3.701 \(\int \frac{\sec (c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=252 \[ \frac{a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{\left (-a^2 b^2 (11 A+10 C)+a^4 C-2 b^4 (2 A+3 C)\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}-\frac{a \left (a^2 (-C)+5 A b^2+6 b^2 C\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]

[Out]

(a*(a^2*(2*A + C) + b^2*(3*A + 4*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a +
b)^(7/2)*d) - ((A*b^2 + a^2*C)*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) - (a*(5*A*b^2 - a^2*C
+ 6*b^2*C)*Tan[c + d*x])/(6*b*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) + ((a^4*C - 2*b^4*(2*A + 3*C) - a^2*b^2*
(11*A + 10*C))*Tan[c + d*x])/(6*b*(a^2 - b^2)^3*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.553413, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4081, 4003, 12, 3831, 2659, 208} \[ \frac{a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{\left (-a^2 b^2 (11 A+10 C)+a^4 C-2 b^4 (2 A+3 C)\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}-\frac{a \left (a^2 (-C)+5 A b^2+6 b^2 C\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^4,x]

[Out]

(a*(a^2*(2*A + C) + b^2*(3*A + 4*C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*(a +
b)^(7/2)*d) - ((A*b^2 + a^2*C)*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) - (a*(5*A*b^2 - a^2*C
+ 6*b^2*C)*Tan[c + d*x])/(6*b*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) + ((a^4*C - 2*b^4*(2*A + 3*C) - a^2*b^2*
(11*A + 10*C))*Tan[c + d*x])/(6*b*(a^2 - b^2)^3*d*(a + b*Sec[c + d*x]))

Rule 4081

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

Rule 4003

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a*B
)*(m + 2)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] &
& LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{\int \frac{\sec (c+d x) \left (-3 a b (A+C)+\left (2 A b^2-a^2 C+3 b^2 C\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{a \left (5 A b^2-a^2 C+6 b^2 C\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\int \frac{\sec (c+d x) \left (2 \left (b^3 (2 A+3 C)+\frac{1}{2} a^2 (6 A b+4 b C)\right )-a \left (5 A b^2-\left (a^2-6 b^2\right ) C\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b \left (a^2-b^2\right )^2}\\ &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{a \left (5 A b^2-a^2 C+6 b^2 C\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\int -\frac{3 a b \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b \left (a^2-b^2\right )^3}\\ &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{a \left (5 A b^2-a^2 C+6 b^2 C\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\left (a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{a \left (5 A b^2-a^2 C+6 b^2 C\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\left (a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^3}\\ &=-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{a \left (5 A b^2-a^2 C+6 b^2 C\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\left (a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b \left (a^2-b^2\right )^3 d}\\ &=\frac{a \left (2 a^2 A+3 A b^2+a^2 C+4 b^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{a \left (5 A b^2-a^2 C+6 b^2 C\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 6.16409, size = 438, normalized size = 1.74 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (\frac{2 b \sec (c) \left (a^2 C+A b^2\right ) (b \sin (c)-a \sin (d x))}{a^5-a^3 b^2}-\frac{6 i a (\cos (c)-i \sin (c)) \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right ) (a \cos (c+d x)+b)^3 \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{\left (a^2-b^2\right )^{7/2} \sqrt{(\cos (c)-i \sin (c))^2}}+\frac{\sec (c) (a \cos (c+d x)+b)^2 \left (3 \sin (c) \left (a^4 b^2 (9 A+4 C)-6 a^2 A b^4+a^6 C+2 A b^6\right )-a b \sin (d x) \left (a^2 b^2 (2 C-5 A)+a^4 (18 A+13 C)+2 A b^4\right )\right )}{\left (a^3-a b^2\right )^3}+\frac{\sec (c) (a \cos (c+d x)+b) \left (\sin (c) \left (-11 a^2 A b^3-5 a^4 b C+6 A b^5\right )+a \sin (d x) \left (a^2 b^2 (9 A+2 C)+3 a^4 C-4 A b^4\right )\right )}{a^3 \left (a^2-b^2\right )^2}\right )}{3 d (a+b \sec (c+d x))^4 (A \cos (2 (c+d x))+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^4,x]

