3.1484 \(\int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=103 \[ \frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac{3 a \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b \tan ^4(c+d x)}{4 d}-\frac{b \tan ^2(c+d x)}{2 d}-\frac{b \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.120409, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2834, 2611, 3770, 3473, 3475} \[ \frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac{3 a \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b \tan ^4(c+d x)}{4 d}-\frac{b \tan ^2(c+d x)}{2 d}-\frac{b \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + b*Sin[c + d*x])*Tan[c + d*x]^4,x]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) - (b*Log[Cos[c + d*x]])/d - (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (b*Tan[c
+ d*x]^2)/(2*d) + (a*Sec[c + d*x]*Tan[c + d*x]^3)/(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 3770

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

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3475

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

Rubi steps

\begin{align*} \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx &=a \int \sec (c+d x) \tan ^4(c+d x) \, dx+b \int \tan ^5(c+d x) \, dx\\ &=\frac{a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac{b \tan ^4(c+d x)}{4 d}-\frac{1}{4} (3 a) \int \sec (c+d x) \tan ^2(c+d x) \, dx-b \int \tan ^3(c+d x) \, dx\\ &=-\frac{3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac{b \tan ^2(c+d x)}{2 d}+\frac{a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac{b \tan ^4(c+d x)}{4 d}+\frac{1}{8} (3 a) \int \sec (c+d x) \, dx+b \int \tan (c+d x) \, dx\\ &=\frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{b \log (\cos (c+d x))}{d}-\frac{3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac{b \tan ^2(c+d x)}{2 d}+\frac{a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac{b \tan ^4(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.309409, size = 106, normalized size = 1.03 \[ \frac{a \tan ^3(c+d x) \sec (c+d x)}{d}-\frac{a \left (6 \tan (c+d x) \sec ^3(c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{8 d}-\frac{b \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + b*Sin[c + d*x])*Tan[c + d*x]^4,x]

[Out]

(a*Sec[c + d*x]*Tan[c + d*x]^3)/d - (b*(4*Log[Cos[c + d*x]] + 2*Tan[c + d*x]^2 - Tan[c + d*x]^4))/(4*d) - (a*(
6*Sec[c + d*x]^3*Tan[c + d*x] - 3*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*Tan[c + d*x])))/(8*d)

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Maple [A]  time = 0.049, size = 133, normalized size = 1.3 \begin{align*}{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{3\,a\sin \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 \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{b\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.00186, size = 135, normalized size = 1.31 \begin{align*} \frac{{\left (3 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac{2 \,{\left (5 \, a \sin \left (d x + c\right )^{3} + 8 \, b \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right ) - 6 \, 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)^4*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/16*((3*a - 8*b)*log(sin(d*x + c) + 1) - (3*a + 8*b)*log(sin(d*x + c) - 1) + 2*(5*a*sin(d*x + c)^3 + 8*b*sin(
d*x + c)^2 - 3*a*sin(d*x + c) - 6*b)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

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Fricas [A]  time = 1.98223, size = 270, normalized size = 2.62 \begin{align*} \frac{{\left (3 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 16 \, b \cos \left (d x + c\right )^{2} - 2 \,{\left (5 \, 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)^4*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.2443, size = 135, normalized size = 1.31 \begin{align*} \frac{{\left (3 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (3 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (6 \, b \sin \left (d x + c\right )^{4} + 5 \, a \sin \left (d x + c\right )^{3} - 4 \, b \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right )\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)^4*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*((3*a - 8*b)*log(abs(sin(d*x + c) + 1)) - (3*a + 8*b)*log(abs(sin(d*x + c) - 1)) + 2*(6*b*sin(d*x + c)^4
+ 5*a*sin(d*x + c)^3 - 4*b*sin(d*x + c)^2 - 3*a*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d