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

Optimal. Leaf size=73 \[ \frac{\tan ^3(c+d x) (4 a+3 b \sec (c+d x))}{12 d}-\frac{\tan (c+d x) (8 a+3 b \sec (c+d x))}{8 d}+a x+\frac{3 b \tanh ^{-1}(\sin (c+d x))}{8 d} \]

[Out]

a*x + (3*b*ArcTanh[Sin[c + d*x]])/(8*d) - ((8*a + 3*b*Sec[c + d*x])*Tan[c + d*x])/(8*d) + ((4*a + 3*b*Sec[c +
d*x])*Tan[c + d*x]^3)/(12*d)

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Rubi [A]  time = 0.065274, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3881, 3770} \[ \frac{\tan ^3(c+d x) (4 a+3 b \sec (c+d x))}{12 d}-\frac{\tan (c+d x) (8 a+3 b \sec (c+d x))}{8 d}+a x+\frac{3 b \tanh ^{-1}(\sin (c+d x))}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*x + (3*b*ArcTanh[Sin[c + d*x]])/(8*d) - ((8*a + 3*b*Sec[c + d*x])*Tan[c + d*x])/(8*d) + ((4*a + 3*b*Sec[c +
d*x])*Tan[c + d*x]^3)/(12*d)

Rule 3881

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(e*(e*Cot[
c + d*x])^(m - 1)*(a*m + b*(m - 1)*Csc[c + d*x]))/(d*m*(m - 1)), x] - Dist[e^2/m, Int[(e*Cot[c + d*x])^(m - 2)
*(a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]

Rule 3770

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

Rubi steps

\begin{align*} \int (a+b \sec (c+d x)) \tan ^4(c+d x) \, dx &=\frac{(4 a+3 b \sec (c+d x)) \tan ^3(c+d x)}{12 d}-\frac{1}{4} \int (4 a+3 b \sec (c+d x)) \tan ^2(c+d x) \, dx\\ &=-\frac{(8 a+3 b \sec (c+d x)) \tan (c+d x)}{8 d}+\frac{(4 a+3 b \sec (c+d x)) \tan ^3(c+d x)}{12 d}+\frac{1}{8} \int (8 a+3 b \sec (c+d x)) \, dx\\ &=a x-\frac{(8 a+3 b \sec (c+d x)) \tan (c+d x)}{8 d}+\frac{(4 a+3 b \sec (c+d x)) \tan ^3(c+d x)}{12 d}+\frac{1}{8} (3 b) \int \sec (c+d x) \, dx\\ &=a x+\frac{3 b \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{(8 a+3 b \sec (c+d x)) \tan (c+d x)}{8 d}+\frac{(4 a+3 b \sec (c+d x)) \tan ^3(c+d x)}{12 d}\\ \end{align*}

Mathematica [A]  time = 0.577929, size = 79, normalized size = 1.08 \[ \frac{\tan (c+d x) \sec ^3(c+d x) (-(32 a \cos (c+d x)+16 a \cos (3 (c+d x))+15 b \cos (2 (c+d x))+3 b))+48 a \tan ^{-1}(\tan (c+d x))+18 b \tanh ^{-1}(\sin (c+d x))}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*a*ArcTan[Tan[c + d*x]] + 18*b*ArcTanh[Sin[c + d*x]] - (3*b + 32*a*Cos[c + d*x] + 15*b*Cos[2*(c + d*x)] + 1
6*a*Cos[3*(c + d*x)])*Sec[c + d*x]^3*Tan[c + d*x])/(48*d)

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Maple [A]  time = 0.04, size = 127, normalized size = 1.7 \begin{align*}{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{a\tan \left ( dx+c \right ) }{d}}+ax+{\frac{ac}{d}}+{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{3\,\sin \left ( dx+c \right ) b}{8\,d}}+{\frac{3\,b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.47086, size = 138, normalized size = 1.89 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a + 3 \, b{\left (\frac{2 \,{\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*tan(d*x+c)^4,x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a + 3*b*(2*(5*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d
*x + c)^4 - 2*sin(d*x + c)^2 + 1) + 3*log(sin(d*x + c) + 1) - 3*log(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 0.923648, size = 302, normalized size = 4.14 \begin{align*} \frac{48 \, a d x \cos \left (d x + c\right )^{4} + 9 \, b \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, b \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (32 \, a \cos \left (d x + c\right )^{3} + 15 \, b \cos \left (d x + c\right )^{2} - 8 \, a \cos \left (d x + c\right ) - 6 \, b\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*tan(d*x+c)^4,x, algorithm="fricas")

[Out]

1/48*(48*a*d*x*cos(d*x + c)^4 + 9*b*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 9*b*cos(d*x + c)^4*log(-sin(d*x + c
) + 1) - 2*(32*a*cos(d*x + c)^3 + 15*b*cos(d*x + c)^2 - 8*a*cos(d*x + c) - 6*b)*sin(d*x + c))/(d*cos(d*x + c)^
4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (c + d x \right )}\right ) \tan ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*tan(d*x+c)**4,x)

[Out]

Integral((a + b*sec(c + d*x))*tan(c + d*x)**4, x)

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Giac [B]  time = 2.32488, size = 232, normalized size = 3.18 \begin{align*} \frac{24 \,{\left (d x + c\right )} a + 9 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (24 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 9 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 104 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 33 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 104 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 33 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*tan(d*x+c)^4,x, algorithm="giac")

[Out]

1/24*(24*(d*x + c)*a + 9*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 9*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(24
*a*tan(1/2*d*x + 1/2*c)^7 - 9*b*tan(1/2*d*x + 1/2*c)^7 - 104*a*tan(1/2*d*x + 1/2*c)^5 + 33*b*tan(1/2*d*x + 1/2
*c)^5 + 104*a*tan(1/2*d*x + 1/2*c)^3 + 33*b*tan(1/2*d*x + 1/2*c)^3 - 24*a*tan(1/2*d*x + 1/2*c) - 9*b*tan(1/2*d
*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d