3.288 \(\int (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=86 \[ \frac{\left (2 a^2 B+4 a A b+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 A x+\frac{b (3 a B+2 A b) \tan (c+d x)}{2 d}+\frac{b B \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0807489, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3918, 3770, 3767, 8} \[ \frac{\left (2 a^2 B+4 a A b+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 A x+\frac{b (3 a B+2 A b) \tan (c+d x)}{2 d}+\frac{b B \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.262533, size = 67, normalized size = 0.78 \[ \frac{\left (2 a^2 B+4 a A b+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+2 a^2 A d x+b \tan (c+d x) (4 a B+2 A b+b B \sec (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.032, size = 133, normalized size = 1.6 \begin{align*}{a}^{2}Ax+{\frac{A{a}^{2}c}{d}}+{\frac{B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{Aab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{Bab\tan \left ( dx+c \right ) }{d}}+{\frac{A{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{B{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.965132, size = 170, normalized size = 1.98 \begin{align*} \frac{4 \,{\left (d x + c\right )} A a^{2} - B b^{2}{\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 )} + 4 \, B a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 8 \, A a b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 8 \, B a b \tan \left (d x + c\right ) + 4 \, A b^{2} \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(d*x + c)*A*a^2 - B*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c)
- 1)) + 4*B*a^2*log(sec(d*x + c) + tan(d*x + c)) + 8*A*a*b*log(sec(d*x + c) + tan(d*x + c)) + 8*B*a*b*tan(d*x
+ c) + 4*A*b^2*tan(d*x + c))/d

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Fricas [A]  time = 0.506255, size = 335, normalized size = 3.9 \begin{align*} \frac{4 \, A a^{2} d x \cos \left (d x + c\right )^{2} +{\left (2 \, B a^{2} + 4 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, B a^{2} + 4 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (B b^{2} + 2 \,{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*A*a^2*d*x*cos(d*x + c)^2 + (2*B*a^2 + 4*A*a*b + B*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*B*a^2
+ 4*A*a*b + B*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(B*b^2 + 2*(2*B*a*b + A*b^2)*cos(d*x + c))*sin(d*
x + c))/(d*cos(d*x + c)^2)

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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 )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.20235, size = 259, normalized size = 3.01 \begin{align*} \frac{2 \,{\left (d x + c\right )} A a^{2} +{\left (2 \, B a^{2} + 4 \, A a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (2 \, B a^{2} + 4 \, A a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (4 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B 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 )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(d*x + c)*A*a^2 + (2*B*a^2 + 4*A*a*b + B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*B*a^2 + 4*A*a*b +
B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(4*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 2*A*b^2*tan(1/2*d*x + 1/2*c)^
3 - B*b^2*tan(1/2*d*x + 1/2*c)^3 - 4*B*a*b*tan(1/2*d*x + 1/2*c) - 2*A*b^2*tan(1/2*d*x + 1/2*c) - B*b^2*tan(1/2
*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d