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

Optimal. Leaf size=263 \[ -\frac{\left (4 a^2 A b+a^3 (-B)-4 a b^2 B+A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{\left (2 a^3 A b-10 a^2 b^2 B+a^4 B+13 a A b^3-6 b^4 B\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\left (2 a^2 A b+a^3 B-6 a b^2 B+3 A b^3\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac{a (A b-a B) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]

[Out]

-(((4*a^2*A*b + A*b^3 - a^3*B - 4*a*b^2*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)
*(a + b)^(7/2)*d)) + (a*(A*b - a*B)*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) + ((2*a^2*A*b + 3
*A*b^3 + a^3*B - 6*a*b^2*B)*Tan[c + d*x])/(6*b*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) + ((2*a^3*A*b + 13*a*A*
b^3 + a^4*B - 10*a^2*b^2*B - 6*b^4*B)*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 = 0.615023, antiderivative size = 263, 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 = {4009, 4003, 12, 3831, 2659, 208} \[ -\frac{\left (4 a^2 A b+a^3 (-B)-4 a b^2 B+A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{\left (2 a^3 A b-10 a^2 b^2 B+a^4 B+13 a A b^3-6 b^4 B\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\left (2 a^2 A b+a^3 B-6 a b^2 B+3 A b^3\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac{a (A b-a B) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((4*a^2*A*b + A*b^3 - a^3*B - 4*a*b^2*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)
*(a + b)^(7/2)*d)) + (a*(A*b - a*B)*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) + ((2*a^2*A*b + 3
*A*b^3 + a^3*B - 6*a*b^2*B)*Tan[c + d*x])/(6*b*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) + ((2*a^3*A*b + 13*a*A*
b^3 + a^4*B - 10*a^2*b^2*B - 6*b^4*B)*Tan[c + d*x])/(6*b*(a^2 - b^2)^3*d*(a + b*Sec[c + d*x]))

Rule 4009

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

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 ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx &=\frac{a (A b-a B) \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 b (A b-a B)+\left (2 a A b+a^2 B-3 b^2 B\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )}\\ &=\frac{a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\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 b \left (5 a A b-2 a^2 B-3 b^2 B\right )-\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b \left (a^2-b^2\right )^2}\\ &=\frac{a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\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 b \left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b \left (a^2-b^2\right )^3}\\ &=\frac{a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac{a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^3}\\ &=\frac{a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\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{\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\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{a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.15707, size = 252, normalized size = 0.96 \[ \frac{\frac{2 \sin (c+d x) \left (a \left (-10 a^2 A b^2-6 a^4 A+13 a^3 b B+2 a b^3 B+A b^4\right ) \cos (2 (c+d x))-6 \left (9 a^2 A b^3+2 a^4 A b-9 a^3 b^2 B+a^5 B-2 a b^4 B-A b^5\right ) \cos (c+d x)-14 a^3 A b^2-6 a^5 A+22 a^2 b^3 B+11 a^4 b B-25 a A b^4+12 b^5 B\right )}{(a \cos (c+d x)+b)^3}+\frac{24 \left (-4 a^2 A b+a^3 B+4 a b^2 B-A b^3\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{24 d \left (b^2-a^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((24*(-4*a^2*A*b - A*b^3 + a^3*B + 4*a*b^2*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 -
b^2] + (2*(-6*a^5*A - 14*a^3*A*b^2 - 25*a*A*b^4 + 11*a^4*b*B + 22*a^2*b^3*B + 12*b^5*B - 6*(2*a^4*A*b + 9*a^2
*A*b^3 - A*b^5 + a^5*B - 9*a^3*b^2*B - 2*a*b^4*B)*Cos[c + d*x] + a*(-6*a^4*A - 10*a^2*A*b^2 + A*b^4 + 13*a^3*b
*B + 2*a*b^3*B)*Cos[2*(c + d*x)])*Sin[c + d*x])/(b + a*Cos[c + d*x])^3)/(24*(-a^2 + b^2)^3*d)

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Maple [A]  time = 0.089, size = 388, 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 ( 2\,A{a}^{3}+2\,A{a}^{2}b+6\,Aa{b}^{2}+A{b}^{3}-B{a}^{3}-6\,B{a}^{2}b-2\,Ba{b}^{2}-2\,B{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 ( 3\,A{a}^{3}+7\,Aa{b}^{2}-7\,B{a}^{2}b-3\,B{b}^{3} \right ) \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 ( 2\,A{a}^{3}-2\,A{a}^{2}b+6\,Aa{b}^{2}-A{b}^{3}+B{a}^{3}-6\,B{a}^{2}b+2\,Ba{b}^{2}-2\,B{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{4\,A{a}^{2}b+A{b}^{3}-B{a}^{3}-4\,Ba{b}^{2}}{{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)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x)

[Out]

