3.1.85 \(\int \sec ^2(c+d x) (a+a \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [85]

Optimal. Leaf size=117 \[ \frac {a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (3 A+2 C) \tan (c+d x)}{3 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4162, 4132, 3853, 3855, 4131, 3852, 8} \begin {gather*} \frac {a (3 A+2 C) \tan (c+d x)}{3 d}+\frac {a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*A + 3*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(3*A + 2*C)*Tan[c + d*x])/(3*d) + (a*(4*A + 3*C)*Sec[c + d*x]
*Tan[c + d*x])/(8*d) + (a*C*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (a*C*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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

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

Rule 4131

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

Rule 4132

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

Rule 4162

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 2))), x] + Dist[1
/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + b*(C*(n + 1) + A*(n + 2))*Csc[e + f*x] + a*C*(n + 2)*Csc[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int \sec ^2(c+d x) \left (4 a A+a (4 A+3 C) \sec (c+d x)+4 a C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int \sec ^2(c+d x) \left (4 a A+4 a C \sec ^2(c+d x)\right ) \, dx+\frac {1}{4} (a (4 A+3 C)) \int \sec ^3(c+d x) \, dx\\ &=\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{3} (a (3 A+2 C)) \int \sec ^2(c+d x) \, dx+\frac {1}{8} (a (4 A+3 C)) \int \sec (c+d x) \, dx\\ &=\frac {a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {(a (3 A+2 C)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (3 A+2 C) \tan (c+d x)}{3 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]
time = 0.47, size = 75, normalized size = 0.64 \begin {gather*} \frac {a \left (3 (4 A+3 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (24 (A+C)+3 (4 A+3 C) \sec (c+d x)+6 C \sec ^3(c+d x)+8 C \tan ^2(c+d x)\right )\right )}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3*(4*A + 3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(24*(A + C) + 3*(4*A + 3*C)*Sec[c + d*x] + 6*C*Sec[c +
d*x]^3 + 8*C*Tan[c + d*x]^2)))/(24*d)

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Maple [A]
time = 0.64, size = 118, normalized size = 1.01

method	result	size
derivativedivides	\(\frac {A a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A a \tan \left (d x +c \right )-a C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(118\)
default	\(\frac {A a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A a \tan \left (d x +c \right )-a C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(118\)
norman	\(\frac {-\frac {a \left (4 A +3 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (12 A +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (60 A +49 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {a \left (84 A +31 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {a \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(163\)
risch	\(-\frac {i a \left (12 A \,{\mathrm e}^{7 i \left (d x +c \right )}+9 C \,{\mathrm e}^{7 i \left (d x +c \right )}-24 A \,{\mathrm e}^{6 i \left (d x +c \right )}+12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+33 C \,{\mathrm e}^{5 i \left (d x +c \right )}-72 A \,{\mathrm e}^{4 i \left (d x +c \right )}-48 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-33 C \,{\mathrm e}^{3 i \left (d x +c \right )}-72 A \,{\mathrm e}^{2 i \left (d x +c \right )}-64 C \,{\mathrm e}^{2 i \left (d x +c \right )}-12 \,{\mathrm e}^{i \left (d x +c \right )} A -9 C \,{\mathrm e}^{i \left (d x +c \right )}-24 A -16 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(265\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 152, normalized size = 1.30 \begin {gather*} \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a - 3 \, C a {\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 \, A a {\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 \tan \left (d x + c\right )}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a - 3*C*a*(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*A*a*(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*tan(d*x + c))/d

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Fricas [A]
time = 2.56, size = 129, normalized size = 1.10 \begin {gather*} \frac {3 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (3 \, A + 2 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, C a \cos \left (d x + c\right ) + 6 \, C a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(4*A + 3*C)*a*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*A + 3*C)*a*cos(d*x + c)^4*log(-sin(d*x + c)
+ 1) + 2*(8*(3*A + 2*C)*a*cos(d*x + c)^3 + 3*(4*A + 3*C)*a*cos(d*x + c)^2 + 8*C*a*cos(d*x + c) + 6*C*a)*sin(d*
x + c))/(d*cos(d*x + c)^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(A*sec(c + d*x)**2, x) + Integral(A*sec(c + d*x)**3, x) + Integral(C*sec(c + d*x)**4, x) + Integral
(C*sec(c + d*x)**5, x))

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Giac [A]
time = 0.48, size = 188, normalized size = 1.61 \begin {gather*} \frac {3 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 60 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 49 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 84 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 31 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 39 \, C a \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}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(4*A*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*A*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) - 2*(12*A*a*tan(1/2*d*x + 1/2*c)^7 + 9*C*a*tan(1/2*d*x + 1/2*c)^7 - 60*A*a*tan(1/2*d*x + 1/2*c)^5 - 49*C*a
*tan(1/2*d*x + 1/2*c)^5 + 84*A*a*tan(1/2*d*x + 1/2*c)^3 + 31*C*a*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*tan(1/2*d*x +
1/2*c) - 39*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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Mupad [B]
time = 5.07, size = 166, normalized size = 1.42 \begin {gather*} \frac {\left (-A\,a-\frac {3\,C\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (5\,A\,a+\frac {49\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-7\,A\,a-\frac {31\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+\frac {13\,C\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,C\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(3*A*a + (13*C*a)/4) - tan(c/2 + (d*x)/2)^7*(A*a + (3*C*a)/4) - tan(c/2 + (d*x)/2)^3*(7*A*
a + (31*C*a)/12) + tan(c/2 + (d*x)/2)^5*(5*A*a + (49*C*a)/12))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/
2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (a*atanh(tan(c/2 + (d*x)/2))*(4*A + 3*C))/(4*d)

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