3.1486 \(\int \sec ^3(c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=74 \[ -\frac{a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac{a \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b \tan ^4(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.139193, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2834, 2611, 3768, 3770, 2607, 30} \[ -\frac{a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac{a \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b \tan ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2834

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

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0246122, size = 74, normalized size = 1. \[ -\frac{a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac{a \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b \tan ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 100, normalized size = 1.4 \begin{align*}{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{a\sin \left ( dx+c \right ) }{8\,d}}-{\frac{a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{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*sin(d*x+c)^2*(a+b*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.965348, size = 116, normalized size = 1.57 \begin{align*} -\frac{a \log \left (\sin \left (d x + c\right ) + 1\right ) - a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (a \sin \left (d x + c\right )^{3} + 4 \, b \sin \left (d x + c\right )^{2} + 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*sin(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*(a*log(sin(d*x + c) + 1) - a*log(sin(d*x + c) - 1) - 2*(a*sin(d*x + c)^3 + 4*b*sin(d*x + c)^2 + 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 = 2.02741, size = 240, normalized size = 3.24 \begin{align*} -\frac{a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, b \cos \left (d x + c\right )^{2} + 2 \,{\left (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*sin(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.26667, size = 105, normalized size = 1.42 \begin{align*} -\frac{a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (a \sin \left (d x + c\right )^{3} + 4 \, b \sin \left (d x + c\right )^{2} + 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*sin(d*x+c)^2*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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