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

Optimal. Leaf size=61 \[ \frac{\sec ^4(c+d x) (a \sin (c+d x)+b)}{4 d}+\frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 a \tan (c+d x) \sec (c+d x)}{8 d} \]

[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)

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Rubi [A]  time = 0.0438092, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2668, 639, 199, 206} \[ \frac{\sec ^4(c+d x) (a \sin (c+d x)+b)}{4 d}+\frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 a \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[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 2668

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]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(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 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

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

Rubi steps

\begin{align*} \int \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx &=\frac{b^5 \operatorname{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 ) \operatorname{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) \operatorname{Subst}\left (\int \frac{1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac{3 a \tanh ^{-1}(\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]  time = 0.161547, size = 68, normalized size = 1.11 \[ \frac{a \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 a \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{8 d}+\frac{b \sec ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 74, normalized size = 1.2 \begin{align*}{\frac{a\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,a\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{b}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.964794, size = 105, normalized size = 1.72 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 2.27274, size = 219, normalized size = 3.59 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.14051, size = 95, normalized size = 1.56 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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