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

Optimal. Leaf size=120 \[ \frac{a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac{a^2 (4 A+3 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 A x+\frac{(3 B+2 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{6 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

[Out]

a^2*A*x + (a^2*(4*A + 3*B + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a^2*(2*A + 3*B + 2*C)*Tan[c + d*x])/(2*d) + (
C*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(3*d) + ((3*B + 2*C)*(a^2 + a^2*Sec[c + d*x])*Tan[c + d*x])/(6*d)

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Rubi [A]  time = 0.157222, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4054, 3917, 3914, 3767, 8, 3770} \[ \frac{a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac{a^2 (4 A+3 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 A x+\frac{(3 B+2 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{6 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*A*x + (a^2*(4*A + 3*B + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a^2*(2*A + 3*B + 2*C)*Tan[c + d*x])/(2*d) + (
C*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(3*d) + ((3*B + 2*C)*(a^2 + a^2*Sec[c + d*x])*Tan[c + d*x])/(6*d)

Rule 4054

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

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

Mathematica [B]  time = 5.68545, size = 542, normalized size = 4.52 \[ \frac{a^2 \cos ^4(c+d x) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{4 (3 A+6 B+5 C) \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 (3 A+6 B+5 C) \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{6 (4 A+3 B+2 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{6 (4 A+3 B+2 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+12 A x+\frac{(3 B+7 C) \cos \left (\frac{c}{2}\right )-(3 B+5 C) \sin \left (\frac{c}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{(3 B+5 C) \sin \left (\frac{c}{2}\right )+(3 B+7 C) \cos \left (\frac{c}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 C \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 C \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}\right )}{24 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cos[c + d*x]^4*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(12*A*x -
(6*(4*A + 3*B + 2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (6*(4*A + 3*B + 2*C)*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]])/d + (2*C*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) +
((3*B + 7*C)*Cos[c/2] - (3*B + 5*C)*Sin[c/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^
2) + (4*(3*A + 6*B + 5*C)*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*C
*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - ((3*B + 7*C)*Cos[c/2] + (3*
B + 5*C)*Sin[c/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(3*A + 6*B + 5*C)*Si
n[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(24*(A + 2*C + 2*B*Cos[c + d*x]
+ A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.056, size = 193, normalized size = 1.6 \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}}+{\frac{5\,{a}^{2}C\tan \left ( dx+c \right ) }{3\,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}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \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}}+{\frac{{a}^{2}C\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+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),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))+5/3/d*a^2*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*C*sec(d*x+c)*tan(d*x+c)+1/d*a^2*C*ln(sec(d*x+c)+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)+1/3/d*a^2*C*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 0.951897, size = 284, normalized size = 2.37 \begin{align*} \frac{12 \,{\left (d x + c\right )} A a^{2} + 4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 3 \, 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 )} - 6 \, C 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 )} + 24 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, B a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, A a^{2} \tan \left (d x + c\right ) + 24 \, B a^{2} \tan \left (d x + c\right ) + 12 \, C a^{2} \tan \left (d x + c\right )}{12 \, 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)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/12*(12*(d*x + c)*A*a^2 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 3*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)) - 6*C*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(si
n(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*A*a^2*log(sec(d*x + c) + tan(d*x + c)) + 12*B*a^2*log(sec(d*x +
c) + tan(d*x + c)) + 12*A*a^2*tan(d*x + c) + 24*B*a^2*tan(d*x + c) + 12*C*a^2*tan(d*x + c))/d

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Fricas [A]  time = 0.526347, size = 379, normalized size = 3.16 \begin{align*} \frac{12 \, A a^{2} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, A + 3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (4 \, A + 3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (3 \, A + 6 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, C a^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \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)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(12*A*a^2*d*x*cos(d*x + c)^3 + 3*(4*A + 3*B + 2*C)*a^2*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(4*A + 3*
B + 2*C)*a^2*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(3*A + 6*B + 5*C)*a^2*cos(d*x + c)^2 + 3*(B + 2*C)*a
^2*cos(d*x + c) + 2*C*a^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*} 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 + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\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)+C*sec(d*x+c)**2),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) + Integral(C*sec(c + d*x)**2, x) + I
ntegral(2*C*sec(c + d*x)**3, x) + Integral(C*sec(c + d*x)**4, x))

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Giac [B]  time = 1.29302, size = 338, normalized size = 2.82 \begin{align*} \frac{6 \,{\left (d x + c\right )} A a^{2} + 3 \,{\left (4 \, A a^{2} + 3 \, B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (4 \, A a^{2} + 3 \, B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, C 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 )}^{3}}}{6 \, 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)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*A*a^2 + 3*(4*A*a^2 + 3*B*a^2 + 2*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*A*a^2 + 3*B
*a^2 + 2*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 9*B*a^2*tan(1/2*d*x +
1/2*c)^5 + 6*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 12*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 24*B*a^2*tan(1/2*d*x + 1/2*c)^3
- 16*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) + 15*B*a^2*tan(1/2*d*x + 1/2*c) + 18*C*a^2*t
an(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d