3.256 \(\int \frac{\sec ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=108 \[ -\frac{a^3 \log (a \cos (c+d x)+b)}{b^2 d \left (a^2-b^2\right )}+\frac{a \log (\cos (c+d x))}{b^2 d}+\frac{\log (1-\cos (c+d x))}{2 d (a+b)}+\frac{\log (\cos (c+d x)+1)}{2 d (a-b)}+\frac{\sec (c+d x)}{b d} \]

[Out]

Log[1 - Cos[c + d*x]]/(2*(a + b)*d) + (a*Log[Cos[c + d*x]])/(b^2*d) + Log[1 + Cos[c + d*x]]/(2*(a - b)*d) - (a
^3*Log[b + a*Cos[c + d*x]])/(b^2*(a^2 - b^2)*d) + Sec[c + d*x]/(b*d)

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Rubi [A]  time = 0.284128, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4397, 2837, 12, 894} \[ -\frac{a^3 \log (a \cos (c+d x)+b)}{b^2 d \left (a^2-b^2\right )}+\frac{a \log (\cos (c+d x))}{b^2 d}+\frac{\log (1-\cos (c+d x))}{2 d (a+b)}+\frac{\log (\cos (c+d x)+1)}{2 d (a-b)}+\frac{\sec (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]

[Out]

Log[1 - Cos[c + d*x]]/(2*(a + b)*d) + (a*Log[Cos[c + d*x]])/(b^2*d) + Log[1 + Cos[c + d*x]]/(2*(a - b)*d) - (a
^3*Log[b + a*Cos[c + d*x]])/(b^2*(a^2 - b^2)*d) + Sec[c + d*x]/(b*d)

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

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

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.292099, size = 92, normalized size = 0.85 \[ \frac{\frac{a^3 \log (a \cos (c+d x)+b)}{b^4-a^2 b^2}+\frac{a \log (\cos (c+d x))}{b^2}+\frac{\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a+b}+\frac{\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a-b}+\frac{\sec (c+d x)}{b}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]

[Out]

(Log[Cos[(c + d*x)/2]]/(a - b) + (a*Log[Cos[c + d*x]])/b^2 + (a^3*Log[b + a*Cos[c + d*x]])/(-(a^2*b^2) + b^4)
+ Log[Sin[(c + d*x)/2]]/(a + b) + Sec[c + d*x]/b)/d

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Maple [A]  time = 0.132, size = 110, normalized size = 1. \begin{align*}{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{d \left ( 2\,a-2\,b \right ) }}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{d \left ( 2\,a+2\,b \right ) }}+{\frac{a\ln \left ( \cos \left ( dx+c \right ) \right ) }{{b}^{2}d}}+{\frac{1}{db\cos \left ( dx+c \right ) }}-{\frac{{a}^{3}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) \left ( a-b \right ){b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

1/d/(2*a-2*b)*ln(cos(d*x+c)+1)+1/d/(2*a+2*b)*ln(-1+cos(d*x+c))+a*ln(cos(d*x+c))/b^2/d+1/d/b/cos(d*x+c)-1/d*a^3
/(a+b)/(a-b)/b^2*ln(b+a*cos(d*x+c))

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Maxima [A]  time = 1.0753, size = 213, normalized size = 1.97 \begin{align*} -\frac{\frac{a^{3} \log \left (a + b - \frac{{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} b^{2} - b^{4}} - \frac{a \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b^{2}} - \frac{a \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b^{2}} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} - \frac{2}{b - \frac{b \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

-(a^3*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^2*b^2 - b^4) - a*log(sin(d*x + c)/(cos(d*x +
c) + 1) + 1)/b^2 - a*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/b^2 - log(sin(d*x + c)/(cos(d*x + c) + 1))/(a +
b) - 2/(b - b*sin(d*x + c)^2/(cos(d*x + c) + 1)^2))/d

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Fricas [A]  time = 0.911594, size = 360, normalized size = 3.33 \begin{align*} -\frac{2 \, a^{3} \cos \left (d x + c\right ) \log \left (a \cos \left (d x + c\right ) + b\right ) - 2 \, a^{2} b + 2 \, b^{3} - 2 \,{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) -{\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right ) \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (a b^{2} - b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(2*a^3*cos(d*x + c)*log(a*cos(d*x + c) + b) - 2*a^2*b + 2*b^3 - 2*(a^3 - a*b^2)*cos(d*x + c)*log(-cos(d*x
+ c)) - (a*b^2 + b^3)*cos(d*x + c)*log(1/2*cos(d*x + c) + 1/2) - (a*b^2 - b^3)*cos(d*x + c)*log(-1/2*cos(d*x
+ c) + 1/2))/((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 ^{3}{\left (c + d x \right )}}{a \sin{\left (c + d x \right )} + b \tan{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

Integral(sec(c + d*x)**3/(a*sin(c + d*x) + b*tan(c + d*x)), x)

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Giac [A]  time = 1.26384, size = 257, normalized size = 2.38 \begin{align*} -\frac{\frac{2 \, a^{3} \log \left ({\left | -a - b - \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{2} b^{2} - b^{4}} - \frac{\log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b} - \frac{2 \, a \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{b^{2}} + \frac{2 \,{\left (a - 2 \, b + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{b^{2}{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(2*a^3*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
))/(a^2*b^2 - b^4) - log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b) - 2*a*log(abs(-(cos(d*x + c) -
1)/(cos(d*x + c) + 1) - 1))/b^2 + 2*(a - 2*b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/(b^2*((cos(d*x + c) -
1)/(cos(d*x + c) + 1) + 1)))/d