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

Optimal. Leaf size=149 \[ a^3 x (b B-a C)+\frac {b^2 \left (a^2 (-C)+9 a b B+2 b^2 C\right ) \tan (c+d x)}{3 d}+\frac {b \left (-4 a^3 C+6 a^2 b B+2 a b^2 C+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b^3 (2 a C+3 b B) \tan (c+d x) \sec (c+d x)}{6 d}+\frac {b^2 C \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

[Out]

a^3*(B*b-C*a)*x+1/2*b*(6*B*a^2*b+B*b^3-4*C*a^3+2*C*a*b^2)*arctanh(sin(d*x+c))/d+1/3*b^2*(9*B*a*b-C*a^2+2*C*b^2
)*tan(d*x+c)/d+1/6*b^3*(3*B*b+2*C*a)*sec(d*x+c)*tan(d*x+c)/d+1/3*b^2*C*(a+b*sec(d*x+c))^2*tan(d*x+c)/d

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Rubi [A]  time = 0.31, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4041, 3918, 4048, 3770, 3767, 8} \[ \frac {b^2 \left (a^2 (-C)+9 a b B+2 b^2 C\right ) \tan (c+d x)}{3 d}+\frac {b \left (6 a^2 b B-4 a^3 C+2 a b^2 C+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^3 x (b B-a C)+\frac {b^3 (2 a C+3 b B) \tan (c+d x) \sec (c+d x)}{6 d}+\frac {b^2 C \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*(b*B - a*C)*x + (b*(6*a^2*b*B + b^3*B - 4*a^3*C + 2*a*b^2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (b^2*(9*a*b*B
- a^2*C + 2*b^2*C)*Tan[c + d*x])/(3*d) + (b^3*(3*b*B + 2*a*C)*Sec[c + d*x]*Tan[c + d*x])/(6*d) + (b^2*C*(a + b
*Sec[c + d*x])^2*Tan[c + d*x])/(3*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3770

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

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 4041

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

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]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^2 \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx &=\frac {\int (a+b \sec (c+d x))^3 \left (b^2 (b B-a C)+b^3 C \sec (c+d x)\right ) \, dx}{b^2}\\ &=\frac {b^2 C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {\int (a+b \sec (c+d x)) \left (3 a^2 b^2 (b B-a C)+b^3 \left (6 a b B-3 a^2 C+2 b^2 C\right ) \sec (c+d x)+b^4 (3 b B+2 a C) \sec ^2(c+d x)\right ) \, dx}{3 b^2}\\ &=\frac {b^3 (3 b B+2 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {b^2 C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {\int \left (6 a^3 b^2 (b B-a C)+3 b^3 \left (6 a^2 b B+b^3 B-4 a^3 C+2 a b^2 C\right ) \sec (c+d x)+2 b^4 \left (9 a b B-a^2 C+2 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx}{6 b^2}\\ &=a^3 (b B-a C) x+\frac {b^3 (3 b B+2 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {b^2 C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{3} \left (b^2 \left (9 a b B-a^2 C+2 b^2 C\right )\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{2} \left (b \left (6 a^2 b B+b^3 B-4 a^3 C+2 a b^2 C\right )\right ) \int \sec (c+d x) \, dx\\ &=a^3 (b B-a C) x+\frac {b \left (6 a^2 b B+b^3 B-4 a^3 C+2 a b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b^3 (3 b B+2 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {b^2 C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac {\left (b^2 \left (9 a b B-a^2 C+2 b^2 C\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=a^3 (b B-a C) x+\frac {b \left (6 a^2 b B+b^3 B-4 a^3 C+2 a b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b^2 \left (9 a b B-a^2 C+2 b^2 C\right ) \tan (c+d x)}{3 d}+\frac {b^3 (3 b B+2 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {b^2 C (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.93, size = 114, normalized size = 0.77 \[ \frac {6 a^3 d x (b B-a C)+3 b \left (-4 a^3 C+6 a^2 b B+2 a b^2 C+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))+3 b^3 \tan (c+d x) \sec (c+d x) (2 (3 a B+b C) \cos (c+d x)+2 a C+b B)+2 b^4 C \tan ^3(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 199, normalized size = 1.34 \[ -\frac {12 \, {\left (C a^{4} - B a^{3} b\right )} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, C a^{3} b - 6 \, B a^{2} b^{2} - 2 \, C a b^{3} - B b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, C a^{3} b - 6 \, B a^{2} b^{2} - 2 \, C a b^{3} - B b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, C b^{4} + 2 \, {\left (9 \, B a b^{3} + 2 \, C b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*(C*a^4 - B*a^3*b)*d*x*cos(d*x + c)^3 + 3*(4*C*a^3*b - 6*B*a^2*b^2 - 2*C*a*b^3 - B*b^4)*cos(d*x + c)^
3*log(sin(d*x + c) + 1) - 3*(4*C*a^3*b - 6*B*a^2*b^2 - 2*C*a*b^3 - B*b^4)*cos(d*x + c)^3*log(-sin(d*x + c) + 1
) - 2*(2*C*b^4 + 2*(9*B*a*b^3 + 2*C*b^4)*cos(d*x + c)^2 + 3*(2*C*a*b^3 + B*b^4)*cos(d*x + c))*sin(d*x + c))/(d
*cos(d*x + c)^3)

