3.811 \(\int \frac{B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=164 \[ \frac{\left (2 a^2 B-3 a b C+b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac{(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]

[Out]

((2*a^2*B + b^2*B - 3*a*b*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)
*d) - ((b*B - a*C)*Tan[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - ((3*a*b*B - a^2*C - 2*b^2*C)*Tan[c
+ d*x])/(2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.264986, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {4060, 12, 3831, 2659, 208} \[ \frac{\left (2 a^2 B-3 a b C+b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac{(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*B + b^2*B - 3*a*b*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)
*d) - ((b*B - a*C)*Tan[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - ((3*a*b*B - a^2*C - 2*b^2*C)*Tan[c
+ d*x])/(2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

Rule 4060

Int[((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))/(a*f*(m + 1
)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m +
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, e, f, A, B, C}, x] && 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{B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx &=-\frac{(b B-a C) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{-2 a (a B-b C) \sec (c+d x)+a (b B-a C) \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=-\frac{(b B-a C) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 a b B-a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{a^2 \left (2 a^2 B+b^2 B-3 a b C\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{(b B-a C) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 a b B-a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (2 a^2 B+b^2 B-3 a b C\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{(b B-a C) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 a b B-a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (2 a^2 B+b^2 B-3 a b C\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=-\frac{(b B-a C) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 a b B-a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (2 a^2 B+b^2 B-3 a b C\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 )^2 d}\\ &=\frac{\left (2 a^2 B+b^2 B-3 a b C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac{(b B-a C) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 a b B-a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.840085, size = 172, normalized size = 1.05 \[ \frac{\frac{\left (-4 a^2 b B+2 a^3 C+a b^2 C+b^3 B\right ) \sin (c+d x)}{a (a-b)^2 (a+b)^2 (a \cos (c+d x)+b)}-\frac{2 \left (2 a^2 B-3 a b C+b^2 B\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{b (b B-a C) \sin (c+d x)}{a (a-b) (a+b) (a \cos (c+d x)+b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.091, size = 236, 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 ) ^{2}} \left ( -1/2\,{\frac{ \left ( 4\,Bab+B{b}^{2}-2\,{a}^{2}C-abC-2\,{b}^{2}C \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ \left ( a-b \right ) \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+1/2\,{\frac{ \left ( 4\,Bab-B{b}^{2}-2\,{a}^{2}C+abC-2\,{b}^{2}C \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( a+b \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }} \right ) }+{\frac{2\,B{a}^{2}+B{b}^{2}-3\,abC}{{a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4}}{\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((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.651, size = 1631, normalized size = 9.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((2*B*a^2*b^2 - 3*C*a*b^3 + B*b^4 + (2*B*a^4 - 3*C*a^3*b + B*a^2*b^2)*cos(d*x + c)^2 + 2*(2*B*a^3*b - 3*C
*a^2*b^2 + B*a*b^3)*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*(C*a^4*b - 3*B*a^3*b^2 + C*a^2*b^3 + 3*B*a*b^4 - 2*C*b^5 + (2*C*a^5 - 4*B*a^4*b - C*a^3*b^2 + 5*B*a^2*b
^3 - C*a*b^4 - B*b^5)*cos(d*x + c))*sin(d*x + c))/((a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^2 +
2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d), 1/2*((2
*B*a^2*b^2 - 3*C*a*b^3 + B*b^4 + (2*B*a^4 - 3*C*a^3*b + B*a^2*b^2)*cos(d*x + c)^2 + 2*(2*B*a^3*b - 3*C*a^2*b^2
+ B*a*b^3)*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^4*b - 3*B*a^3*b^2 + C*a^2*b^3 + 3*B*a*b^4 - 2*C*b^5 + (2*C*a^5 - 4*B*a^4*b - C*a^3*b^2 + 5*B*a^2
*b^3 - C*a*b^4 - B*b^5)*cos(d*x + c))*sin(d*x + c))/((a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^2
+ 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)/(a + b*sec(c + d*x))**3, x)

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Giac [B]  time = 1.42175, size = 539, normalized size = 3.29 \begin{align*} \frac{\frac{{\left (2 \, B a^{2} - 3 \, C a b + B b^{2}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{2 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*B*a^2 - 3*C*a*b + B*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^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) - (2*C*a^3*tan(1/2*
d*x + 1/2*c)^3 - 4*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 3*B*a*b^2*tan(1/2*d*x + 1
/2*c)^3 + C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + B*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*C*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*C
*a^3*tan(1/2*d*x + 1/2*c) + 4*B*a^2*b*tan(1/2*d*x + 1/2*c) - C*a^2*b*tan(1/2*d*x + 1/2*c) + 3*B*a*b^2*tan(1/2*
d*x + 1/2*c) - C*a*b^2*tan(1/2*d*x + 1/2*c) - B*b^3*tan(1/2*d*x + 1/2*c) - 2*C*b^3*tan(1/2*d*x + 1/2*c))/((a^4
- 2*a^2*b^2 + b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^2))/d