3.442 \(\int \frac{\sec ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=177 \[ -\frac{b \left (a^2+3 b^2\right )}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{4 a b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{(a+3 b) \log (1-\sin (c+d x))}{4 d (a+b)^3}+\frac{(a-3 b) \log (\sin (c+d x)+1)}{4 d (a-b)^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.20884, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2668, 741, 801} \[ -\frac{b \left (a^2+3 b^2\right )}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{4 a b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{(a+3 b) \log (1-\sin (c+d x))}{4 d (a+b)^3}+\frac{(a-3 b) \log (\sin (c+d x)+1)}{4 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 3*b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^3*d) + ((a - 3*b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^3*d) + (4*a*
b^3*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^3*d) - (b*(a^2 + 3*b^2))/(2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])) -
(Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/(2*(a^2 - b^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 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

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 ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{(a+x)^2 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{b \operatorname{Subst}\left (\int \frac{a^2-3 b^2+2 a x}{(a+x)^2 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{b \operatorname{Subst}\left (\int \left (\frac{(a-b) (a+3 b)}{2 b (a+b)^2 (b-x)}+\frac{a^2+3 b^2}{(a-b) (a+b) (a+x)^2}+\frac{8 a b^2}{(a-b)^2 (a+b)^2 (a+x)}+\frac{(a-3 b) (a+b)}{2 (a-b)^2 b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\frac{(a+3 b) \log (1-\sin (c+d x))}{4 (a+b)^3 d}+\frac{(a-3 b) \log (1+\sin (c+d x))}{4 (a-b)^3 d}+\frac{4 a b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac{b \left (a^2+3 b^2\right )}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.73611, size = 222, normalized size = 1.25 \[ \frac{-b \left (-a^2-3 b^2\right ) \left (\frac{1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac{\log (\sin (c+d x)+1)}{2 b (a-b)^2}-\frac{2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}\right )+\frac{a ((a-b) \log (1-\sin (c+d x))-(a+b) \log (\sin (c+d x)+1)+2 b \log (a+b \sin (c+d x)))}{(a-b) (a+b)}+\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{a+b \sin (c+d x)}}{2 d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.108, size = 192, normalized size = 1.1 \begin{align*} -{\frac{{b}^{3}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( a+b\sin \left ( dx+c \right ) \right ) }}+4\,{\frac{a{b}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}-{\frac{1}{4\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) a}{4\,d \left ( a+b \right ) ^{3}}}-{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) b}{4\,d \left ( a+b \right ) ^{3}}}-{\frac{1}{4\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) a}{4\,d \left ( a-b \right ) ^{3}}}-{\frac{3\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) b}{4\,d \left ( a-b \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+b*sin(d*x+c))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.97815, size = 371, normalized size = 2.1 \begin{align*} \frac{\frac{16 \, a b^{3} \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{{\left (a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{{\left (a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{2 \,{\left (2 \, a^{2} b + 2 \, b^{3} -{\left (a^{2} b + 3 \, b^{3}\right )} \sin \left (d x + c\right )^{2} -{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )\right )}}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} -{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{3} -{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{2} +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/4*(16*a*b^3*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + (a - 3*b)*log(sin(d*x + c) + 1)/(a
^3 - 3*a^2*b + 3*a*b^2 - b^3) - (a + 3*b)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*(2*a^2*b +
2*b^3 - (a^2*b + 3*b^3)*sin(d*x + c)^2 - (a^3 - a*b^2)*sin(d*x + c))/(a^5 - 2*a^3*b^2 + a*b^4 - (a^4*b - 2*a^
2*b^3 + b^5)*sin(d*x + c)^3 - (a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c)^2 + (a^4*b - 2*a^2*b^3 + b^5)*sin(d*x + c
)))/d

________________________________________________________________________________________

Fricas [B]  time = 3.82479, size = 867, normalized size = 4.9 \begin{align*} -\frac{2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} + 2 \,{\left (a^{4} b + 2 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} - 16 \,{\left (a b^{4} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a^{2} b^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) -{\left ({\left (a^{4} b - 6 \, a^{2} b^{3} - 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) +{\left (a^{5} - 6 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left ({\left (a^{4} b - 6 \, a^{2} b^{3} + 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) +{\left (a^{5} - 6 \, a^{3} b^{2} + 8 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) +{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a^4*b - 4*a^2*b^3 + 2*b^5 + 2*(a^4*b + 2*a^2*b^3 - 3*b^5)*cos(d*x + c)^2 - 16*(a*b^4*cos(d*x + c)^2*si
n(d*x + c) + a^2*b^3*cos(d*x + c)^2)*log(b*sin(d*x + c) + a) - ((a^4*b - 6*a^2*b^3 - 8*a*b^4 - 3*b^5)*cos(d*x
+ c)^2*sin(d*x + c) + (a^5 - 6*a^3*b^2 - 8*a^2*b^3 - 3*a*b^4)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((a^4*b
- 6*a^2*b^3 + 8*a*b^4 - 3*b^5)*cos(d*x + c)^2*sin(d*x + c) + (a^5 - 6*a^3*b^2 + 8*a^2*b^3 - 3*a*b^4)*cos(d*x +
c)^2)*log(-sin(d*x + c) + 1) - 2*(a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c))/((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^
7)*d*cos(d*x + c)^2*sin(d*x + c) + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)**3/(a + b*sin(c + d*x))**2, x)

________________________________________________________________________________________

Giac [A]  time = 1.1665, size = 329, normalized size = 1.86 \begin{align*} \frac{\frac{16 \, a b^{4} \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{{\left (a - 3 \, b\right )} \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{{\left (a + 3 \, b\right )} \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{2 \,{\left (a^{2} b \sin \left (d x + c\right )^{2} + 3 \, b^{3} \sin \left (d x + c\right )^{2} + a^{3} \sin \left (d x + c\right ) - a b^{2} \sin \left (d x + c\right ) - 2 \, a^{2} b - 2 \, b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - a\right )}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(16*a*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) + (a - 3*b)*log(abs(sin(d*x +
c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (a + 3*b)*log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^
3) - 2*(a^2*b*sin(d*x + c)^2 + 3*b^3*sin(d*x + c)^2 + a^3*sin(d*x + c) - a*b^2*sin(d*x + c) - 2*a^2*b - 2*b^3)
/((a^4 - 2*a^2*b^2 + b^4)*(b*sin(d*x + c)^3 + a*sin(d*x + c)^2 - b*sin(d*x + c) - a)))/d