3.338 \(\int \frac{\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=274 \[ \frac{\left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}+\frac{a \left (a^2 A b-4 a^3 B+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac{\left (-10 a^2 A b^3+a^4 A b-5 a^3 b^2 B+2 a^5 B+18 a b^4 B-6 A b^5\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))} \]

[Out]

((a^3*A + 4*a*A*b^2 - 3*a^2*b*B - 2*b^3*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)
*(a + b)^(7/2)*d) - (a^2*(A*b - a*B)*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) + (a*(a^2*A*b
- 6*A*b^3 - 4*a^3*B + 9*a*b^2*B)*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) + ((a^4*A*b - 10
*a^2*A*b^3 - 6*A*b^5 + 2*a^5*B - 5*a^3*b^2*B + 18*a*b^4*B)*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^3*d*(a + b*Sec[c +
d*x]))

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Rubi [A]  time = 0.699892, antiderivative size = 274, normalized size of antiderivative = 1., 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 = {4028, 4080, 4003, 12, 3831, 2659, 208} \[ \frac{\left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}+\frac{a \left (a^2 A b-4 a^3 B+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac{\left (-10 a^2 A b^3+a^4 A b-5 a^3 b^2 B+2 a^5 B+18 a b^4 B-6 A b^5\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^4,x]

[Out]

((a^3*A + 4*a*A*b^2 - 3*a^2*b*B - 2*b^3*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)
*(a + b)^(7/2)*d) - (a^2*(A*b - a*B)*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^3) + (a*(a^2*A*b
- 6*A*b^3 - 4*a^3*B + 9*a*b^2*B)*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) + ((a^4*A*b - 10
*a^2*A*b^3 - 6*A*b^5 + 2*a^5*B - 5*a^3*b^2*B + 18*a*b^4*B)*Tan[c + d*x])/(6*b^2*(a^2 - b^2)^3*d*(a + b*Sec[c +
d*x]))

Rule 4028

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(a^2*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2))
, x] + Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[a*b*(A*b - a*B)*(m
+ 1) - (A*b - a*B)*(a^2 + b^2*(m + 1))*Csc[e + f*x] + b*B*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 4080

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((A*b^2 - a*b*B + 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[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, 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 ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx &=-\frac{a^2 (A b-a B) \tan (c+d x)}{3 b^2 \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 b-a B)-\left (a^2-3 b^2\right ) (A b-a B) \sec (c+d x)-3 b \left (a^2-b^2\right ) B \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac{a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \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^2 \left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right )+b \left (a^3 A b-6 a A b^3+2 a^4 B-3 a^2 b^2 B+6 b^4 B\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\int -\frac{3 b^3 \left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 B\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )^3}\\ &=-\frac{a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 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^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 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^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\left (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 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 (a^3 A+4 a A b^2-3 a^2 b B-2 b^3 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^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a \left (a^2 A b-6 A b^3-4 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\left (a^4 A b-10 a^2 A b^3-6 A b^5+2 a^5 B-5 a^3 b^2 B+18 a b^4 B\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.23161, size = 226, normalized size = 0.82 \[ \frac{\frac{\left (-13 a^2 A b+4 a^3 B+11 a b^2 B-2 A b^3\right ) \sin (c+d x)}{(a-b)^3 (a+b)^3 (a \cos (c+d x)+b)}+\frac{\left (3 a^2 A-5 a b B+2 A b^2\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a \cos (c+d x)+b)^2}-\frac{6 \left (a^3 A-3 a^2 b B+4 a A b^2-2 b^3 B\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac{2 (a B-A b) \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)^3}}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^4,x]

[Out]

((-6*(a^3*A + 4*a*A*b^2 - 3*a^2*b*B - 2*b^3*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^
2)^(7/2) + (2*(-(A*b) + a*B)*Sin[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])^3) + ((3*a^2*A + 2*A*b^2 - 5*
a*b*B)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])^2) + ((-13*a^2*A*b - 2*A*b^3 + 4*a^3*B + 11*a*b
^2*B)*Sin[c + d*x])/((a - b)^3*(a + b)^3*(b + a*Cos[c + d*x])))/(6*d)

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Maple [A]  time = 0.094, size = 375, normalized size = 1.4 \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 ( A{a}^{3}+6\,A{a}^{2}b+2\,Aa{b}^{2}+2\,A{b}^{3}-2\,B{a}^{3}-3\,B{a}^{2}b-6\,Ba{b}^{2} \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 ( 7\,A{a}^{2}b+3\,A{b}^{3}-B{a}^{3}-9\,Ba{b}^{2} \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 ( A{a}^{3}-6\,A{a}^{2}b+2\,Aa{b}^{2}-2\,A{b}^{3}+2\,B{a}^{3}-3\,B{a}^{2}b+6\,Ba{b}^{2} \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{a}^{3}+4\,Aa{b}^{2}-3\,B{a}^{2}b-2\,B{b}^{3}}{{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)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x)

