\(\int (a+b \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\) [656]

Optimal result

Integrand size = 25, antiderivative size = 167 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^3 A x+\frac {b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \tan (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d} \]

[Out]

a^3*A*x+1/8*b*(12*a^2*(2*A+C)+b^2*(4*A+3*C))*arctanh(sin(d*x+c))/d+1/2*a*(6*A*b^2+(a^2+4*b^2)*C)*tan(d*x+c)/d+
1/8*b*(2*C*a^2+b^2*(4*A+3*C))*sec(d*x+c)*tan(d*x+c)/d+1/4*a*C*(a+b*sec(d*x+c))^2*tan(d*x+c)/d+1/4*C*(a+b*sec(d
*x+c))^3*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4142, 4141, 4133, 3855, 3852, 8} \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^3 A x+\frac {b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \tan (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^2}{4 d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]

[In]

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

[Out]

a^3*A*x + (b*(12*a^2*(2*A + C) + b^2*(4*A + 3*C))*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(6*A*b^2 + (a^2 + 4*b^2)*C
)*Tan[c + d*x])/(2*d) + (b*(2*a^2*C + b^2*(4*A + 3*C))*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*C*(a + b*Sec[c +
d*x])^2*Tan[c + d*x])/(4*d) + (C*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*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 4133

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 4141

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), I
nt[(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 4142

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

Rubi steps \begin{align*} \text {integral}& = \frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+b \sec (c+d x))^2 \left (4 a A+b (4 A+3 C) \sec (c+d x)+3 a C \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{12} \int (a+b \sec (c+d x)) \left (12 a^2 A+3 a b (8 A+5 C) \sec (c+d x)+3 \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{24} \int \left (24 a^3 A+3 b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \sec (c+d x)+12 a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \sec ^2(c+d x)\right ) \, dx \\ & = a^3 A x+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{2} \left (a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right )\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right )\right ) \int \sec (c+d x) \, dx \\ & = a^3 A x+\frac {b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\left (a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d} \\ & = a^3 A x+\frac {b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \tan (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.79 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 a^3 A d x+b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))+\left (8 a \left (3 A b^2+\left (a^2+3 b^2\right ) C\right )+b \left (4 A b^2+3 \left (4 a^2+b^2\right ) C\right ) \sec (c+d x)+2 b^3 C \sec ^3(c+d x)\right ) \tan (c+d x)+8 a b^2 C \tan ^3(c+d x)}{8 d} \]

[In]

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

[Out]

(8*a^3*A*d*x + b*(12*a^2*(2*A + C) + b^2*(4*A + 3*C))*ArcTanh[Sin[c + d*x]] + (8*a*(3*A*b^2 + (a^2 + 3*b^2)*C)
+ b*(4*A*b^2 + 3*(4*a^2 + b^2)*C)*Sec[c + d*x] + 2*b^3*C*Sec[c + d*x]^3)*Tan[c + d*x] + 8*a*b^2*C*Tan[c + d*x
]^3)/(8*d)

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.10

method	result	size
parts	\(a^{3} A x +\frac {\left (A \,b^{3}+3 a^{2} b C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (3 a A \,b^{2}+a^{3} C \right ) \tan \left (d x +c \right )}{d}+\frac {C \,b^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {3 C a \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(183\)
derivativedivides	\(\frac {a^{3} A \left (d x +c \right )+a^{3} C \tan \left (d x +c \right )+3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{2} b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \tan \left (d x +c \right ) a \,b^{2}-3 C a \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,b^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(209\)
default	\(\frac {a^{3} A \left (d x +c \right )+a^{3} C \tan \left (d x +c \right )+3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{2} b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \tan \left (d x +c \right ) a \,b^{2}-3 C a \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,b^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(209\)
parallelrisch	\(\frac {-96 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (\left (\frac {A}{6}+\frac {C}{8}\right ) b^{2}+a^{2} \left (A +\frac {C}{2}\right )\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+96 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (\left (\frac {A}{6}+\frac {C}{8}\right ) b^{2}+a^{2} \left (A +\frac {C}{2}\right )\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+32 a^{3} A x d \cos \left (2 d x +2 c \right )+8 a^{3} A x d \cos \left (4 d x +4 c \right )+48 a \left (b^{2} \left (A +\frac {4 C}{3}\right )+\frac {C \,a^{2}}{3}\right ) \sin \left (2 d x +2 c \right )+8 \left (\left (A +\frac {3 C}{4}\right ) b^{2}+3 C \,a^{2}\right ) b \sin \left (3 d x +3 c \right )+24 a \left (b^{2} \left (A +\frac {2 C}{3}\right )+\frac {C \,a^{2}}{3}\right ) \sin \left (4 d x +4 c \right )+8 b \left (\left (A +\frac {11 C}{4}\right ) b^{2}+3 C \,a^{2}\right ) \sin \left (d x +c \right )+24 a^{3} A x d}{8 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(298\)
norman	\(\frac {a^{3} A x +a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {\left (24 a A \,b^{2}-4 A \,b^{3}+8 a^{3} C -12 a^{2} b C +24 C a \,b^{2}-5 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (24 a A \,b^{2}+4 A \,b^{3}+8 a^{3} C +12 a^{2} b C +24 C a \,b^{2}+5 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (72 a A \,b^{2}-4 A \,b^{3}+24 a^{3} C -12 a^{2} b C +40 C a \,b^{2}+3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}-\frac {\left (72 a A \,b^{2}+4 A \,b^{3}+24 a^{3} C +12 a^{2} b C +40 C a \,b^{2}-3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}-4 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {b \left (24 a^{2} A +4 A \,b^{2}+12 C \,a^{2}+3 C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {b \left (24 a^{2} A +4 A \,b^{2}+12 C \,a^{2}+3 C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(404\)
risch	\(a^{3} A x -\frac {i \left (4 A \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+12 C \,a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+3 C \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-24 A a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-8 C \,a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+4 A \,b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+12 C \,a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+11 C \,b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-72 A a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-24 C \,a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-48 C a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-4 A \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-12 C \,a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-11 C \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-72 A a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-24 C \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-64 C a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 A \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-12 C \,a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-3 C \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-24 a A \,b^{2}-8 a^{3} C -16 C a \,b^{2}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} A}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,a^{2}}{2 d}-\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} A}{d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,a^{2}}{2 d}+\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(537\)

