\(\int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx\) [1504]

Optimal result

Integrand size = 29, antiderivative size = 144 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (a^3-9 a b^2+8 b^3\right ) \log (1+\sin (c+d x))}{16 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d} \]

[Out]

1/16*(a^3-9*a*b^2-8*b^3)*ln(1-sin(d*x+c))/d-1/16*(a^3-9*a*b^2+8*b^3)*ln(1+sin(d*x+c))/d-1/8*sec(d*x+c)^2*(a+b*
sin(d*x+c))*(5*a*b+(a^2+4*b^2)*sin(d*x+c))/d+1/4*sec(d*x+c)^3*(a+b*sin(d*x+c))^3*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2916, 12, 1659, 833, 647, 31} \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (a^3-9 a b^2+8 b^3\right ) \log (\sin (c+d x)+1)}{16 d}-\frac {\sec ^2(c+d x) \left (\left (a^2+4 b^2\right ) \sin (c+d x)+5 a b\right ) (a+b \sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d} \]

[In]

Int[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^3*Tan[c + d*x]^2,x]

[Out]

((a^3 - 9*a*b^2 - 8*b^3)*Log[1 - Sin[c + d*x]])/(16*d) - ((a^3 - 9*a*b^2 + 8*b^3)*Log[1 + Sin[c + d*x]])/(16*d
) - (Sec[c + d*x]^2*(a + b*Sin[c + d*x])*(5*a*b + (a^2 + 4*b^2)*Sin[c + d*x]))/(8*d) + (Sec[c + d*x]^3*(a + b*
Sin[c + d*x])^3*Tan[c + d*x])/(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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
- 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 1659

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, Simp[(d + e*x)^m*(a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c*(p + 1))), x] + Dist[1/(2*a*c*(p
+ 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*c*(p + 1)*(d + e*x)*Q - a*e*g*m + c*d*f*(2*p
+ 3) + c*e*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 2916

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

Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {x^2 (a+x)^3}{b^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^3 \text {Subst}\left (\int \frac {x^2 (a+x)^3}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac {b \text {Subst}\left (\int \frac {(a+x)^2 \left (-a b^2-4 b^2 x\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = -\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\text {Subst}\left (\int \frac {a b^2 \left (a^2-9 b^2\right )-8 b^4 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b d} \\ & = -\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\left (a^3-9 a b^2-8 b^3\right ) \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {\left (a^3-9 a b^2+8 b^3\right ) \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d} \\ & = \frac {\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (a^3-9 a b^2+8 b^3\right ) \log (1+\sin (c+d x))}{16 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.97 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))-\left (a^3-9 a b^2+8 b^3\right ) \log (1+\sin (c+d x))+\frac {(a+b)^3}{(-1+\sin (c+d x))^2}+\frac {(a+b)^2 (a+7 b)}{-1+\sin (c+d x)}-\frac {(a-b)^3}{(1+\sin (c+d x))^2}+\frac {(a-7 b) (a-b)^2}{1+\sin (c+d x)}}{16 d} \]

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^3*Tan[c + d*x]^2,x]

[Out]

((a^3 - 9*a*b^2 - 8*b^3)*Log[1 - Sin[c + d*x]] - (a^3 - 9*a*b^2 + 8*b^3)*Log[1 + Sin[c + d*x]] + (a + b)^3/(-1
+ Sin[c + d*x])^2 + ((a + b)^2*(a + 7*b))/(-1 + Sin[c + d*x]) - (a - b)^3/(1 + Sin[c + d*x])^2 + ((a - 7*b)*(
a - b)^2)/(1 + Sin[c + d*x]))/(16*d)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.42

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{2} b \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{3} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(204\)
default	\(\frac {a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{2} b \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{3} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(204\)
parallelrisch	\(\frac {16 b^{3} \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \left (a^{2}-a b -8 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-2 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right ) \left (a^{2}+a b -8 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (-3 a^{2} b -b^{3}\right ) \cos \left (2 d x +2 c \right )+3 \left (a^{2} b +b^{3}\right ) \cos \left (4 d x +4 c \right )+\left (-a^{3}-15 a \,b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (7 a^{3}+9 a \,b^{2}\right ) \sin \left (d x +c \right )+9 a^{2} b +b^{3}}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(268\)
risch	\(i x \,b^{3}+\frac {2 i b^{3} c}{d}-\frac {{\mathrm e}^{i \left (d x +c \right )} \left (-i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-15 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+7 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+9 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+24 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-7 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-9 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+i a^{3}+15 i a \,b^{2}+24 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{3}}{8 d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{d}\)	\(373\)
norman	\(\frac {\frac {\left (12 a^{2} b +2 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (12 a^{2} b +2 b^{3}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (36 a^{2} b +16 b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (36 a^{2} b +16 b^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (11 a^{2}+45 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (a^{2}-9 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (a^{2}-9 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (5 a^{2}+3 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a \left (5 a^{2}+3 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a \left (31 a^{2}+105 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (31 a^{2}+105 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {b^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (a^{3}-9 a \,b^{2}-8 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}-\frac {\left (a^{3}-9 a \,b^{2}+8 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(454\)

