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

Optimal. Leaf size=226 \[ -\frac{a b \left (a^2+11 b^2\right )}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{b \left (a^2+2 b^2\right )}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac{2 b^3 \left (5 a^2+b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}-\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{(a+4 b) \log (1-\sin (c+d x))}{4 d (a+b)^4}+\frac{(a-4 b) \log (\sin (c+d x)+1)}{4 d (a-b)^4} \]

[Out]

-((a + 4*b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^4*d) + ((a - 4*b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^4*d) + (2*b^
3*(5*a^2 + b^2)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^4*d) - (b*(a^2 + 2*b^2))/(2*(a^2 - b^2)^2*d*(a + b*Sin[c
+ d*x])^2) - (Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) - (a*b*(a^2 + 11*
b^2))/(2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x]))

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Rubi [A]  time = 0.276535, antiderivative size = 226, 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} \[ -\frac{a b \left (a^2+11 b^2\right )}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{b \left (a^2+2 b^2\right )}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac{2 b^3 \left (5 a^2+b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}-\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{(a+4 b) \log (1-\sin (c+d x))}{4 d (a+b)^4}+\frac{(a-4 b) \log (\sin (c+d x)+1)}{4 d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 4*b)*Log[1 - Sin[c + d*x]])/(4*(a + b)^4*d) + ((a - 4*b)*Log[1 + Sin[c + d*x]])/(4*(a - b)^4*d) + (2*b^
3*(5*a^2 + b^2)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^4*d) - (b*(a^2 + 2*b^2))/(2*(a^2 - b^2)^2*d*(a + b*Sin[c
+ d*x])^2) - (Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) - (a*b*(a^2 + 11*
b^2))/(2*(a^2 - b^2)^3*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))^3} \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{(a+x)^3 \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))^2}+\frac{b \operatorname{Subst}\left (\int \frac{a^2-4 b^2+3 a x}{(a+x)^3 \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))^2}+\frac{b \operatorname{Subst}\left (\int \left (\frac{(a-b) (a+4 b)}{2 b (a+b)^3 (b-x)}+\frac{2 \left (a^2+2 b^2\right )}{(a-b) (a+b) (a+x)^3}+\frac{a \left (a^2+11 b^2\right )}{(a-b)^2 (a+b)^2 (a+x)^2}+\frac{4 \left (5 a^2 b^2+b^4\right )}{(a-b)^3 (a+b)^3 (a+x)}+\frac{(a-4 b) (a+b)}{2 (a-b)^3 b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d}\\ &=-\frac{(a+4 b) \log (1-\sin (c+d x))}{4 (a+b)^4 d}+\frac{(a-4 b) \log (1+\sin (c+d x))}{4 (a-b)^4 d}+\frac{2 b^3 \left (5 a^2+b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4 d}-\frac{b \left (a^2+2 b^2\right )}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\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))^2}-\frac{a b \left (a^2+11 b^2\right )}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 4.10495, size = 283, normalized size = 1.25 \[ \frac{b \left (a^2+2 b^2\right ) \left (\frac{1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{2 \left (3 a^2+b^2\right ) \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac{4 a}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}-\frac{\log (1-\sin (c+d x))}{b (a+b)^3}+\frac{\log (\sin (c+d x)+1)}{b (a-b)^3}\right )+\frac{3}{2} a \left (\frac{2 b}{\left (b^2-a^2\right ) (a+b \sin (c+d x))}+\frac{4 a b \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}+\frac{\log (1-\sin (c+d x))}{(a+b)^2}-\frac{\log (\sin (c+d x)+1)}{(a-b)^2}\right )+\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{(a+b \sin (c+d x))^2}}{2 d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.141, size = 258, normalized size = 1.1 \begin{align*} -{\frac{{b}^{3}}{2\,d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}-4\,{\frac{a{b}^{3}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3} \left ( a+b\sin \left ( dx+c \right ) \right ) }}+10\,{\frac{{b}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ){a}^{2}}{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}}+2\,{\frac{{b}^{5}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4}}}-{\frac{1}{4\,d \left ( a+b \right ) ^{3} \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 ) ^{4}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) b}{d \left ( a+b \right ) ^{4}}}-{\frac{1}{4\,d \left ( a-b \right ) ^{3} \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 ) ^{4}}}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) b}{d \left ( a-b \right ) ^{4}}} \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))^3,x)

[Out]

-1/2/d*b^3/(a+b)^2/(a-b)^2/(a+b*sin(d*x+c))^2-4/d*a*b^3/(a+b)^3/(a-b)^3/(a+b*sin(d*x+c))+10/d*b^3/(a+b)^4/(a-b
)^4*ln(a+b*sin(d*x+c))*a^2+2/d*b^5/(a+b)^4/(a-b)^4*ln(a+b*sin(d*x+c))-1/4/d/(a+b)^3/(sin(d*x+c)-1)-1/4/d/(a+b)
^4*ln(sin(d*x+c)-1)*a-1/d/(a+b)^4*ln(sin(d*x+c)-1)*b-1/4/d/(a-b)^3/(1+sin(d*x+c))+1/4/d/(a-b)^4*ln(1+sin(d*x+c
))*a-1/d/(a-b)^4*ln(1+sin(d*x+c))*b

