\(\int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [1353]

Optimal result

Integrand size = 27, antiderivative size = 142 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 a^3 b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}+\frac {a \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {a^2 \sec (c+d x) (a-b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {b \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d} \]

[Out]

-2*a^3*b*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)/d+1/3*a*sec(d*x+c)^3/(a^2-b^2)/d-a^2
*sec(d*x+c)*(a-b*sin(d*x+c))/(a^2-b^2)^2/d-1/3*b*tan(d*x+c)^3/(a^2-b^2)/d

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2981, 2686, 30, 2687, 2945, 12, 2739, 632, 210} \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b \tan ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac {a \sec ^3(c+d x)}{3 d \left (a^2-b^2\right )}-\frac {a^2 \sec (c+d x) (a-b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac {2 a^3 b \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}} \]

[In]

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

[Out]

(-2*a^3*b*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(5/2)*d) + (a*Sec[c + d*x]^3)/(3*(a^2
- b^2)*d) - (a^2*Sec[c + d*x]*(a - b*Sin[c + d*x]))/((a^2 - b^2)^2*d) - (b*Tan[c + d*x]^3)/(3*(a^2 - b^2)*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 210

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

Rule 632

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 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2739

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 2945

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]

Rule 2981

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[a*(d^2/(a^2 - b^2)), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[b*(d/(a^2 - b^2)), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[a^2*(d^2/(g^2*(a^2 - b^2
))), Int[(g*Cos[e + f*x])^(p + 2)*((d*Sin[e + f*x])^(n - 2)/(a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {a \int \sec ^3(c+d x) \tan (c+d x) \, dx}{a^2-b^2}-\frac {a^2 \int \frac {\sec (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx}{a^2-b^2}-\frac {b \int \sec ^2(c+d x) \tan ^2(c+d x) \, dx}{a^2-b^2} \\ & = -\frac {a^2 \sec (c+d x) (a-b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {a^2 \int \frac {a b}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}+\frac {a \text {Subst}\left (\int x^2 \, dx,x,\sec (c+d x)\right )}{\left (a^2-b^2\right ) d}-\frac {b \text {Subst}\left (\int x^2 \, dx,x,\tan (c+d x)\right )}{\left (a^2-b^2\right ) d} \\ & = \frac {a \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {a^2 \sec (c+d x) (a-b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {b \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {\left (a^3 b\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2} \\ & = \frac {a \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {a^2 \sec (c+d x) (a-b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {b \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {\left (2 a^3 b\right ) \text {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 )^2 d} \\ & = \frac {a \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {a^2 \sec (c+d x) (a-b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {b \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {\left (4 a^3 b\right ) \text {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 )^2 d} \\ & = -\frac {2 a^3 b \arctan \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 )^{5/2} d}+\frac {a \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {a^2 \sec (c+d x) (a-b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {b \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.30 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {48 a^3 b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {\sec ^3(c+d x) \left (-4 a^3-8 a b^2+3 a \left (5 a^2+b^2\right ) \cos (c+d x)-12 a^3 \cos (2 (c+d x))+5 a^3 \cos (3 (c+d x))+a b^2 \cos (3 (c+d x))+6 b^3 \sin (c+d x)+8 a^2 b \sin (3 (c+d x))-2 b^3 \sin (3 (c+d x))\right )}{(a-b)^2 (a+b)^2}}{24 d} \]

[In]

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

[Out]

((-48*a^3*b*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (Sec[c + d*x]^3*(-4*a^3 - 8*
a*b^2 + 3*a*(5*a^2 + b^2)*Cos[c + d*x] - 12*a^3*Cos[2*(c + d*x)] + 5*a^3*Cos[3*(c + d*x)] + a*b^2*Cos[3*(c + d
*x)] + 6*b^3*Sin[c + d*x] + 8*a^2*b*Sin[3*(c + d*x)] - 2*b^3*Sin[3*(c + d*x)]))/((a - b)^2*(a + b)^2))/(24*d)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.44

method	result	size
derivativedivides	\(\frac {-\frac {8}{\left (16 a -16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (16 a -16 b \right )}-\frac {a}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a^{3} b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {16}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (16 a +16 b \right )}-\frac {8}{\left (16 a +16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(205\)
default	\(\frac {-\frac {8}{\left (16 a -16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (16 a -16 b \right )}-\frac {a}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a^{3} b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {16}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (16 a +16 b \right )}-\frac {8}{\left (16 a +16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(205\)
risch	\(-\frac {2 i \left (-3 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-2 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-4 i b^{2} a \,{\mathrm e}^{3 i \left (d x +c \right )}-6 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}-6 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 a^{2} b +b^{3}\right )}{3 d \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {i b \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {i b \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(313\)

[In]

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

[Out]

1/d*(-8/(16*a-16*b)/(tan(1/2*d*x+1/2*c)+1)^2+16/3/(tan(1/2*d*x+1/2*c)+1)^3/(16*a-16*b)-1/2*a/(a-b)^2/(tan(1/2*
d*x+1/2*c)+1)-2*a^3*b/(a-b)^2/(a+b)^2/(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))
-16/3/(tan(1/2*d*x+1/2*c)-1)^3/(16*a+16*b)-8/(16*a+16*b)/(tan(1/2*d*x+1/2*c)-1)^2+1/2*a/(a+b)^2/(tan(1/2*d*x+1
/2*c)-1))

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 465, normalized size of antiderivative = 3.27 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} a^{3} b \cos \left (d x + c\right )^{3} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{5} + 4 \, a^{3} b^{2} - 2 \, a b^{4} + 6 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} - {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, \frac {3 \, \sqrt {a^{2} - b^{2}} a^{3} b \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + a^{5} - 2 \, a^{3} b^{2} + a b^{4} - 3 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} - {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3/(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 de

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.60 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{3} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{3} - a b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 17.48 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.61 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2\,\left (2\,a^3+a\,b^2\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,a^2\,b-2\,b^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {2\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^4-2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,a^3\,b\,\mathrm {atan}\left (\frac {\frac {a^3\,b\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {2\,a^4\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{2\,a^3\,b}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]

[In]

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

[Out]

((2*(a*b^2 + 2*a^3))/(3*(a^4 + b^4 - 2*a^2*b^2)) - (4*a^3*tan(c/2 + (d*x)/2)^2)/(a^4 + b^4 - 2*a^2*b^2) + (4*t
an(c/2 + (d*x)/2)^3*(5*a^2*b - 2*b^3))/(3*(a^4 + b^4 - 2*a^2*b^2)) - (2*a^2*b*tan(c/2 + (d*x)/2))/(a^4 + b^4 -
2*a^2*b^2) + (2*a*b^2*tan(c/2 + (d*x)/2)^4)/(a^4 + b^4 - 2*a^2*b^2) - (2*a^2*b*tan(c/2 + (d*x)/2)^5)/(a^4 + b
^4 - 2*a^2*b^2))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1)) - (2*a^3*b*a
tan(((a^3*b*(2*a^4*b + 2*b^5 - 4*a^2*b^3))/((a + b)^(5/2)*(a - b)^(5/2)) + (2*a^4*b*tan(c/2 + (d*x)/2)*(a^4 +
b^4 - 2*a^2*b^2))/((a + b)^(5/2)*(a - b)^(5/2)))/(2*a^3*b)))/(d*(a + b)^(5/2)*(a - b)^(5/2))