[Out]

1/d*(-2*(-1/2*(A*a^3+6*A*a^2*b+2*A*a*b^2+2*A*b^3-2*B*a^3-3*B*a^2*b-6*B*a*b^2)/(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*(7*A*a^2*b+3*A*b^3-B*a^3-9*B*a*b^2)/(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*(A*a^3-6*A*a^2*b+2*A*a*b^2-2*A*b^3+2*B*a^3-3*B*a^2*b+6*B*a*b^2)/(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*a^3+4*A*a*b^2-3*B*a^2*b-2*B*b^3)/(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)^3*(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 = 0.799262, size = 2707, normalized size = 9.88 \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)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

[1/12*(3*(A*a^3*b^3 - 3*B*a^2*b^4 + 4*A*a*b^5 - 2*B*b^6 + (A*a^6 - 3*B*a^5*b + 4*A*a^4*b^2 - 2*B*a^3*b^3)*cos(
d*x + c)^3 + 3*(A*a^5*b - 3*B*a^4*b^2 + 4*A*a^3*b^3 - 2*B*a^2*b^4)*cos(d*x + c)^2 + 3*(A*a^4*b^2 - 3*B*a^3*b^3
+ 4*A*a^2*b^4 - 2*B*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*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*(2*B*a^7 + A*a^6*b - 7*B*a^5*b^2 - 11*A*a^4*b^3 + 23*B*a^3*b^4 + 4*A*a^2*b^5 - 18*B*a*b^6 + 6*A
*b^7 + (4*B*a^7 - 13*A*a^6*b + 7*B*a^5*b^2 + 11*A*a^4*b^3 - 11*B*a^3*b^4 + 2*A*a^2*b^5)*cos(d*x + c)^2 + 3*(A*
a^7 + B*a^6*b - 10*A*a^5*b^2 + 8*B*a^4*b^3 + 7*A*a^3*b^4 - 9*B*a^2*b^5 + 2*A*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), 1/6*(3*(A*a^3*b^3 - 3*B*a^2*b^4 + 4*A*a*b^5 - 2*
B*b^6 + (A*a^6 - 3*B*a^5*b + 4*A*a^4*b^2 - 2*B*a^3*b^3)*cos(d*x + c)^3 + 3*(A*a^5*b - 3*B*a^4*b^2 + 4*A*a^3*b^
3 - 2*B*a^2*b^4)*cos(d*x + c)^2 + 3*(A*a^4*b^2 - 3*B*a^3*b^3 + 4*A*a^2*b^4 - 2*B*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))) + (2*B*a^7 + A*a^6*b - 7*B*
a^5*b^2 - 11*A*a^4*b^3 + 23*B*a^3*b^4 + 4*A*a^2*b^5 - 18*B*a*b^6 + 6*A*b^7 + (4*B*a^7 - 13*A*a^6*b + 7*B*a^5*b
^2 + 11*A*a^4*b^3 - 11*B*a^3*b^4 + 2*A*a^2*b^5)*cos(d*x + c)^2 + 3*(A*a^7 + B*a^6*b - 10*A*a^5*b^2 + 8*B*a^4*b
^3 + 7*A*a^3*b^4 - 9*B*a^2*b^5 + 2*A*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 + B \sec{\left (c + d x \right )}\right ) \sec ^{3}{\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)**3*(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)**3/(a + b*sec(c + d*x))**4, x)

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Giac [B]  time = 1.55525, size = 936, normalized size = 3.42 \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)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/3*(3*(A*a^3 - 3*B*a^2*b + 4*A*a*b^2 - 2*B*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)) + (3*A*a^5*tan(1/2*d*x + 1/2*c)^5 - 6*B*a^5*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^4*b*tan(1/2*d*x + 1/2
*c)^5 + 3*B*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 - 6*B*a^3*b^2*tan(1/2*d*x + 1/2
*c)^5 + 12*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 27*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 6*A*a*b^4*tan(1/2*d*x + 1/
2*c)^5 - 18*B*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 + 4*B*a^5*tan(1/2*d*x + 1/2*c)^3 -
28*A*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 32*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 16*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^3
- 36*B*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^5*tan(1/2*d*x + 1/2*c)^3 - 3*A*a^5*tan(1/2*d*x + 1/2*c) - 6*B*a^
5*tan(1/2*d*x + 1/2*c) + 12*A*a^4*b*tan(1/2*d*x + 1/2*c) - 3*B*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) - 6*B*a^3*b^2*tan(1/2*d*x + 1/2*c) + 12*A*a^2*b^3*tan(1/2*d*x + 1/2*c) - 27*B*a^2*b^3*tan(1/2*
d*x + 1/2*c) + 6*A*a*b^4*tan(1/2*d*x + 1/2*c) - 18*B*a*b^4*tan(1/2*d*x + 1/2*c) + 6*A*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