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

Optimal. Leaf size=82 \[ \frac{a^2 (2 A+3 B) \tan (c+d x)}{2 d}+\frac{a^2 (4 A+3 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 A x+\frac{B \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0836368, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3917, 3914, 3767, 8, 3770} \[ \frac{a^2 (2 A+3 B) \tan (c+d x)}{2 d}+\frac{a^2 (4 A+3 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 A x+\frac{B \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3917

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 - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 3914

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

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]

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

Mathematica [B]  time = 1.25495, size = 307, normalized size = 3.74 \[ \frac{a^2 \cos ^3(c+d x) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 (A+B \sec (c+d x)) \left (\frac{4 (A+2 B) \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 (A+2 B) \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{2 (4 A+3 B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{2 (4 A+3 B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+4 A x+\frac{B}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{B}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{16 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cos[c + d*x]^3*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(A + B*Sec[c + d*x])*(4*A*x - (2*(4*A + 3*B)*Log[C
os[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(4*A + 3*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + B/(d*(Co
s[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (4*(A + 2*B)*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2]
- Sin[(c + d*x)/2])) - B/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(A + 2*B)*Sin[(d*x)/2])/(d*(Cos[c/2]
+ Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(16*(B + A*Cos[c + d*x]))

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Maple [A]  time = 0.038, size = 113, normalized size = 1.4 \begin{align*}{a}^{2}Ax+{\frac{A{a}^{2}c}{d}}+{\frac{3\,B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{B{a}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}A\tan \left ( dx+c \right ) }{d}}+{\frac{B{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.979652, size = 173, normalized size = 2.11 \begin{align*} \frac{4 \,{\left (d x + c\right )} A a^{2} - B a^{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 )} + 8 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, B a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, A a^{2} \tan \left (d x + c\right ) + 8 \, B a^{2} \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 0.496223, size = 297, normalized size = 3.62 \begin{align*} \frac{4 \, A a^{2} d x \cos \left (d x + c\right )^{2} +{\left (4 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (4 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) + B a^{2}\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+a*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 + (4*A + 3*B)*a^2*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (4*A + 3*B)*a^2*cos(d
*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*(A + 2*B)*a^2*cos(d*x + c) + B*a^2)*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*} a^{2} \left (\int A\, dx + \int 2 A \sec{\left (c + d x \right )}\, dx + \int A \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sec{\left (c + d x \right )}\, dx + \int 2 B \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sec ^{3}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A, x) + Integral(2*A*sec(c + d*x), x) + Integral(A*sec(c + d*x)**2, x) + Integral(B*sec(c + d*x
), x) + Integral(2*B*sec(c + d*x)**2, x) + Integral(B*sec(c + d*x)**3, x))

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Giac [B]  time = 1.4353, size = 208, normalized size = 2.54 \begin{align*} \frac{2 \,{\left (d x + c\right )} A a^{2} +{\left (4 \, A a^{2} + 3 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (4 \, A a^{2} + 3 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (2 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, B a^{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+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*(d*x + c)*A*a^2 + (4*A*a^2 + 3*B*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (4*A*a^2 + 3*B*a^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1)) - 2*(2*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 3*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 2*A*a^2*tan(
1/2*d*x + 1/2*c) - 5*B*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d