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

Optimal. Leaf size=106 \[ \frac{b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{3 d}+\frac{a b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+a^4 x-\frac{a b^3 \tan (c+d x) \sec (c+d x)}{3 d}-\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.172336, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4042, 3918, 4048, 3770, 3767, 8} \[ \frac{b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{3 d}+\frac{a b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+a^4 x-\frac{a b^3 \tan (c+d x) \sec (c+d x)}{3 d}-\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4042

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

Rule 3918

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2*c
*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

Rule 3770

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

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 (a+b \sec (c+d x))^2 \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx &=-\int (-a+b \sec (c+d x)) (a+b \sec (c+d x))^3 \, dx\\ &=-\frac{b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac{1}{3} \int (a+b \sec (c+d x)) \left (-3 a^3-b \left (3 a^2-2 b^2\right ) \sec (c+d x)+2 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=-\frac{a b^3 \sec (c+d x) \tan (c+d x)}{3 d}-\frac{b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-6 a^4-6 a b \left (2 a^2-b^2\right ) \sec (c+d x)-2 b^2 \left (a^2-2 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^4 x-\frac{a b^3 \sec (c+d x) \tan (c+d x)}{3 d}-\frac{b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{3} \left (b^2 \left (a^2-2 b^2\right )\right ) \int \sec ^2(c+d x) \, dx+\left (a b \left (2 a^2-b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=a^4 x+\frac{a b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a b^3 \sec (c+d x) \tan (c+d x)}{3 d}-\frac{b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac{\left (b^2 \left (a^2-2 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=a^4 x+\frac{a b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{3 d}-\frac{a b^3 \sec (c+d x) \tan (c+d x)}{3 d}-\frac{b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.238483, size = 86, normalized size = 0.81 \[ \frac{2 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}+a^4 x-\frac{a b^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a b^3 \tan (c+d x) \sec (c+d x)}{d}-\frac{b^4 \left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 118, normalized size = 1.1 \begin{align*}{a}^{4}x+{\frac{{a}^{4}c}{d}}+2\,{\frac{{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{a{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}-{\frac{a{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{2\,{b}^{4}\tan \left ( dx+c \right ) }{3\,d}}-{\frac{{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.99441, size = 142, normalized size = 1.34 \begin{align*} \frac{6 \,{\left (d x + c\right )} a^{4} - 2 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b^{4} + 3 \, a b^{3}{\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 )} + 12 \, a^{3} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(d*x + c)*a^4 - 2*(tan(d*x + c)^3 + 3*tan(d*x + c))*b^4 + 3*a*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1)
- log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 12*a^3*b*log(sec(d*x + c) + tan(d*x + c)))/d

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Fricas [A]  time = 0.528129, size = 323, normalized size = 3.05 \begin{align*} \frac{6 \, a^{4} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \, b^{4} \cos \left (d x + c\right )^{2} + 3 \, a b^{3} \cos \left (d x + c\right ) + b^{4}\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((a+b*sec(d*x+c))^2*(a^2-b^2*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/6*(6*a^4*d*x*cos(d*x + c)^3 + 3*(2*a^3*b - a*b^3)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*a^3*b - a*b^3)
*cos(d*x + c)^3*log(-sin(d*x + c) + 1) - 2*(2*b^4*cos(d*x + c)^2 + 3*a*b^3*cos(d*x + c) + b^4)*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 ) \left (a + b \sec{\left (c + d x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a - b*sec(c + d*x))*(a + b*sec(c + d*x))**3, x)

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Giac [A]  time = 1.17904, size = 227, normalized size = 2.14 \begin{align*} \frac{3 \,{\left (d x + c\right )} a^{4} + 3 \,{\left (2 \, a^{3} b - a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, a^{3} b - a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, b^{4} \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((a+b*sec(d*x+c))^2*(a^2-b^2*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)*a^4 + 3*(2*a^3*b - a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*a^3*b - a*b^3)*log(abs(ta
n(1/2*d*x + 1/2*c) - 1)) - 2*(3*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*b^4*tan(1/2*d*x + 1/2*c)^5 + 2*b^4*tan(1/2*d*
x + 1/2*c)^3 - 3*a*b^3*tan(1/2*d*x + 1/2*c) - 3*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d