[In]

int((a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

a^3*A*x+(A*b^3+3*C*a^2*b)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(3*A*a*b^2+C*a^3)/d*tan(
d*x+c)+C*b^3/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+3*A*a^2*b/d*ln(s
ec(d*x+c)+tan(d*x+c))-3*C*a*b^2/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.19 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {16 \, A a^{3} d x \cos \left (d x + c\right )^{4} + {\left (12 \, {\left (2 \, A + C\right )} a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (12 \, {\left (2 \, A + C\right )} a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, C a b^{2} \cos \left (d x + c\right ) + 2 \, C b^{3} + 8 \, {\left (C a^{3} + {\left (3 \, A + 2 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (12 \, C a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate((a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/16*(16*A*a^3*d*x*cos(d*x + c)^4 + (12*(2*A + C)*a^2*b + (4*A + 3*C)*b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1
) - (12*(2*A + C)*a^2*b + (4*A + 3*C)*b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*C*a*b^2*cos(d*x + c) +
2*C*b^3 + 8*(C*a^3 + (3*A + 2*C)*a*b^2)*cos(d*x + c)^3 + (12*C*a^2*b + (4*A + 3*C)*b^3)*cos(d*x + c)^2)*sin(d
*x + c))/(d*cos(d*x + c)^4)

Sympy [F]

\[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{3}\, dx \]

[In]

integrate((a+b*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.52 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {16 \, {\left (d x + c\right )} A a^{3} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b^{2} - C b^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, A b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{2} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 16 \, C a^{3} \tan \left (d x + c\right ) + 48 \, A a b^{2} \tan \left (d x + c\right )}{16 \, d} \]

[In]

integrate((a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/16*(16*(d*x + c)*A*a^3 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a*b^2 - C*b^3*(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)) - 12*C*a
^2*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*A*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*A*a^2*b*log(sec(d*x + c) + t
an(d*x + c)) + 16*C*a^3*tan(d*x + c) + 48*A*a*b^2*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 526 vs. \(2 (157) = 314\).

Time = 0.36 (sec) , antiderivative size = 526, normalized size of antiderivative = 3.15 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, {\left (d x + c\right )} A a^{3} + {\left (24 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 3 \, C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (24 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 3 \, 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 (8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, 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 )}^{4}}}{8 \, d} \]

[In]

integrate((a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/8*(8*(d*x + c)*A*a^3 + (24*A*a^2*b + 12*C*a^2*b + 4*A*b^3 + 3*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2
4*A*a^2*b + 12*C*a^2*b + 4*A*b^3 + 3*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(8*C*a^3*tan(1/2*d*x + 1/2*
c)^7 - 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 24*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24*C*a*b^2*tan(1/2*d*x + 1/2*c)
^7 - 4*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 5*C*b^3*tan(1/2*d*x + 1/2*c)^7 - 24*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 12*C*
a^2*b*tan(1/2*d*x + 1/2*c)^5 - 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*A*b^3
*tan(1/2*d*x + 1/2*c)^5 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^5 + 24*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*b*tan(1/
2*d*x + 1/2*c)^3 + 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*A*b^3*tan(1/2*d*x
+ 1/2*c)^3 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^3 - 8*C*a^3*tan(1/2*d*x + 1/2*c) - 12*C*a^2*b*tan(1/2*d*x + 1/2*c)
- 24*A*a*b^2*tan(1/2*d*x + 1/2*c) - 24*C*a*b^2*tan(1/2*d*x + 1/2*c) - 4*A*b^3*tan(1/2*d*x + 1/2*c) - 5*C*b^3*t
an(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

