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

Optimal. Leaf size=145 \[ \frac{2 a b}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{b}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{b \left (3 a^2+b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)^3}+\frac{\log (\sin (c+d x)+1)}{2 d (a-b)^3} \]

[Out]

-Log[1 - Sin[c + d*x]]/(2*(a + b)^3*d) + Log[1 + Sin[c + d*x]]/(2*(a - b)^3*d) - (b*(3*a^2 + b^2)*Log[a + b*Si
n[c + d*x]])/((a^2 - b^2)^3*d) + b/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) + (2*a*b)/((a^2 - b^2)^2*d*(a + b*
Sin[c + d*x]))

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Rubi [A]  time = 0.153961, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2668, 710, 801} \[ \frac{2 a b}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{b}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{b \left (3 a^2+b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)^3}+\frac{\log (\sin (c+d x)+1)}{2 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[1 - Sin[c + d*x]]/(2*(a + b)^3*d) + Log[1 + Sin[c + d*x]]/(2*(a - b)^3*d) - (b*(3*a^2 + b^2)*Log[a + b*Si
n[c + d*x]])/((a^2 - b^2)^3*d) + b/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) + (2*a*b)/((a^2 - b^2)^2*d*(a + b*
Sin[c + d*x]))

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 710

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

Mathematica [A]  time = 0.549629, size = 135, normalized size = 0.93 \[ \frac{b \left (\frac{1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{2 \left (3 a^2+b^2\right ) \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac{4 a}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}-\frac{\log (1-\sin (c+d x))}{b (a+b)^3}+\frac{\log (\sin (c+d x)+1)}{b (a-b)^3}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(-(Log[1 - Sin[c + d*x]]/(b*(a + b)^3)) + Log[1 + Sin[c + d*x]]/((a - b)^3*b) - (2*(3*a^2 + b^2)*Log[a + b*
Sin[c + d*x]])/((a - b)^3*(a + b)^3) + 1/((a^2 - b^2)*(a + b*Sin[c + d*x])^2) + (4*a)/((a - b)^2*(a + b)^2*(a
+ b*Sin[c + d*x]))))/(2*d)

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Maple [A]  time = 0.119, size = 166, normalized size = 1.1 \begin{align*}{\frac{b}{2\,d \left ( a+b \right ) \left ( a-b \right ) \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{ab}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-3\,{\frac{b\ln \left ( a+b\sin \left ( dx+c \right ) \right ){a}^{2}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}-{\frac{{b}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{2\,d \left ( a+b \right ) ^{3}}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{2\, \left ( a-b \right ) ^{3}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.971178, size = 301, normalized size = 2.08 \begin{align*} -\frac{\frac{2 \,{\left (3 \, a^{2} b + b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac{4 \, a b^{2} \sin \left (d x + c\right ) + 5 \, a^{2} b - b^{3}}{a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} +{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sin \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )} - \frac{\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(3*a^2*b + b^3)*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - (4*a*b^2*sin(d*x + c) +
5*a^2*b - b^3)/(a^6 - 2*a^4*b^2 + a^2*b^4 + (a^4*b^2 - 2*a^2*b^4 + b^6)*sin(d*x + c)^2 + 2*(a^5*b - 2*a^3*b^3
+ a*b^5)*sin(d*x + c)) - log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + log(sin(d*x + c) - 1)/(a^3 +
3*a^2*b + 3*a*b^2 + b^3))/d

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Fricas [B]  time = 3.65521, size = 999, normalized size = 6.89 \begin{align*} -\frac{5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5} - 2 \,{\left (3 \, a^{4} b + 4 \, a^{2} b^{3} + b^{5} -{\left (3 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) +{\left (a^{5} + 3 \, a^{4} b + 4 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5} -{\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (a^{5} - 3 \, a^{4} b + 4 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5} -{\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \,{\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} - 2 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \sin \left (d x + c\right ) -{\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(5*a^4*b - 6*a^2*b^3 + b^5 - 2*(3*a^4*b + 4*a^2*b^3 + b^5 - (3*a^2*b^3 + b^5)*cos(d*x + c)^2 + 2*(3*a^3*b
^2 + a*b^4)*sin(d*x + c))*log(b*sin(d*x + c) + a) + (a^5 + 3*a^4*b + 4*a^3*b^2 + 4*a^2*b^3 + 3*a*b^4 + b^5 - (
a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cos(d*x + c)^2 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*sin(d*x + c))*
log(sin(d*x + c) + 1) - (a^5 - 3*a^4*b + 4*a^3*b^2 - 4*a^2*b^3 + 3*a*b^4 - b^5 - (a^3*b^2 - 3*a^2*b^3 + 3*a*b^
4 - b^5)*cos(d*x + c)^2 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*sin(d*x + c))*log(-sin(d*x + c) + 1) + 4*(
a^3*b^2 - a*b^4)*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 - 2*(a^7*b - 3*a^5*b^
3 + 3*a^3*b^5 - a*b^7)*d*sin(d*x + c) - (a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16443, size = 327, normalized size = 2.26 \begin{align*} -\frac{\frac{2 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac{\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{9 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} + 3 \, b^{5} \sin \left (d x + c\right )^{2} + 22 \, a^{3} b^{2} \sin \left (d x + c\right ) + 2 \, a b^{4} \sin \left (d x + c\right ) + 14 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(3*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - log(abs(sin(d*x
+ c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - (9*
a^2*b^3*sin(d*x + c)^2 + 3*b^5*sin(d*x + c)^2 + 22*a^3*b^2*sin(d*x + c) + 2*a*b^4*sin(d*x + c) + 14*a^4*b - 3*
a^2*b^3 + b^5)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(b*sin(d*x + c) + a)^2))/d