3.897 \(\int (a+b \sec (c+d x))^3 (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=214 \[ \frac{b^2 \left (34 a^2 b B-15 a^3 C+12 a b^2 C+4 b^3 B\right ) \tan (c+d x)}{6 d}+\frac{b \left (8 a^2 b^2 C+32 a^3 b B-24 a^4 C+16 a b^3 B+3 b^4 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^3 \left (-6 a^2 C+32 a b B+9 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{24 d}+a^4 x (b B-a C)+\frac{b^2 (3 a C+4 b B) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{b^2 C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]

[Out]

a^4*(b*B - a*C)*x + (b*(32*a^3*b*B + 16*a*b^3*B - 24*a^4*C + 8*a^2*b^2*C + 3*b^4*C)*ArcTanh[Sin[c + d*x]])/(8*
d) + (b^2*(34*a^2*b*B + 4*b^3*B - 15*a^3*C + 12*a*b^2*C)*Tan[c + d*x])/(6*d) + (b^3*(32*a*b*B - 6*a^2*C + 9*b^
2*C)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + (b^2*(4*b*B + 3*a*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) + (b
^2*C*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.476074, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4041, 3918, 4056, 4048, 3770, 3767, 8} \[ \frac{b^2 \left (34 a^2 b B-15 a^3 C+12 a b^2 C+4 b^3 B\right ) \tan (c+d x)}{6 d}+\frac{b \left (8 a^2 b^2 C+32 a^3 b B-24 a^4 C+16 a b^3 B+3 b^4 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^3 \left (-6 a^2 C+32 a b B+9 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{24 d}+a^4 x (b B-a C)+\frac{b^2 (3 a C+4 b B) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{b^2 C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^3*(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2),x]

[Out]

a^4*(b*B - a*C)*x + (b*(32*a^3*b*B + 16*a*b^3*B - 24*a^4*C + 8*a^2*b^2*C + 3*b^4*C)*ArcTanh[Sin[c + d*x]])/(8*
d) + (b^2*(34*a^2*b*B + 4*b^3*B - 15*a^3*C + 12*a*b^2*C)*Tan[c + d*x])/(6*d) + (b^3*(32*a*b*B - 6*a^2*C + 9*b^
2*C)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + (b^2*(4*b*B + 3*a*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) + (b
^2*C*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*d)

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 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 4056

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/(m + 1), Int
[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) + a
*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 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]

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))^3 \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))^4 \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))^3 \tan (c+d x)}{4 d}+\frac{\int (a+b \sec (c+d x))^2 \left (4 a^2 b^2 (b B-a C)+b^3 \left (8 a b B-4 a^2 C+3 b^2 C\right ) \sec (c+d x)+b^4 (4 b B+3 a C) \sec ^2(c+d x)\right ) \, dx}{4 b^2}\\ &=\frac{b^2 (4 b B+3 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b^2 C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{\int (a+b \sec (c+d x)) \left (12 a^3 b^2 (b B-a C)+b^3 \left (36 a^2 b B+8 b^3 B-24 a^3 C+15 a b^2 C\right ) \sec (c+d x)+b^4 \left (32 a b B-6 a^2 C+9 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx}{12 b^2}\\ &=\frac{b^3 \left (32 a b B-6 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{b^2 (4 b B+3 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b^2 C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{\int \left (24 a^4 b^2 (b B-a C)+3 b^3 \left (32 a^3 b B+16 a b^3 B-24 a^4 C+8 a^2 b^2 C+3 b^4 C\right ) \sec (c+d x)+4 b^4 \left (34 a^2 b B+4 b^3 B-15 a^3 C+12 a b^2 C\right ) \sec ^2(c+d x)\right ) \, dx}{24 b^2}\\ &=a^4 (b B-a C) x+\frac{b^3 \left (32 a b B-6 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{b^2 (4 b B+3 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b^2 C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{6} \left (b^2 \left (34 a^2 b B+4 b^3 B-15 a^3 C+12 a b^2 C\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (b \left (32 a^3 b B+16 a b^3 B-24 a^4 C+8 a^2 b^2 C+3 b^4 C\right )\right ) \int \sec (c+d x) \, dx\\ &=a^4 (b B-a C) x+\frac{b \left (32 a^3 b B+16 a b^3 B-24 a^4 C+8 a^2 b^2 C+3 b^4 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^3 \left (32 a b B-6 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{b^2 (4 b B+3 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b^2 C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (b^2 \left (34 a^2 b B+4 b^3 B-15 a^3 C+12 a b^2 C\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=a^4 (b B-a C) x+\frac{b \left (32 a^3 b B+16 a b^3 B-24 a^4 C+8 a^2 b^2 C+3 b^4 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^2 \left (34 a^2 b B+4 b^3 B-15 a^3 C+12 a b^2 C\right ) \tan (c+d x)}{6 d}+\frac{b^3 \left (32 a b B-6 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{b^2 (4 b B+3 a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b^2 C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.27378, size = 170, normalized size = 0.79 \[ \frac{3 b \left (8 a^2 b^2 C+32 a^3 b B-24 a^4 C+16 a b^3 B+3 b^4 C\right ) \tanh ^{-1}(\sin (c+d x))+3 b^2 \tan (c+d x) \left (b \left (8 a^2 C+16 a b B+3 b^2 C\right ) \sec (c+d x)+8 \left (6 a^2 b B-2 a^3 C+3 a b^2 C+b^3 B\right )+2 b^3 C \sec ^3(c+d x)\right )+24 a^4 d x (b B-a C)+8 b^4 (3 a C+b B) \tan ^3(c+d x)}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^3*(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2),x]

