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

Optimal. Leaf size=97 \[ \frac{b \left (-2 a^2 C+4 a b B+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (b B-a C)+\frac{b^2 (a C+2 b B) \tan (c+d x)}{2 d}+\frac{b^2 C \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.16879, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 46, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.109, Rules used = {4041, 3918, 3770, 3767, 8} \[ \frac{b \left (-2 a^2 C+4 a b B+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (b B-a C)+\frac{b^2 (a C+2 b B) \tan (c+d x)}{2 d}+\frac{b^2 C \tan (c+d x) (a+b \sec (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*(b*B - a*C)*x + (b*(4*a*b*B - 2*a^2*C + b^2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (b^2*(2*b*B + a*C)*Tan[c + d
*x])/(2*d) + (b^2*C*(a + b*Sec[c + d*x])*Tan[c + d*x])/(2*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 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)) \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))^2 \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)) \tan (c+d x)}{2 d}+\frac{\int \left (2 a^2 b^2 (b B-a C)+b^3 \left (4 a b B-2 a^2 C+b^2 C\right ) \sec (c+d x)+b^4 (2 b B+a C) \sec ^2(c+d x)\right ) \, dx}{2 b^2}\\ &=a^2 (b B-a C) x+\frac{b^2 C (a+b \sec (c+d x)) \tan (c+d x)}{2 d}+\frac{1}{2} \left (b^2 (2 b B+a C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (b \left (4 a b B-2 a^2 C+b^2 C\right )\right ) \int \sec (c+d x) \, dx\\ &=a^2 (b B-a C) x+\frac{b \left (4 a b B-2 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^2 C (a+b \sec (c+d x)) \tan (c+d x)}{2 d}-\frac{\left (b^2 (2 b B+a C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^2 (b B-a C) x+\frac{b \left (4 a b B-2 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b^2 (2 b B+a C) \tan (c+d x)}{2 d}+\frac{b^2 C (a+b \sec (c+d x)) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.562134, size = 77, normalized size = 0.79 \[ \frac{b \left (-2 a^2 C+4 a b B+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))+2 a^2 d x (b B-a C)+b^2 \tan (c+d x) (2 a C+2 b B+b C \sec (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 157, normalized size = 1.6 \begin{align*} B{a}^{2}bx+{\frac{B{a}^{2}bc}{d}}-{a}^{3}Cx-{\frac{C{a}^{3}c}{d}}+2\,{\frac{Ba{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{Ca{b}^{2}\tan \left ( dx+c \right ) }{d}}-{\frac{{a}^{2}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{B{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{C{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{C{b}^{3}\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))*(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^2*b*x+1/d*B*a^2*b*c-a^3*C*x-1/d*C*a^3*c+2/d*B*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+1/d*C*a*b^2*tan(d*x+c)-1/d*a
^2*b*C*ln(sec(d*x+c)+tan(d*x+c))+1/d*B*b^3*tan(d*x+c)+1/2/d*C*b^3*sec(d*x+c)*tan(d*x+c)+1/2/d*C*b^3*ln(sec(d*x
+c)+tan(d*x+c))

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Maxima [A]  time = 1.02709, size = 192, normalized size = 1.98 \begin{align*} -\frac{4 \,{\left (d x + c\right )} C a^{3} - 4 \,{\left (d x + c\right )} B a^{2} b + C 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 )} + 4 \, C a^{2} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 8 \, B a b^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 4 \, C a b^{2} \tan \left (d x + c\right ) - 4 \, B b^{3} \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))*(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]

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

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Fricas [A]  time = 0.551722, size = 363, normalized size = 3.74 \begin{align*} -\frac{4 \,{\left (C a^{3} - B a^{2} b\right )} d x \cos \left (d x + c\right )^{2} +{\left (2 \, C a^{2} b - 4 \, B a b^{2} - C b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, C a^{2} b - 4 \, B a b^{2} - C b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (C b^{3} + 2 \,{\left (C a b^{2} + B b^{3}\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))*(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]

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

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Giac [B]  time = 1.21328, size = 288, normalized size = 2.97 \begin{align*} -\frac{2 \,{\left (C a^{3} - B a^{2} b\right )}{\left (d x + c\right )} +{\left (2 \, C a^{2} b - 4 \, B a b^{2} - C b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (2 \, C a^{2} b - 4 \, B a b^{2} - C b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (2 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C b^{3} \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))*(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]

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