a^5*b^2 - 17*a^3*b^4 + a*b^
6)*cos(d*x + c)^2 + 3*((2*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 - 2*(2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)*sin(d*x +
c) - (2*a^5*b + 5*a^3*b^3 + 3*a*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b
^2)*cos(d*x + c))) - (2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 - (11*a^4*b^3 - 7*a^2*b^5 - 4*b^7)*cos(d*x + c)^
2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3
+ 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*
a^2*b^8 + b^10)*d*cos(d*x + c))]

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giac [A]  time = 0.32, size = 365, normalized size = 1.79 \[ \frac {\frac {3 \, {\left (2 \, a^{3} b + 3 \, a b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3} - 3 \, a b^{2}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} + \frac {7 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 13 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 17 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{4} b^{2} + a^{2} b^{4}}{{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

(3*(2*a^3*b + 3*a*b^3)*(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^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + 2*(3*a^2*b*tan(1/2*d*x + 1/2*c) + b^3*tan(1/
2*d*x + 1/2*c) - a^3 - 3*a*b^2)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/2*d*x + 1/2*c)^2 - 1)) + (7*a^3*b^
3*tan(1/2*d*x + 1/2*c)^3 + 6*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 + 13*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 + 2*b^6*tan(1/
2*d*x + 1/2*c)^2 + 17*a^3*b^3*tan(1/2*d*x + 1/2*c) + 4*a*b^5*tan(1/2*d*x + 1/2*c) + 6*a^4*b^2 + a^2*b^4)/((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)^2))/d

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maple [B]  time = 0.51, size = 643, normalized size = 3.15 \[ -\frac {1}{d \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{d \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {7 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{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 )^{2}}+\frac {6 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{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 )^{2}}+\frac {13 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{4}}{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 )^{2}}+\frac {2 b^{6} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )^{2} a}+\frac {17 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{3}}{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 )^{2}}+\frac {4 b^{5} \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 )^{2}}+\frac {6 a^{3} b^{2}}{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 )^{2}}+\frac {b^{4} a}{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 )^{2}}+\frac {6 a^{3} b \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 {9 a \,b^{3} \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}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)/(a+b*sin(d*x+c))^3,x)

[Out]

-1/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)+1/d/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)+7/d*a^2/(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)^2*tan(1/2*d*x+1/2*c)^3*b^3+6/d*a^3/(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)^2*tan(1/2*d*x+1/2*c)^2*b^2+13/d*a/(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)^2*tan(1/2*d*x+1/2*c)^2*b^4+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)^2/a*tan(1/2*d*x+1/2*c)^2+17/d*a^2/(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)^2*t
an(1/2*d*x+1/2*c)*b^3+4/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)^2*tan(1/2*d*x+
1/2*c)+6/d*a^3/(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)^2*b^2+1/d*b^4/(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)^2*a+6/d*a^3/(a-b)^3/(a+b)^3*b/(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))+9/d*a/(a-b)^3/(a+b)^3*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))

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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)^2*sin(d*x+c)/(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 = 18.11, size = 624, normalized size = 3.06 \[ \frac {3\,a\,b\,\mathrm {atan}\left (\frac {\frac {3\,a\,b\,\left (2\,a^2+3\,b^2\right )\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{2\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+3\,b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}}{6\,a^3\,b+9\,a\,b^3}\right )\,\left (2\,a^2+3\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {2\,a^5+12\,a^3\,b^2+a\,b^4}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^4\,b+7\,a^2\,b^3+6\,b^5\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-2\,a^6+24\,a^4\,b^2+21\,a^2\,b^4+2\,b^6\right )}{a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^4+39\,a^2\,b^2+4\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^6-2\,a^4\,b^2+14\,a^2\,b^4+b^6\right )}{a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {3\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^2+3\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-a^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+4\,b^2\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)/(cos(c + d*x)^2*(a + b*sin(c + d*x))^3),x)

[Out]

(3*a*b*atan(((3*a*b*(2*a^2 + 3*b^2)*(2*a^6*b - 2*b^7 + 6*a^2*b^5 - 6*a^4*b^3))/(2*(a + b)^(7/2)*(a - b)^(7/2))
+ (3*a^2*b*tan(c/2 + (d*x)/2)*(2*a^2 + 3*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/((a + b)^(7/2)*(a - b)^(7/
2)))/(9*a*b^3 + 6*a^3*b))*(2*a^2 + 3*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2)) - ((a*b^4 + 2*a^5 + 12*a^3*b^2)/(a^
6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (2*tan(c/2 + (d*x)/2)^3*(2*a^4*b + 6*b^5 + 7*a^2*b^3))/(a^6 - b^6 + 3*a^2*b
^4 - 3*a^4*b^2) - (tan(c/2 + (d*x)/2)^4*(2*b^6 - 2*a^6 + 21*a^2*b^4 + 24*a^4*b^2))/(a*(a^6 - b^6 + 3*a^2*b^4 -
3*a^4*b^2)) + (b*tan(c/2 + (d*x)/2)*(2*a^4 + 4*b^4 + 39*a^2*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (2*ta
n(c/2 + (d*x)/2)^2*(2*a^6 + b^6 + 14*a^2*b^4 - 2*a^4*b^2))/(a*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (3*a^2*b*
tan(c/2 + (d*x)/2)^5*(2*a^2 + 3*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/(d*(a^2*tan(c/2 + (d*x)/2)^6 - a^2
- tan(c/2 + (d*x)/2)^2*(a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^4*(a^2 + 4*b^2) + 4*a*b*tan(c/2 + (d*x)/2)^5 - 4*a*b
*tan(c/2 + (d*x)/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)/(a+b*sin(d*x+c))**3,x)

[Out]

Integral(sin(c + d*x)*sec(c + d*x)**2/(a + b*sin(c + d*x))**3, x)

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