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giac [B]  time = 0.30, size = 301, normalized size = 2.02 \[ -\frac {6 \, {\left (C a^{4} - B a^{3} b\right )} {\left (d x + c\right )} + 3 \, {\left (4 \, C a^{3} b - 6 \, B a^{2} b^{2} - 2 \, C a b^{3} - B b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, C a^{3} b - 6 \, B a^{2} b^{2} - 2 \, C a b^{3} - B b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (18 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(C*a^4 - B*a^3*b)*(d*x + c) + 3*(4*C*a^3*b - 6*B*a^2*b^2 - 2*C*a*b^3 - B*b^4)*log(abs(tan(1/2*d*x + 1/
2*c) + 1)) - 3*(4*C*a^3*b - 6*B*a^2*b^2 - 2*C*a*b^3 - B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(18*B*a*b^
3*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^4*tan(1/2
*d*x + 1/2*c)^5 - 36*B*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 4*C*b^4*tan(1/2*d*x + 1/2*c)^3 + 18*B*a*b^3*tan(1/2*d*x
+ 1/2*c) + 6*C*a*b^3*tan(1/2*d*x + 1/2*c) + 3*B*b^4*tan(1/2*d*x + 1/2*c) + 6*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(
1/2*d*x + 1/2*c)^2 - 1)^3)/d

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maple [A]  time = 1.14, size = 228, normalized size = 1.53 \[ B x \,a^{3} b +\frac {B \,a^{3} b c}{d}-a^{4} C x -\frac {C \,a^{4} c}{d}+\frac {3 a^{2} b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {2 a^{3} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 B a \,b^{3} \tan \left (d x +c \right )}{d}+\frac {C a \,b^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {C a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B \,b^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {B \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 C \,b^{4} \tan \left (d x +c \right )}{3 d}+\frac {C \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 204, normalized size = 1.37 \[ -\frac {12 \, {\left (d x + c\right )} C a^{4} - 12 \, {\left (d x + c\right )} B a^{3} b - 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{4} + 6 \, C a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 3 \, B b^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{3} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 36 \, B a^{2} b^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 36 \, B a b^{3} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*(d*x + c)*C*a^4 - 12*(d*x + c)*B*a^3*b - 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*b^4 + 6*C*a*b^3*(2*si
n(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 3*B*b^4*(2*sin(d*x + c)/(si
n(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*C*a^3*b*log(sec(d*x + c) + tan(d*x + c
)) - 36*B*a^2*b^2*log(sec(d*x + c) + tan(d*x + c)) - 36*B*a*b^3*tan(d*x + c))/d

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mupad [B]  time = 6.72, size = 576, normalized size = 3.87 \[ \frac {\frac {B\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {C\,b^4\,\sin \left (3\,c+3\,d\,x\right )}{6}+\frac {C\,b^4\,\sin \left (c+d\,x\right )}{2}+\frac {3\,B\,a\,b^3\,\sin \left (c+d\,x\right )}{4}-\frac {3\,C\,a^4\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}-\frac {B\,b^4\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{4}+\frac {3\,B\,a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {C\,a\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{2}-\frac {C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{2}-\frac {B\,b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{4}+\frac {B\,a^3\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{2}-\frac {B\,a^2\,b^2\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,9{}\mathrm {i}}{2}-\frac {C\,a\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{2}+C\,a^3\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}-\frac {B\,a^2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{2}+\frac {3\,B\,a^3\,b\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}-\frac {C\,a\,b^3\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{2}+C\,a^3\,b\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^2*((B*b^2)/cos(c + d*x) - C*a^2 + (C*b^2)/cos(c + d*x)^2 + B*a*b),x)

[Out]

((B*b^4*sin(2*c + 2*d*x))/4 + (C*b^4*sin(3*c + 3*d*x))/6 + (C*b^4*sin(c + d*x))/2 + (3*B*a*b^3*sin(c + d*x))/4
- (3*C*a^4*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - (B*b^4*cos(c + d*x)*atan((sin(c/2 +
(d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/4 + (3*B*a*b^3*sin(3*c + 3*d*x))/4 + (C*a*b^3*sin(2*c + 2*d*x))/2 - (C*a^
4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 - (B*b^4*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/
2 + (d*x)/2))*cos(3*c + 3*d*x)*1i)/4 + (B*a^3*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/
2 - (B*a^2*b^2*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*9i)/2 - (C*a*b^3*atan((sin(c/2 +
(d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*1i)/2 + C*a^3*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)
/2))*cos(3*c + 3*d*x)*1i - (B*a^2*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*3i)/2
+ (3*B*a^3*b*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - (C*a*b^3*cos(c + d*x)*atan((sin(c/2
+ (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/2 + C*a^3*b*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/
2))*3i)/(d*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int C a^{4}\, dx - \int \left (- B a^{3} b\right )\, dx - \int \left (- B b^{4} \sec ^{3}{\left (c + d x \right )}\right )\, dx - \int \left (- C b^{4} \sec ^{4}{\left (c + d x \right )}\right )\, dx - \int \left (- 3 B a b^{3} \sec ^{2}{\left (c + d x \right )}\right )\, dx - \int \left (- 3 B a^{2} b^{2} \sec {\left (c + d x \right )}\right )\, dx - \int \left (- 2 C a b^{3} \sec ^{3}{\left (c + d x \right )}\right )\, dx - \int 2 C a^{3} b \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(C*a**4, x) - Integral(-B*a**3*b, x) - Integral(-B*b**4*sec(c + d*x)**3, x) - Integral(-C*b**4*sec(c
+ d*x)**4, x) - Integral(-3*B*a*b**3*sec(c + d*x)**2, x) - Integral(-3*B*a**2*b**2*sec(c + d*x), x) - Integral
(-2*C*a*b**3*sec(c + d*x)**3, x) - Integral(2*C*a**3*b*sec(c + d*x), x)

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