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2)*(((-6*I)*a*(a^2*(2*A + C) + b^2*(3*A + 4*C))*ArcTa
n[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2]
)]*(b + a*Cos[c + d*x])^3*(Cos[c] - I*Sin[c]))/((a^2 - b^2)^(7/2)*Sqrt[(Cos[c] - I*Sin[c])^2]) + (2*b*(A*b^2 +
a^2*C)*Sec[c]*(b*Sin[c] - a*Sin[d*x]))/(a^5 - a^3*b^2) + ((b + a*Cos[c + d*x])*Sec[c]*((-11*a^2*A*b^3 + 6*A*b
^5 - 5*a^4*b*C)*Sin[c] + a*(-4*A*b^4 + 3*a^4*C + a^2*b^2*(9*A + 2*C))*Sin[d*x]))/(a^3*(a^2 - b^2)^2) + ((b + a
*Cos[c + d*x])^2*Sec[c]*(3*(-6*a^2*A*b^4 + 2*A*b^6 + a^6*C + a^4*b^2*(9*A + 4*C))*Sin[c] - a*b*(2*A*b^4 + a^2*
b^2*(-5*A + 2*C) + a^4*(18*A + 13*C))*Sin[d*x]))/(a^3 - a*b^2)^3))/(3*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*
Sec[c + d*x])^4)

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Maple [A]  time = 0.102, size = 373, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) ^{3}} \left ( -1/2\,{\frac{ \left ( 6\,A{a}^{2}b+3\,Aa{b}^{2}+2\,A{b}^{3}+{a}^{3}C+6\,{a}^{2}bC+2\,Ca{b}^{2}+2\,C{b}^{3} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{ \left ( a-b \right ) \left ({a}^{3}+3\,{a}^{2}b+3\,a{b}^{2}+{b}^{3} \right ) }}+2/3\,{\frac{ \left ( 9\,{a}^{2}A+A{b}^{2}+7\,{a}^{2}C+3\,{b}^{2}C \right ) b \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }}-1/2\,{\frac{ \left ( 6\,A{a}^{2}b-3\,Aa{b}^{2}+2\,A{b}^{3}-{a}^{3}C+6\,{a}^{2}bC-2\,Ca{b}^{2}+2\,C{b}^{3} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( a+b \right ) \left ({a}^{3}-3\,{a}^{2}b+3\,a{b}^{2}-{b}^{3} \right ) }} \right ) }+{\frac{a \left ( 2\,{a}^{2}A+3\,A{b}^{2}+{a}^{2}C+4\,{b}^{2}C \right ) }{{a}^{6}-3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}-{b}^{6}}{\it Artanh} \left ({(a-b)\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x)

[Out]

1/d*(-2*(-1/2*(6*A*a^2*b+3*A*a*b^2+2*A*b^3+C*a^3+6*C*a^2*b+2*C*a*b^2+2*C*b^3)/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*
tan(1/2*d*x+1/2*c)^5+2/3*(9*A*a^2+A*b^2+7*C*a^2+3*C*b^2)*b/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^
3-1/2*(6*A*a^2*b-3*A*a*b^2+2*A*b^3-C*a^3+6*C*a^2*b-2*C*a*b^2+2*C*b^3)/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*
d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3+a*(2*A*a^2+3*A*b^2+C*a^2+4*C*b^2)/(a^6-3*a^4
*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.750736, size = 2473, normalized size = 9.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