[Out]

(24*a^4*(b*B - a*C)*d*x + 3*b*(32*a^3*b*B + 16*a*b^3*B - 24*a^4*C + 8*a^2*b^2*C + 3*b^4*C)*ArcTanh[Sin[c + d*x
]] + 3*b^2*(8*(6*a^2*b*B + b^3*B - 2*a^3*C + 3*a*b^2*C) + b*(16*a*b*B + 8*a^2*C + 3*b^2*C)*Sec[c + d*x] + 2*b^
3*C*Sec[c + d*x]^3)*Tan[c + d*x] + 8*b^4*(b*B + 3*a*C)*Tan[c + d*x]^3)/(24*d)

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Maple [A]  time = 0.059, size = 360, normalized size = 1.7 \begin{align*} B{a}^{4}bx+{\frac{B{a}^{4}bc}{d}}-{a}^{5}Cx-{\frac{C{a}^{5}c}{d}}+4\,{\frac{B{a}^{3}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{{a}^{3}{b}^{2}C\tan \left ( dx+c \right ) }{d}}-3\,{\frac{{a}^{4}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{B{a}^{2}{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{C{a}^{2}{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{C{a}^{2}{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{Ba{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{Ba{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{{b}^{4}Ca\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{4}Ca\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{2\,B{b}^{5}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B{b}^{5}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{C{b}^{5}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,C{b}^{5}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,C{b}^{5}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*a^4*b*x+1/d*B*a^4*b*c-a^5*C*x-1/d*C*a^5*c+4/d*B*a^3*b^2*ln(sec(d*x+c)+tan(d*x+c))-2/d*a^3*b^2*C*tan(d*x+c)-3
/d*a^4*b*C*ln(sec(d*x+c)+tan(d*x+c))+6/d*B*a^2*b^3*tan(d*x+c)+1/d*C*a^2*b^3*sec(d*x+c)*tan(d*x+c)+1/d*C*a^2*b^
3*ln(sec(d*x+c)+tan(d*x+c))+2/d*B*a*b^4*sec(d*x+c)*tan(d*x+c)+2/d*B*a*b^4*ln(sec(d*x+c)+tan(d*x+c))+2/d*b^4*C*
a*tan(d*x+c)+1/d*b^4*C*a*tan(d*x+c)*sec(d*x+c)^2+2/3/d*B*b^5*tan(d*x+c)+1/3/d*B*b^5*tan(d*x+c)*sec(d*x+c)^2+1/
4/d*C*b^5*tan(d*x+c)*sec(d*x+c)^3+3/8/d*C*b^5*sec(d*x+c)*tan(d*x+c)+3/8/d*C*b^5*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.10925, size = 432, normalized size = 2.02 \begin{align*} -\frac{48 \,{\left (d x + c\right )} C a^{5} - 48 \,{\left (d x + c\right )} B a^{4} b - 48 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b^{4} - 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b^{5} + 3 \, C b^{5}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{2} 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 )} + 48 \, B a 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 )} + 144 \, C a^{4} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 192 \, B a^{3} b^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 96 \, C a^{3} b^{2} \tan \left (d x + c\right ) - 288 \, B a^{2} b^{3} \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(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]

-1/48*(48*(d*x + c)*C*a^5 - 48*(d*x + c)*B*a^4*b - 48*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a*b^4 - 16*(tan(d*x
+ c)^3 + 3*tan(d*x + c))*B*b^5 + 3*C*b^5*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x +
c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) + 24*C*a^2*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)) + 48*B*a*b^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(
sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 144*C*a^4*b*log(sec(d*x + c) + tan(d*x + c)) - 192*B*a^3*b^2*log(
sec(d*x + c) + tan(d*x + c)) + 96*C*a^3*b^2*tan(d*x + c) - 288*B*a^2*b^3*tan(d*x + c))/d

