3.122 \(\int \frac{\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=133 \[ \frac{(3 A+4 C) \tan ^3(c+d x)}{3 a d}+\frac{(3 A+4 C) \tan (c+d x)}{a d}-\frac{(2 A+3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}-\frac{(2 A+3 C) \tan (c+d x) \sec (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.178874, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4085, 3787, 3768, 3770, 3767} \[ \frac{(3 A+4 C) \tan ^3(c+d x)}{3 a d}+\frac{(3 A+4 C) \tan (c+d x)}{a d}-\frac{(2 A+3 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}-\frac{(2 A+3 C) \tan (c+d x) \sec (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4085

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(
2*m + 1)), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*C*n + A*b*(
2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x]
&& EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3768

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 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]

Rubi steps

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

Mathematica [B]  time = 6.51847, size = 1090, normalized size = 8.2 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(2*A + 3*C)*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(A + C*Sec[c + d
*x]^2))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) - (2*(2*A + 3*C)*Cos[c/2 + (d*x)/2]^2*Cos[c +
d*x]*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(A + C*Sec[c + d*x]^2))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a
+ a*Sec[c + d*x])) + (4*Cos[c/2 + (d*x)/2]*Cos[c + d*x]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*(A*Sin[(d*x)/2] + C*S
in[(d*x)/2]))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (2*C*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]
*(A + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] - Sin
[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) - (2*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*(A + C*Sec[c + d*x]
^2)*(C*Cos[c/2] - 2*C*Sin[c/2]))/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] - Sin[c/2]
)*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (4*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*(A + C*Sec[c + d*x]^2)*(
3*A*Sin[(d*x)/2] + 5*C*Sin[(d*x)/2]))/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] - Sin
[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (2*C*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*(A + C*Sec[c + d*x]
^2)*Sin[(d*x)/2])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d
*x)/2] + Sin[c/2 + (d*x)/2])^3) + (2*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*(A + C*Sec[c + d*x]^2)*(C*Cos[c/2] + 2*
C*Sin[c/2]))/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2
] + Sin[c/2 + (d*x)/2])^2) + (4*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*(A + C*Sec[c + d*x]^2)*(3*A*Sin[(d*x)/2] + 5
*C*Sin[(d*x)/2]))/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d
*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [B]  time = 0.061, size = 294, normalized size = 2.2 \begin{align*}{\frac{A}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{3\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{C}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{3\,C}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{A}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{5\,C}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{A}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{C}{3\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{C}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{3\,C}{2\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{A}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{5\,C}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{A}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a/d*A*tan(1/2*d*x+1/2*c)+1/a/d*C*tan(1/2*d*x+1/2*c)-1/3/a/d*C/(tan(1/2*d*x+1/2*c)+1)^3+1/a/d/(tan(1/2*d*x+1/
2*c)+1)^2*C-3/2/a/d*ln(tan(1/2*d*x+1/2*c)+1)*C-1/a/d*ln(tan(1/2*d*x+1/2*c)+1)*A-5/2/a/d/(tan(1/2*d*x+1/2*c)+1)
*C-1/a/d/(tan(1/2*d*x+1/2*c)+1)*A-1/3/a/d*C/(tan(1/2*d*x+1/2*c)-1)^3-1/a/d/(tan(1/2*d*x+1/2*c)-1)^2*C+3/2/a/d*
ln(tan(1/2*d*x+1/2*c)-1)*C+1/a/d*ln(tan(1/2*d*x+1/2*c)-1)*A-5/2/a/d/(tan(1/2*d*x+1/2*c)-1)*C-1/a/d/(tan(1/2*d*
x+1/2*c)-1)*A

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Maxima [B]  time = 0.949656, size = 439, normalized size = 3.3 \begin{align*} \frac{C{\left (\frac{2 \,{\left (\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a - \frac{3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac{9 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac{9 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac{6 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 6 \, A{\left (\frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac{2 \, \sin \left (d x + c\right )}{{\left (a - \frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} - \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(C*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a - 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a + 6*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 6*A*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a
- log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - 2*sin(d*x + c)/((a - a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(c
os(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 0.519058, size = 429, normalized size = 3.23 \begin{align*} -\frac{3 \,{\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (6 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} - C \cos \left (d x + c\right ) + 2 \, C\right )} \sin \left (d x + c\right )}{12 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*((2*A + 3*C)*cos(d*x + c)^4 + (2*A + 3*C)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 3*((2*A + 3*C)*cos(
d*x + c)^4 + (2*A + 3*C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(3*A + 4*C)*cos(d*x + c)^3 + (6*A + 7*C
)*cos(d*x + c)^2 - C*cos(d*x + c) + 2*C)*sin(d*x + c))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{3}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{5}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sec(c + d*x)**3/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**5/(sec(c + d*x) + 1), x))/a

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Giac [A]  time = 1.2242, size = 250, normalized size = 1.88 \begin{align*} -\frac{\frac{3 \,{\left (2 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{3 \,{\left (2 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{6 \,{\left (A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a} + \frac{2 \,{\left (6 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, C \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 )}^{3} a}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(2*A + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 3*(2*A + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a
- 6*(A*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a + 2*(6*A*tan(1/2*d*x + 1/2*c)^5 + 15*C*tan(1/2*d*x + 1
/2*c)^5 - 12*A*tan(1/2*d*x + 1/2*c)^3 - 16*C*tan(1/2*d*x + 1/2*c)^3 + 6*A*tan(1/2*d*x + 1/2*c) + 9*C*tan(1/2*d
*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a))/d