1/d*(2*(-1/2*(2*A*a^3+2*A*a^2*b+6*A*a*b^2+A*b^3-B*a^3-6*B*a^2*b-2*B*a*b^2-2*B*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*(3*A*a^3+7*A*a*b^2-7*B*a^2*b-3*B*b^3)/(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*(2*A*a^3-2*A*a^2*b+6*A*a*b^2-A*b^3+B*a^3-6*B*a^2*b+2*B*a*b^2-2*B*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-(4*A*a^2*b+A*b^3-B*a^3-4*B*a*
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)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.844989, size = 2715, normalized size = 10.32 \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)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

[1/12*(3*(B*a^3*b^3 - 4*A*a^2*b^4 + 4*B*a*b^5 - A*b^6 + (B*a^6 - 4*A*a^5*b + 4*B*a^4*b^2 - A*a^3*b^3)*cos(d*x
+ c)^3 + 3*(B*a^5*b - 4*A*a^4*b^2 + 4*B*a^3*b^3 - A*a^2*b^4)*cos(d*x + c)^2 + 3*(B*a^4*b^2 - 4*A*a^3*b^3 + 4*B
*a^2*b^4 - A*a*b^5)*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*s
qrt(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*(B*a^6*b + 2*A*a^5*b^2 - 11*B*a^4*b^3 + 11*A*a^3*b^4 + 4*B*a^2*b^5 - 13*A*a*b^6 + 6*B*b^7 + (6*A*a^7 -
13*B*a^6*b + 4*A*a^5*b^2 + 11*B*a^4*b^3 - 11*A*a^3*b^4 + 2*B*a^2*b^5 + A*a*b^6)*cos(d*x + c)^2 + 3*(B*a^7 + 2*
A*a^6*b - 10*B*a^5*b^2 + 7*A*a^4*b^3 + 7*B*a^3*b^4 - 10*A*a^2*b^5 + 2*B*a*b^6 + A*b^7)*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), 1/6*(3*(B*a^3*b^3 - 4*A*a^2*b^4 + 4*B*a*b^5 -
A*b^6 + (B*a^6 - 4*A*a^5*b + 4*B*a^4*b^2 - A*a^3*b^3)*cos(d*x + c)^3 + 3*(B*a^5*b - 4*A*a^4*b^2 + 4*B*a^3*b^3
- A*a^2*b^4)*cos(d*x + c)^2 + 3*(B*a^4*b^2 - 4*A*a^3*b^3 + 4*B*a^2*b^4 - A*a*b^5)*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))) + (B*a^6*b + 2*A*a^5*b^2 - 11*B*
a^4*b^3 + 11*A*a^3*b^4 + 4*B*a^2*b^5 - 13*A*a*b^6 + 6*B*b^7 + (6*A*a^7 - 13*B*a^6*b + 4*A*a^5*b^2 + 11*B*a^4*b
^3 - 11*A*a^3*b^4 + 2*B*a^2*b^5 + A*a*b^6)*cos(d*x + c)^2 + 3*(B*a^7 + 2*A*a^6*b - 10*B*a^5*b^2 + 7*A*a^4*b^3
+ 7*B*a^3*b^4 - 10*A*a^2*b^5 + 2*B*a*b^6 + A*b^7)*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 + B \sec{\left (c + d x \right )}\right ) \sec ^{2}{\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)**2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**4,x)

[Out]

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

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Giac [B]  time = 1.58856, size = 980, normalized size = 3.73 \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)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/3*(3*(B*a^3 - 4*A*a^2*b + 4*B*a*b^2 - A*b^3)*(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)) - (6*A*a^5*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^5*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^4*b*tan(1/2*d*x + 1/2*c)
^5 - 12*B*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 27*B*a^3*b^2*tan(1/2*d*x + 1/2*
c)^5 - 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 12*A*a*b^4*tan(1/2*d*x + 1/
2*c)^5 + 6*B*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*A*b^5*tan(1/2*d*x + 1/2*c)^5 - 6*B*b^5*tan(1/2*d*x + 1/2*c)^5 -
12*A*a^5*tan(1/2*d*x + 1/2*c)^3 + 28*B*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 16*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 16
*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 28*A*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 12*B*b^5*tan(1/2*d*x + 1/2*c)^3 + 6*A*
a^5*tan(1/2*d*x + 1/2*c) + 3*B*a^5*tan(1/2*d*x + 1/2*c) + 6*A*a^4*b*tan(1/2*d*x + 1/2*c) - 12*B*a^4*b*tan(1/2*
d*x + 1/2*c) + 12*A*a^3*b^2*tan(1/2*d*x + 1/2*c) - 27*B*a^3*b^2*tan(1/2*d*x + 1/2*c) + 27*A*a^2*b^3*tan(1/2*d*
x + 1/2*c) - 12*B*a^2*b^3*tan(1/2*d*x + 1/2*c) + 12*A*a*b^4*tan(1/2*d*x + 1/2*c) - 6*B*a*b^4*tan(1/2*d*x + 1/2
*c) - 3*A*b^5*tan(1/2*d*x + 1/2*c) - 6*B*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