[-1/12*(3*((2*A + C)*a^3*b^3 + (3*A + 4*C)*a*b^5 + ((2*A + C)*a^6 + (3*A + 4*C)*a^4*b^2)*cos(d*x + c)^3 + 3*((
2*A + C)*a^5*b + (3*A + 4*C)*a^3*b^3)*cos(d*x + c)^2 + 3*((2*A + C)*a^4*b^2 + (3*A + 4*C)*a^2*b^4)*cos(d*x + c
))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c)
+ a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 2*(C*a^6*b - 11*(A + C)*a^
4*b^3 + (7*A + 4*C)*a^2*b^5 + 2*(2*A + 3*C)*b^7 - ((18*A + 13*C)*a^6*b - (23*A + 11*C)*a^4*b^3 + (7*A - 2*C)*a
^2*b^5 - 2*A*b^7)*cos(d*x + c)^2 + 3*(C*a^7 - (9*A + 10*C)*a^5*b^2 + (8*A + 7*C)*a^3*b^4 + (A + 2*C)*a*b^6)*co
s(d*x + c))*sin(d*x + c))/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d*cos(d*x + c)^3 + 3*(a^10*b -
4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^
8 + a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d), 1/6*(3*((2*A + C)*a^3*b^
3 + (3*A + 4*C)*a*b^5 + ((2*A + C)*a^6 + (3*A + 4*C)*a^4*b^2)*cos(d*x + c)^3 + 3*((2*A + C)*a^5*b + (3*A + 4*C
)*a^3*b^3)*cos(d*x + c)^2 + 3*((2*A + C)*a^4*b^2 + (3*A + 4*C)*a^2*b^4)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(
-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (C*a^6*b - 11*(A + C)*a^4*b^3 + (7*A + 4*
C)*a^2*b^5 + 2*(2*A + 3*C)*b^7 - ((18*A + 13*C)*a^6*b - (23*A + 11*C)*a^4*b^3 + (7*A - 2*C)*a^2*b^5 - 2*A*b^7)
*cos(d*x + c)^2 + 3*(C*a^7 - (9*A + 10*C)*a^5*b^2 + (8*A + 7*C)*a^3*b^4 + (A + 2*C)*a*b^6)*cos(d*x + c))*sin(d
*x + c))/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d*cos(d*x + c)^3 + 3*(a^10*b - 4*a^8*b^3 + 6*a^
6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos
(d*x + c) + (a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**4,x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)/(a + b*sec(c + d*x))**4, x)

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Giac [B]  time = 1.31245, size = 936, normalized size = 3.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

-1/3*(3*(2*A*a^3 + C*a^3 + 3*A*a*b^2 + 4*C*a*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a
*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(-
a^2 + b^2)) - (3*C*a^5*tan(1/2*d*x + 1/2*c)^5 + 18*A*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^4*b*tan(1/2*d*x + 1
/2*c)^5 - 27*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 - 27*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*A*a^2*b^3*tan(1/2*d*x
+ 1/2*c)^5 + 12*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*A*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b^4*tan(1/2*d*x +
1/2*c)^5 + 6*A*b^5*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^5*tan(1/2*d*x + 1/2*c)^5 - 36*A*a^4*b*tan(1/2*d*x + 1/2*c)^3
- 28*C*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 32*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 16*C*a^2*b^3*tan(1/2*d*x + 1/2*c)
^3 + 4*A*b^5*tan(1/2*d*x + 1/2*c)^3 + 12*C*b^5*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^5*tan(1/2*d*x + 1/2*c) + 18*A*a^
4*b*tan(1/2*d*x + 1/2*c) + 12*C*a^4*b*tan(1/2*d*x + 1/2*c) + 27*A*a^3*b^2*tan(1/2*d*x + 1/2*c) + 27*C*a^3*b^2*
tan(1/2*d*x + 1/2*c) + 6*A*a^2*b^3*tan(1/2*d*x + 1/2*c) + 12*C*a^2*b^3*tan(1/2*d*x + 1/2*c) + 3*A*a*b^4*tan(1/
2*d*x + 1/2*c) + 6*C*a*b^4*tan(1/2*d*x + 1/2*c) + 6*A*b^5*tan(1/2*d*x + 1/2*c) + 6*C*b^5*tan(1/2*d*x + 1/2*c))
/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^3))/d