[In]

int(sec(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^3*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8*sin(d*x+c)-1/8*ln(sec(d*x+c)+tan(d*x
+c)))+3/4*a^2*b*sin(d*x+c)^4/cos(d*x+c)^4+3*a*b^2*(1/4*sin(d*x+c)^5/cos(d*x+c)^4-1/8*sin(d*x+c)^5/cos(d*x+c)^2
-1/8*sin(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+b^3*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(
d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 12 \, a^{2} b - 4 \, b^{3} + 8 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{3} + 6 \, a b^{2} - {\left (a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/16*((a^3 - 9*a*b^2 + 8*b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (a^3 - 9*a*b^2 - 8*b^3)*cos(d*x + c)^4*l
og(-sin(d*x + c) + 1) - 12*a^2*b - 4*b^3 + 8*(3*a^2*b + 2*b^3)*cos(d*x + c)^2 - 2*(2*a^3 + 6*a*b^2 - (a^3 + 15
*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**2*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.05 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (a^{3} + 15 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{2} b - 6 \, b^{3} + 4 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/16*((a^3 - 9*a*b^2 + 8*b^3)*log(sin(d*x + c) + 1) - (a^3 - 9*a*b^2 - 8*b^3)*log(sin(d*x + c) - 1) - 2*((a^3
+ 15*a*b^2)*sin(d*x + c)^3 - 6*a^2*b - 6*b^3 + 4*(3*a^2*b + 2*b^3)*sin(d*x + c)^2 + (a^3 - 9*a*b^2)*sin(d*x +
c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.17 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, b^{3} \sin \left (d x + c\right )^{4} + a^{3} \sin \left (d x + c\right )^{3} + 15 \, a b^{2} \sin \left (d x + c\right )^{3} + 12 \, a^{2} b \sin \left (d x + c\right )^{2} - 4 \, b^{3} \sin \left (d x + c\right )^{2} + a^{3} \sin \left (d x + c\right ) - 9 \, a b^{2} \sin \left (d x + c\right ) - 6 \, a^{2} b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/16*((a^3 - 9*a*b^2 + 8*b^3)*log(abs(sin(d*x + c) + 1)) - (a^3 - 9*a*b^2 - 8*b^3)*log(abs(sin(d*x + c) - 1))
- 2*(6*b^3*sin(d*x + c)^4 + a^3*sin(d*x + c)^3 + 15*a*b^2*sin(d*x + c)^3 + 12*a^2*b*sin(d*x + c)^2 - 4*b^3*si
n(d*x + c)^2 + a^3*sin(d*x + c) - 9*a*b^2*sin(d*x + c) - 6*a^2*b)/(sin(d*x + c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 11.82 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.08 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {b^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (-\frac {a^3}{8}+\frac {9\,a\,b^2}{8}+b^3\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {a^3}{8}-\frac {9\,a\,b^2}{8}+b^3\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {9\,a\,b^2}{4}-\frac {a^3}{4}\right )+2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {9\,a\,b^2}{4}-\frac {a^3}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {7\,a^3}{4}+\frac {33\,a\,b^2}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {7\,a^3}{4}+\frac {33\,a\,b^2}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,a^2\,b+8\,b^3\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

int((sin(c + d*x)^2*(a + b*sin(c + d*x))^3)/cos(c + d*x)^5,x)

[Out]

(b^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (log(tan(c/2 + (d*x)/2) - 1)*((9*a*b^2)/8 - a^3/8 + b^3))/d - (log(tan
(c/2 + (d*x)/2) + 1)*(a^3/8 - (9*a*b^2)/8 + b^3))/d - (tan(c/2 + (d*x)/2)*((9*a*b^2)/4 - a^3/4) + 2*b^3*tan(c/
2 + (d*x)/2)^2 + 2*b^3*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^7*((9*a*b^2)/4 - a^3/4) - tan(c/2 + (d*x)/2)^
3*((33*a*b^2)/4 + (7*a^3)/4) - tan(c/2 + (d*x)/2)^5*((33*a*b^2)/4 + (7*a^3)/4) - tan(c/2 + (d*x)/2)^4*(12*a^2*
b + 8*b^3))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^
8 + 1))