Mupad [B] (verification not implemented)

Time = 17.89 (sec) , antiderivative size = 1547, normalized size of antiderivative = 9.26 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

int((A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^3,x)

[Out]

((3*A*a^3*atan((64*A^2*a^6*sin(c/2 + (d*x)/2) + 16*A^2*b^6*sin(c/2 + (d*x)/2) + 9*C^2*b^6*sin(c/2 + (d*x)/2) +
192*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A^2*a^4*b^2*sin(c/2 + (d*x)/2) + 72*C^2*a^2*b^4*sin(c/2 + (d*x)/2) +
144*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*b^6*sin(c/2 + (d*x)/2) + 240*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 576
*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(64*A^2*a^6 + 16*A^2*b^6 + 9*C^2*b^6 + 192*A^2*a^2*b^4 +
576*A^2*a^4*b^2 + 72*C^2*a^2*b^4 + 144*C^2*a^4*b^2 + 24*A*C*b^6 + 240*A*C*a^2*b^4 + 576*A*C*a^4*b^2))))/4 + (3
*A*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 + (9*C*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
)/32 + (A*b^3*sin(3*c + 3*d*x))/8 + (C*a^3*sin(2*c + 2*d*x))/4 + (C*a^3*sin(4*c + 4*d*x))/8 + (3*C*b^3*sin(3*c
+ 3*d*x))/32 + (A*b^3*sin(c + d*x))/8 + (11*C*b^3*sin(c + d*x))/32 + (3*C*a^2*b*sin(c + d*x))/8 + (9*A*a^2*b*
atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (9*C*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 +
(3*A*a*b^2*sin(2*c + 2*d*x))/4 + (3*A*a*b^2*sin(4*c + 4*d*x))/8 + C*a*b^2*sin(2*c + 2*d*x) + (3*C*a^2*b*sin(3
*c + 3*d*x))/8 + (C*a*b^2*sin(4*c + 4*d*x))/4 + A*a^3*atan((64*A^2*a^6*sin(c/2 + (d*x)/2) + 16*A^2*b^6*sin(c/2
+ (d*x)/2) + 9*C^2*b^6*sin(c/2 + (d*x)/2) + 192*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A^2*a^4*b^2*sin(c/2 + (d
*x)/2) + 72*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 144*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*b^6*sin(c/2 + (d*x)/2
) + 240*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(64*A^2*a^6 +
16*A^2*b^6 + 9*C^2*b^6 + 192*A^2*a^2*b^4 + 576*A^2*a^4*b^2 + 72*C^2*a^2*b^4 + 144*C^2*a^4*b^2 + 24*A*C*b^6 +
240*A*C*a^2*b^4 + 576*A*C*a^4*b^2)))*cos(2*c + 2*d*x) + (A*a^3*atan((64*A^2*a^6*sin(c/2 + (d*x)/2) + 16*A^2*b^
6*sin(c/2 + (d*x)/2) + 9*C^2*b^6*sin(c/2 + (d*x)/2) + 192*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A^2*a^4*b^2*sin
(c/2 + (d*x)/2) + 72*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 144*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*b^6*sin(c/2
+ (d*x)/2) + 240*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(64*
A^2*a^6 + 16*A^2*b^6 + 9*C^2*b^6 + 192*A^2*a^2*b^4 + 576*A^2*a^4*b^2 + 72*C^2*a^2*b^4 + 144*C^2*a^4*b^2 + 24*A
*C*b^6 + 240*A*C*a^2*b^4 + 576*A*C*a^4*b^2)))*cos(4*c + 4*d*x))/4 + (A*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 +
(d*x)/2))*cos(2*c + 2*d*x))/2 + (A*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/8 + (3*C
*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/8 + (3*C*b^3*atanh(sin(c/2 + (d*x)/2)/cos(
c/2 + (d*x)/2))*cos(4*c + 4*d*x))/32 + 3*A*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)
+ (3*A*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/4 + (3*C*a^2*b*atanh(sin(c/2 + (d
*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/2 + (3*C*a^2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4
*c + 4*d*x))/8)/(d*(cos(2*c + 2*d*x)/2 + cos(4*c + 4*d*x)/8 + 3/8))