\(\int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\) [391]

Optimal result

Integrand size = 21, antiderivative size = 59 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\sec ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2747, 737, 212} \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{2 d} \]

[In]

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

[Out]

((a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + (Sec[c + d*x]^2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x]))/(2*d)

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 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
+ c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[(2*p + 3)*((c*d^2 + a*e^2)/(2*a*c*(p + 1))), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.92 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (1+\sin (c+d x)))+2 a^3 b \sec ^2(c+d x)-2 \left (a^4-b^4\right ) \sec (c+d x) \tan (c+d x)+\left (-6 a^3 b+4 a b^3\right ) \tan ^2(c+d x)}{4 \left (-a^2+b^2\right ) d} \]

[In]

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

[Out]

((a^2 - b^2)^2*(Log[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]]) + 2*a^3*b*Sec[c + d*x]^2 - 2*(a^4 - b^4)*Sec[c
+ d*x]*Tan[c + d*x] + (-6*a^3*b + 4*a*b^3)*Tan[c + d*x]^2)/(4*(-a^2 + b^2)*d)

Maple [A] (verified)

Time = 1.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.68

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a b}{\cos \left (d x +c \right )^{2}}+b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(99\)
default	\(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a b}{\cos \left (d x +c \right )^{2}}+b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(99\)
parallelrisch	\(\frac {-\left (a -b \right ) \left (a +b \right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a -b \right ) \left (a +b \right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 a b \cos \left (2 d x +2 c \right )+\left (2 a^{2}+2 b^{2}\right ) \sin \left (d x +c \right )+2 a b}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\)	\(120\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-a^{2}-b^{2}+4 i a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{2}}-\frac {a^{2} \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d}+\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) b^{2}}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{2 d}\)	\(171\)
norman	\(\frac {\frac {\left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (a^{2}+b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (a^{2}+b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (a^{2}+b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(228\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.53 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a b + 2 \, {\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.32 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (2 \, a b + {\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

[In]

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

[Out]

1/4*((a^2 - b^2)*log(sin(d*x + c) + 1) - (a^2 - b^2)*log(sin(d*x + c) - 1) - 2*(2*a*b + (a^2 + b^2)*sin(d*x +
c))/(sin(d*x + c)^2 - 1))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.46 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (a^{2} \sin \left (d x + c\right ) + b^{2} \sin \left (d x + c\right ) + 2 \, a b\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.05 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )}{d}-\frac {a\,b+\sin \left (c+d\,x\right )\,\left (\frac {a^2}{2}+\frac {b^2}{2}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )} \]

[In]

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

[Out]

(atanh(sin(c + d*x))*(a^2/2 - b^2/2))/d - (a*b + sin(c + d*x)*(a^2/2 + b^2/2))/(d*(sin(c + d*x)^2 - 1))