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]

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

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Rubi [A]  time = 0.28, antiderivative size = 108, normalized size of antiderivative = 1.00, 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 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]))

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 4397

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

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*}

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Mathematica [A]  time = 0.29, 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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fricas [A]  time = 0.67, size = 147, normalized size = 1.36 \[ -\frac {2 \, a^{3} \cos \left (d x + c\right ) \log \left (a \cos \left (d x + c\right ) + b\right ) - 2 \, a^{2} b + 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - {\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a b^{2} - b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )} \]

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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giac [A]  time = 0.39, size = 190, normalized size = 1.76 \[ -\frac {\frac {2 \, a^{3} \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{2} b^{2} - b^{4}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b} - \frac {2 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{b^{2}} + \frac {2 \, {\left (a - 2 \, b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{b^{2} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}}}{2 \, d} \]

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

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maple [A]  time = 0.20, size = 110, normalized size = 1.02 \[ -\frac {a^{3} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right ) \left (a -b \right ) b^{2}}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{d \left (2 a +2 b \right )}+\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{d \left (2 a -2 b \right )}+\frac {a \ln \left (\cos \left (d x +c \right )\right )}{b^{2} d}+\frac {1}{d b \cos \left (d x +c \right )} \]

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

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maxima [A]  time = 0.44, size = 158, normalized size = 1.46 \[ -\frac {\frac {a^{3} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} b^{2} - b^{4}} - \frac {a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b^{2}} - \frac {a \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b^{2}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} - \frac {2}{b - \frac {b \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{d} \]

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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mupad [B]  time = 0.83, size = 118, normalized size = 1.09 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {2}{b\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}{b^2\,d}-\frac {a^3\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{b^2\,d\,\left (a^2-b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a*sin(c + d*x) + b*tan(c + d*x))),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \]

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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