3.434 \(\int \frac{\sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=84 \[ -\frac{2 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac{\sec (c+d x) (b-a \sin (c+d x))}{d \left (a^2-b^2\right )} \]

[Out]

(-2*b^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(3/2)*d) - (Sec[c + d*x]*(b - a*Sin[c +
d*x]))/((a^2 - b^2)*d)

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Rubi [A]  time = 0.0939213, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2696, 12, 2660, 618, 204} \[ -\frac{2 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac{\sec (c+d x) (b-a \sin (c+d x))}{d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(3/2)*d) - (Sec[c + d*x]*(b - a*Sin[c +
d*x]))/((a^2 - b^2)*d)

Rule 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 12

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.293336, size = 152, normalized size = 1.81 \[ \frac{\sqrt{a^2-b^2} (-a \sin (c+d x)+b (-\cos (c+d x))+b)+2 b^2 \cos (c+d x) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d (b-a) (a+b) \sqrt{a^2-b^2} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*Cos[c + d*x] + Sqrt[a^2 - b^2]*(b - b*Cos[c + d*x] - a
*Sin[c + d*x]))/((-a + b)*(a + b)*Sqrt[a^2 - b^2]*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]))

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Maple [A]  time = 0.046, size = 117, normalized size = 1.4 \begin{align*} -2\,{\frac{1}{d \left ( 2\,a-2\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-2\,{\frac{{b}^{2}}{d \left ( a-b \right ) \left ( a+b \right ) \sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-2\,{\frac{1}{d \left ( 2\,a+2\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+b*sin(d*x+c)),x)

[Out]

-2/d/(2*a-2*b)/(tan(1/2*d*x+1/2*c)+1)-2/d*b^2/(a-b)/(a+b)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2
*b)/(a^2-b^2)^(1/2))-2/d/(2*a+2*b)/(tan(1/2*d*x+1/2*c)-1)

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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*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.84104, size = 684, normalized size = 8.14 \begin{align*} \left [\frac{\sqrt{-a^{2} + b^{2}} b^{2} \cos \left (d x + c\right ) \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \,{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )}, \frac{\sqrt{a^{2} - b^{2}} b^{2} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - a^{2} b + b^{3} +{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(-a^2 + b^2)*b^2*cos(d*x + c)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2
*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a
^2 - b^2)) - 2*a^2*b + 2*b^3 + 2*(a^3 - a*b^2)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*d*cos(d*x + c)), (sqrt(a
^2 - b^2)*b^2*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c) - a^2*b + b^3 + (a^3 -
a*b^2)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*d*cos(d*x + c))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

Integral(sec(c + d*x)**2/(a + b*sin(c + d*x)), x)

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Giac [A]  time = 1.11943, size = 144, normalized size = 1.71 \begin{align*} -\frac{2 \,{\left (\frac{{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} b^{2}}{{\left (a^{2} - b^{2}\right )}^{\frac{3}{2}}} + \frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b}{{\left (a^{2} - b^{2}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*b^2/(a^2
- b^2)^(3/2) + (a*tan(1/2*d*x + 1/2*c) - b)/((a^2 - b^2)*(tan(1/2*d*x + 1/2*c)^2 - 1)))/d