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

Optimal. Leaf size=55 \[ \frac{\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}+\frac{a \log (\cos (c+d x))}{d}-\frac{2 b \sec (c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0661531, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3881, 3884, 3475, 2606, 8} \[ \frac{\tan ^2(c+d x) (3 a+2 b \sec (c+d x))}{6 d}+\frac{a \log (\cos (c+d x))}{d}-\frac{2 b \sec (c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Log[Cos[c + d*x]])/d - (2*b*Sec[c + d*x])/(3*d) + ((3*a + 2*b*Sec[c + d*x])*Tan[c + d*x]^2)/(6*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 3884

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

Rule 3475

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

Rule 2606

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (a+b \sec (c+d x)) \tan ^3(c+d x) \, dx &=\frac{(3 a+2 b \sec (c+d x)) \tan ^2(c+d x)}{6 d}-\frac{1}{3} \int (3 a+2 b \sec (c+d x)) \tan (c+d x) \, dx\\ &=\frac{(3 a+2 b \sec (c+d x)) \tan ^2(c+d x)}{6 d}-a \int \tan (c+d x) \, dx-\frac{1}{3} (2 b) \int \sec (c+d x) \tan (c+d x) \, dx\\ &=\frac{a \log (\cos (c+d x))}{d}+\frac{(3 a+2 b \sec (c+d x)) \tan ^2(c+d x)}{6 d}-\frac{(2 b) \operatorname{Subst}(\int 1 \, dx,x,\sec (c+d x))}{3 d}\\ &=\frac{a \log (\cos (c+d x))}{d}-\frac{2 b \sec (c+d x)}{3 d}+\frac{(3 a+2 b \sec (c+d x)) \tan ^2(c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.155668, size = 55, normalized size = 1. \[ \frac{a \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d}+\frac{b \sec ^3(c+d x)}{3 d}-\frac{b \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.037, size = 104, normalized size = 1.9 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{2}a}{2\,d}}+{\frac{a\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d\cos \left ( dx+c \right ) }}-{\frac{b\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{2\,b\cos \left ( dx+c \right ) }{3\,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)^3,x)

[Out]

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

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Maxima [A]  time = 0.982901, size = 68, normalized size = 1.24 \begin{align*} \frac{6 \, a \log \left (\cos \left (d x + c\right )\right ) - \frac{6 \, b \cos \left (d x + c\right )^{2} - 3 \, a \cos \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{3}}}{6 \, 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)^3,x, algorithm="maxima")

[Out]

1/6*(6*a*log(cos(d*x + c)) - (6*b*cos(d*x + c)^2 - 3*a*cos(d*x + c) - 2*b)/cos(d*x + c)^3)/d

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Fricas [A]  time = 1.09481, size = 149, normalized size = 2.71 \begin{align*} \frac{6 \, a \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) - 6 \, b \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + 2 \, b}{6 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

1/6*(6*a*cos(d*x + c)^3*log(-cos(d*x + c)) - 6*b*cos(d*x + c)^2 + 3*a*cos(d*x + c) + 2*b)/(d*cos(d*x + c)^3)

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Sympy [A]  time = 1.27028, size = 76, normalized size = 1.38 \begin{align*} \begin{cases} - \frac{a \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{a \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac{b \tan ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}}{3 d} - \frac{2 b \sec{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (a + b \sec{\left (c \right )}\right ) \tan ^{3}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*tan(d*x+c)**3,x)

[Out]

Piecewise((-a*log(tan(c + d*x)**2 + 1)/(2*d) + a*tan(c + d*x)**2/(2*d) + b*tan(c + d*x)**2*sec(c + d*x)/(3*d)
- 2*b*sec(c + d*x)/(3*d), Ne(d, 0)), (x*(a + b*sec(c))*tan(c)**3, True))

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Giac [B]  time = 1.75327, size = 242, normalized size = 4.4 \begin{align*} -\frac{6 \, a \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 6 \, a \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{11 \, a + 8 \, b + \frac{45 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{24 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{45 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{11 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}}{6 \, 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)^3,x, algorithm="giac")

[Out]

-1/6*(6*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 6*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) - 1)) + (11*a + 8*b + 45*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 24*b*(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) + 45*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 11*a*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)/((cos
(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^3)/d