\(\int \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\) [381]

Optimal result

Integrand size = 19, antiderivative size = 61 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) (b+a \sin (c+d x))}{4 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2747, 653, 205, 212} \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) (a \sin (c+d x)+b)}{4 d}+\frac {3 a \tan (c+d x) \sec (c+d x)}{8 d} \]

[In]

Int[Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + (Sec[c + d*x]^4*(b + a*Sin[c + d*x]))/(4*d) + (3*a*Sec[c + d*x]*Tan[c + d*
x])/(8*d)

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 653

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.21 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b \sec ^4(c+d x)}{4 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[In]

Integrate[Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + (b*Sec[c + d*x]^4)/(4*d) + (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*Sec[
c + d*x]^3*Tan[c + d*x])/(4*d)

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(\frac {a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(63\)
default	\(\frac {a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(63\)
risch	\(\frac {-3 i a \,{\mathrm e}^{7 i \left (d x +c \right )}-11 i a \,{\mathrm e}^{5 i \left (d x +c \right )}+11 i a \,{\mathrm e}^{3 i \left (d x +c \right )}+16 b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i a \,{\mathrm e}^{i \left (d x +c \right )}}{4 d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{4}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 a \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{8 d}\)	\(123\)
parallelrisch	\(\frac {-6 a \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 a \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+11 a \sin \left (d x +c \right )+3 a \sin \left (3 d x +3 c \right )-4 b \cos \left (2 d x +2 c \right )-b \cos \left (4 d x +4 c \right )+5 b}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(153\)
norman	\(\frac {\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {2 b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(221\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.34 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 \, a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right ) + 4 \, b}{16 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.28 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, a \sin \left (d x + c\right )^{3} - 5 \, a \sin \left (d x + c\right ) - 2 \, b\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.60 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a \sin \left (d x + c\right )^{3} - 5 \, a \sin \left (d x + c\right ) - 2 \, b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.80 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3\,a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{8\,d}+\frac {-\frac {3\,a\,{\sin \left (c+d\,x\right )}^3}{8}+\frac {5\,a\,\sin \left (c+d\,x\right )}{8}+\frac {b}{4}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )} \]

[In]

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

[Out]

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