3.9.99 \(\int (a+b \sec (c+d x)) (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)) \, dx\) [899]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 97, normalized size of antiderivative = 1.00, 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 = {4126, 4003, 3855, 3852, 8} \begin {gather*} \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} \end {gather*}

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 8

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

Rule 3852

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 3855

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

Rule 4003

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 4126

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]

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 ) \text {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 = 0.61, size = 77, normalized size = 0.79 \begin {gather*} \frac {2 a^2 (b B-a C) d x+b \left (4 a b B-2 a^2 C+b^2 C\right ) \tanh ^{-1}(\sin (c+d x))+b^2 (2 b B+2 a C+b C \sec (c+d x)) \tan (c+d x)}{2 d} \end {gather*}

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)

________________________________________________________________________________________

Maple [A]
time = 0.06, size = 129, normalized size = 1.33

method	result	size
derivativedivides	\(\frac {2 a \,b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-a^{2} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+b^{3} B \tan \left (d x +c \right )+C \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} b B \left (d x +c \right )-a^{3} C \left (d x +c \right )+C \,b^{2} a \tan \left (d x +c \right )}{d}\)	\(129\)
default	\(\frac {2 a \,b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-a^{2} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+b^{3} B \tan \left (d x +c \right )+C \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} b B \left (d x +c \right )-a^{3} C \left (d x +c \right )+C \,b^{2} a \tan \left (d x +c \right )}{d}\)	\(129\)
norman	\(\frac {\left (a^{2} b B -a^{3} C \right ) x +\left (-2 a^{2} b B +2 a^{3} C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{2} b B -a^{3} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b^{2} \left (2 b B +2 a C +C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {b^{2} \left (2 b B +2 a C -C b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {b \left (4 a b B -2 a^{2} C +b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {b \left (4 a b B -2 a^{2} C +b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(217\)
risch	\(B \,a^{2} b x -C \,a^{3} x -\frac {i b^{2} \left (C b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-C b \,{\mathrm e}^{i \left (d x +c \right )}-2 b B -2 a C \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2} B}{d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} C}{d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2} B}{d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} C}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\)	\(233\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.30, size = 142, normalized size = 1.46 \begin {gather*} -\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 {gather*}

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

________________________________________________________________________________________

Fricas [A]
time = 2.86, size = 154, normalized size = 1.59 \begin {gather*} -\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 {gather*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (93) = 186\).
time = 0.50, size = 213, normalized size = 2.20 \begin {gather*} -\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 {gather*}

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

________________________________________________________________________________________

Mupad [B]
time = 5.40, size = 221, normalized size = 2.28 \begin {gather*} \frac {\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {C\,b^3\,\sin \left (c+d\,x\right )}{2}+\frac {C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}-\frac {2\,\left (C\,a^3\,\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 )+\frac {C\,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 )\,1{}\mathrm {i}}{2}-B\,a^2\,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 )+B\,a\,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 )\,2{}\mathrm {i}-C\,a^2\,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 )\,1{}\mathrm {i}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________