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

Optimal. Leaf size=123 \[ \frac{b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right )}-\frac{(a+2 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac{(a-2 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.163535, antiderivative size = 123, 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} \[ \frac{b^3 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right )}-\frac{(a+2 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac{(a-2 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.589515, size = 170, normalized size = 1.38 \[ \frac{\frac{4 b^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac{1}{(a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{1}{(a-b) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{2 (a+2 b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{(a+b)^2}+\frac{2 (a-2 b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{(a-b)^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.067, size = 164, normalized size = 1.3 \begin{align*}{\frac{{b}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}-{\frac{1}{d \left ( 4\,a+4\,b \right ) \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 ) ^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) b}{2\,d \left ( a+b \right ) ^{2}}}-{\frac{1}{d \left ( 4\,a-4\,b \right ) \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 ) ^{2}}}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) b}{2\,d \left ( a-b \right ) ^{2}}} \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)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.96123, size = 188, normalized size = 1.53 \begin{align*} \frac{\frac{4 \, b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a - 2 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{{\left (a + 2 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{2 \,{\left (a \sin \left (d x + c\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{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)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 3.26639, size = 366, normalized size = 2.98 \begin{align*} \frac{4 \, b^{3} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) +{\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \,{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{4 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \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)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.17158, size = 239, normalized size = 1.94 \begin{align*} \frac{\frac{4 \, b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} + \frac{{\left (a - 2 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{{\left (a + 2 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac{2 \,{\left (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 ) + a^{2} b - 2 \, b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\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)),x, algorithm="giac")

[Out]

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