3.458 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=80 \[ \frac{\left (3 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac{a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a b \tan (c+d x) \sec (c+d x)}{d}+\frac{b^2 \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0901502, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3788, 3768, 3770, 4046, 3767, 8} \[ \frac{\left (3 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac{a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a b \tan (c+d x) \sec (c+d x)}{d}+\frac{b^2 \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

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 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.22306, size = 71, normalized size = 0.89 \[ \frac{a^2 \tan (c+d x)}{d}+\frac{a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a b \tan (c+d x) \sec (c+d x)}{d}+\frac{b^2 \left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.027, size = 89, normalized size = 1.1 \begin{align*}{\frac{{a}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{ab\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{2\,{b}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{b}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^2,x)

[Out]

a^2*tan(d*x+c)/d+a*b*sec(d*x+c)*tan(d*x+c)/d+1/d*a*b*ln(sec(d*x+c)+tan(d*x+c))+2/3*b^2*tan(d*x+c)/d+1/3*b^2*se
c(d*x+c)^2*tan(d*x+c)/d

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Maxima [A]  time = 1.25119, size = 113, normalized size = 1.41 \begin{align*} \frac{2 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b^{2} - 3 \, a b{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{2} \tan \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.74968, size = 259, normalized size = 3.24 \begin{align*} \frac{3 \, a b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \, a b \cos \left (d x + c\right ) +{\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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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 )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**2,x)

[Out]

Integral((a + b*sec(c + d*x))**2*sec(c + d*x)**2, x)

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Giac [B]  time = 1.37486, size = 240, normalized size = 3. \begin{align*} \frac{3 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, b^{2} \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 )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/3*(3*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*a^2*tan(1/2*d*
x + 1/2*c)^5 - 3*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*a^2*tan(1/2*d*x + 1/2*c)^3 - 2*
b^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*tan(1/2*d*x + 1/2*c) + 3*a*b*tan(1/2*d*x + 1/2*c) + 3*b^2*tan(1/2*d*x + 1/2
*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d