\(\int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx\) [1496]

Optimal result

Integrand size = 29, antiderivative size = 93 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {\left (a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}-\frac {\sec ^2(c+d x) \left (4 a b+\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2916, 12, 1659, 792, 212} \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {\left (a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}-\frac {\sec ^2(c+d x) \left (\left (a^2+3 b^2\right ) \sin (c+d x)+4 a b\right )}{8 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^2}{4 d} \]

[In]

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

[Out]

-1/8*((a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]])/d - (Sec[c + d*x]^2*(4*a*b + (a^2 + 3*b^2)*Sin[c + d*x]))/(8*d) + (
Sec[c + d*x]^3*(a + b*Sin[c + d*x])^2*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 212

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

Rule 792

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

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)^2}{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)^2}{\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))^2 \tan (c+d x)}{4 d}+\frac {b \text {Subst}\left (\int \frac {(a+x) \left (-a b^2-3 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) \left (4 a b+\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}-\frac {\left (b \left (a^2-3 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = -\frac {\left (a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}-\frac {\sec ^2(c+d x) \left (4 a b+\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.91 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {\left (a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {1}{4} \sec ^4(c+d x) \left (16 a b \cos (2 (c+d x))+2 \left (-3 a^2+b^2+\left (a^2+5 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{8 d} \]

[In]

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

[Out]

-1/8*((a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]] + (Sec[c + d*x]^4*(16*a*b*Cos[2*(c + d*x)] + 2*(-3*a^2 + b^2 + (a^2
+ 5*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/4)/d

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.78

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a b \left (\sin ^{4}\left (d x +c \right )\right )}{2 \cos \left (d x +c \right )^{4}}+b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(166\)
default	\(\frac {a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a b \left (\sin ^{4}\left (d x +c \right )\right )}{2 \cos \left (d x +c \right )^{4}}+b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(166\)
parallelrisch	\(\frac {2 \left (a^{2}-3 b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-2 \left (a^{2}-3 b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-8 a b \cos \left (2 d x +2 c \right )+2 \cos \left (4 d x +4 c \right ) a b +\left (-a^{2}-5 b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (7 a^{2}+3 b^{2}\right ) \sin \left (d x +c \right )+6 a b}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(190\)
risch	\(\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+5 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-7 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+16 i a b \,{\mathrm e}^{5 i \left (d x +c \right )}+7 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-a^{2}-5 b^{2}+16 i a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d}\)	\(252\)
norman	\(\frac {\frac {8 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (a^{2}-3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (a^{2}-3 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (9 a^{2}+5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (9 a^{2}+5 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (11 a^{2}+15 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (11 a^{2}+15 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {16 a b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (a^{2}-3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}-\frac {\left (a^{2}-3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(296\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.33 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 16 \, a b \cos \left (d x + c\right )^{2} - 8 \, a b + 2 \, {\left ({\left (a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

-1/16*((a^2 - 3*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (a^2 - 3*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1
) + 16*a*b*cos(d*x + c)^2 - 8*a*b + 2*((a^2 + 5*b^2)*cos(d*x + c)^2 - 2*a^2 - 2*b^2)*sin(d*x + c))/(d*cos(d*x
+ c)^4)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.29 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{2} - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (8 \, a b \sin \left (d x + c\right )^{2} + {\left (a^{2} + 5 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 4 \, a b + {\left (a^{2} - 3 \, b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

[In]

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

[Out]

-1/16*((a^2 - 3*b^2)*log(sin(d*x + c) + 1) - (a^2 - 3*b^2)*log(sin(d*x + c) - 1) - 2*(8*a*b*sin(d*x + c)^2 + (
a^2 + 5*b^2)*sin(d*x + c)^3 - 4*a*b + (a^2 - 3*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 = 0.50 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.33 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (a^{2} \sin \left (d x + c\right )^{3} + 5 \, b^{2} \sin \left (d x + c\right )^{3} + 8 \, a b \sin \left (d x + c\right )^{2} + a^{2} \sin \left (d x + c\right ) - 3 \, b^{2} \sin \left (d x + c\right ) - 4 \, a b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

-1/16*((a^2 - 3*b^2)*log(abs(sin(d*x + c) + 1)) - (a^2 - 3*b^2)*log(abs(sin(d*x + c) - 1)) - 2*(a^2*sin(d*x +
c)^3 + 5*b^2*sin(d*x + c)^3 + 8*a*b*sin(d*x + c)^2 + a^2*sin(d*x + c) - 3*b^2*sin(d*x + c) - 4*a*b)/(sin(d*x +
c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 17.18 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.05 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {\left (\frac {a^2}{4}-\frac {3\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {7\,a^2}{4}+\frac {11\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {7\,a^2}{4}+\frac {11\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {a^2}{4}-\frac {3\,b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {a^2}{4}-\frac {3\,b^2}{4}\right )}{d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^7*(a^2/4 - (3*b^2)/4) + tan(c/2 + (d*x)/2)^3*((7*a^2)/4 + (11*b^2)/4) + tan(c/2 + (d*x)/2)
^5*((7*a^2)/4 + (11*b^2)/4) + tan(c/2 + (d*x)/2)*(a^2/4 - (3*b^2)/4) + 8*a*b*tan(c/2 + (d*x)/2)^4)/(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)) - (atanh(tan(c/
2 + (d*x)/2))*(a^2/4 - (3*b^2)/4))/d