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Fricas [A]  time = 0.59161, size = 633, normalized size = 2.96 \begin{align*} -\frac{48 \,{\left (C a^{5} - B a^{4} b\right )} d x \cos \left (d x + c\right )^{4} + 3 \,{\left (24 \, C a^{4} b - 32 \, B a^{3} b^{2} - 8 \, C a^{2} b^{3} - 16 \, B a b^{4} - 3 \, C b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (24 \, C a^{4} b - 32 \, B a^{3} b^{2} - 8 \, C a^{2} b^{3} - 16 \, B a b^{4} - 3 \, C b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (6 \, C b^{5} - 16 \,{\left (3 \, C a^{3} b^{2} - 9 \, B a^{2} b^{3} - 3 \, C a b^{4} - B b^{5}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (8 \, C a^{2} b^{3} + 16 \, B a b^{4} + 3 \, C b^{5}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (3 \, C a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(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]

-1/48*(48*(C*a^5 - B*a^4*b)*d*x*cos(d*x + c)^4 + 3*(24*C*a^4*b - 32*B*a^3*b^2 - 8*C*a^2*b^3 - 16*B*a*b^4 - 3*C
*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(24*C*a^4*b - 32*B*a^3*b^2 - 8*C*a^2*b^3 - 16*B*a*b^4 - 3*C*b^5
)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 2*(6*C*b^5 - 16*(3*C*a^3*b^2 - 9*B*a^2*b^3 - 3*C*a*b^4 - B*b^5)*cos(
d*x + c)^3 + 3*(8*C*a^2*b^3 + 16*B*a*b^4 + 3*C*b^5)*cos(d*x + c)^2 + 8*(3*C*a*b^4 + B*b^5)*cos(d*x + c))*sin(d
*x + c))/(d*cos(d*x + c)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int C a^{5}\, dx - \int - B a^{4} b\, dx - \int - B b^{5} \sec ^{4}{\left (c + d x \right )}\, dx - \int - C b^{5} \sec ^{5}{\left (c + d x \right )}\, dx - \int - 4 B a b^{4} \sec ^{3}{\left (c + d x \right )}\, dx - \int - 6 B a^{2} b^{3} \sec ^{2}{\left (c + d x \right )}\, dx - \int - 4 B a^{3} b^{2} \sec{\left (c + d x \right )}\, dx - \int - 3 C a b^{4} \sec ^{4}{\left (c + d x \right )}\, dx - \int - 2 C a^{2} b^{3} \sec ^{3}{\left (c + d x \right )}\, dx - \int 2 C a^{3} b^{2} \sec ^{2}{\left (c + d x \right )}\, dx - \int 3 C a^{4} b \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**3*(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**5, x) - Integral(-B*a**4*b, x) - Integral(-B*b**5*sec(c + d*x)**4, x) - Integral(-C*b**5*sec(c
+ d*x)**5, x) - Integral(-4*B*a*b**4*sec(c + d*x)**3, x) - Integral(-6*B*a**2*b**3*sec(c + d*x)**2, x) - Integ
ral(-4*B*a**3*b**2*sec(c + d*x), x) - Integral(-3*C*a*b**4*sec(c + d*x)**4, x) - Integral(-2*C*a**2*b**3*sec(c
+ d*x)**3, x) - Integral(2*C*a**3*b**2*sec(c + d*x)**2, x) - Integral(3*C*a**4*b*sec(c + d*x), x)

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Giac [B]  time = 1.37162, size = 888, normalized size = 4.15 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3*(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]

-1/24*(24*(C*a^5 - B*a^4*b)*(d*x + c) + 3*(24*C*a^4*b - 32*B*a^3*b^2 - 8*C*a^2*b^3 - 16*B*a*b^4 - 3*C*b^5)*log
(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(24*C*a^4*b - 32*B*a^3*b^2 - 8*C*a^2*b^3 - 16*B*a*b^4 - 3*C*b^5)*log(abs(t
an(1/2*d*x + 1/2*c) - 1)) - 2*(48*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^7 - 144*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^7 + 24
*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^7 + 48*B*a*b^4*tan(1/2*d*x + 1/2*c)^7 - 72*C*a*b^4*tan(1/2*d*x + 1/2*c)^7 - 24
*B*b^5*tan(1/2*d*x + 1/2*c)^7 + 15*C*b^5*tan(1/2*d*x + 1/2*c)^7 - 144*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 432*B
*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 24*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 48*B*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 12
0*C*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 40*B*b^5*tan(1/2*d*x + 1/2*c)^5 + 9*C*b^5*tan(1/2*d*x + 1/2*c)^5 + 144*C*a^
3*b^2*tan(1/2*d*x + 1/2*c)^3 - 432*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 24*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 48
*B*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 120*C*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 40*B*b^5*tan(1/2*d*x + 1/2*c)^3 + 9*C*b
^5*tan(1/2*d*x + 1/2*c)^3 - 48*C*a^3*b^2*tan(1/2*d*x + 1/2*c) + 144*B*a^2*b^3*tan(1/2*d*x + 1/2*c) + 24*C*a^2*
b^3*tan(1/2*d*x + 1/2*c) + 48*B*a*b^4*tan(1/2*d*x + 1/2*c) + 72*C*a*b^4*tan(1/2*d*x + 1/2*c) + 24*B*b^5*tan(1/
2*d*x + 1/2*c) + 15*C*b^5*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d