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Maxima [B]  time = 1.02886, size = 591, normalized size = 2.62 \begin{align*} \frac{\frac{8 \,{\left (5 \, a^{2} b^{3} + b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} + \frac{{\left (a - 4 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac{{\left (a + 4 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac{2 \,{\left (3 \, a^{4} b + 10 \, a^{2} b^{3} - b^{5} -{\left (a^{3} b^{2} + 11 \, a b^{4}\right )} \sin \left (d x + c\right )^{3} - 2 \,{\left (a^{4} b + 6 \, a^{2} b^{3} - b^{5}\right )} \sin \left (d x + c\right )^{2} -{\left (a^{5} - 3 \, a^{3} b^{2} - 10 \, a b^{4}\right )} \sin \left (d x + c\right )\right )}}{a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6} -{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \sin \left (d x + c\right )^{4} - 2 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (d x + c\right )^{3} -{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sin \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 )} \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))^3,x, algorithm="maxima")

[Out]

1/4*(8*(5*a^2*b^3 + b^5)*log(b*sin(d*x + c) + a)/(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) + (a - 4*b)*l
og(sin(d*x + c) + 1)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (a + 4*b)*log(sin(d*x + c) - 1)/(a^4 + 4*a^
3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 2*(3*a^4*b + 10*a^2*b^3 - b^5 - (a^3*b^2 + 11*a*b^4)*sin(d*x + c)^3 - 2*(a^
4*b + 6*a^2*b^3 - b^5)*sin(d*x + c)^2 - (a^5 - 3*a^3*b^2 - 10*a*b^4)*sin(d*x + c))/(a^8 - 3*a^6*b^2 + 3*a^4*b^
4 - a^2*b^6 - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*sin(d*x + c)^4 - 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^
7)*sin(d*x + c)^3 - (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*sin(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*
a^3*b^5 - a*b^7)*sin(d*x + c)))/d

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Fricas [B]  time = 5.52251, size = 1577, normalized size = 6.98 \begin{align*} \text{result too large to display} \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))^3,x, algorithm="fricas")

[Out]

1/4*(2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 + 4*(a^6*b + 5*a^4*b^3 - 7*a^2*b^5 + b^7)*cos(d*x + c)^2 + 8*((5*
a^2*b^5 + b^7)*cos(d*x + c)^4 - 2*(5*a^3*b^4 + a*b^6)*cos(d*x + c)^2*sin(d*x + c) - (5*a^4*b^3 + 6*a^2*b^5 + b
^7)*cos(d*x + c)^2)*log(b*sin(d*x + c) + a) + ((a^5*b^2 - 10*a^3*b^4 - 20*a^2*b^5 - 15*a*b^6 - 4*b^7)*cos(d*x
+ c)^4 - 2*(a^6*b - 10*a^4*b^3 - 20*a^3*b^4 - 15*a^2*b^5 - 4*a*b^6)*cos(d*x + c)^2*sin(d*x + c) - (a^7 - 9*a^5
*b^2 - 20*a^4*b^3 - 25*a^3*b^4 - 24*a^2*b^5 - 15*a*b^6 - 4*b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - ((a^5*
b^2 - 10*a^3*b^4 + 20*a^2*b^5 - 15*a*b^6 + 4*b^7)*cos(d*x + c)^4 - 2*(a^6*b - 10*a^4*b^3 + 20*a^3*b^4 - 15*a^2
*b^5 + 4*a*b^6)*cos(d*x + c)^2*sin(d*x + c) - (a^7 - 9*a^5*b^2 + 20*a^4*b^3 - 25*a^3*b^4 + 24*a^2*b^5 - 15*a*b
^6 + 4*b^7)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 - (a^5*b^2 + 10*a^
3*b^4 - 11*a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*
x + c)^4 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)^2*sin(d*x + c) - (a^10 - 3*a^8
*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + c)^2)

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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 )^{3}}\, 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))**3,x)

[Out]

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

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Giac [A]  time = 1.69925, size = 558, normalized size = 2.47 \begin{align*} \frac{\frac{8 \,{\left (5 \, a^{2} b^{4} + b^{6}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}} + \frac{{\left (a - 4 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac{{\left (a + 4 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac{2 \,{\left (10 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} + 2 \, b^{5} \sin \left (d x + c\right )^{2} - a^{5} \sin \left (d x + c\right ) - 2 \, a^{3} b^{2} \sin \left (d x + c\right ) + 3 \, a b^{4} \sin \left (d x + c\right ) + 3 \, a^{4} b - 12 \, a^{2} b^{3} - 3 \, b^{5}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\right )}} - \frac{2 \,{\left (30 \, a^{2} b^{5} \sin \left (d x + c\right )^{2} + 6 \, b^{7} \sin \left (d x + c\right )^{2} + 68 \, a^{3} b^{4} \sin \left (d x + c\right ) + 4 \, a b^{6} \sin \left (d x + c\right ) + 39 \, a^{4} b^{3} - 4 \, a^{2} b^{5} + b^{7}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{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))^3,x, algorithm="giac")

[Out]

1/4*(8*(5*a^2*b^4 + b^6)*log(abs(b*sin(d*x + c) + a))/(a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9) + (a -
4*b)*log(abs(sin(d*x + c) + 1))/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (a + 4*b)*log(abs(sin(d*x + c)
- 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + 2*(10*a^2*b^3*sin(d*x + c)^2 + 2*b^5*sin(d*x + c)^2 - a^5*
sin(d*x + c) - 2*a^3*b^2*sin(d*x + c) + 3*a*b^4*sin(d*x + c) + 3*a^4*b - 12*a^2*b^3 - 3*b^5)/((a^8 - 4*a^6*b^2
+ 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(sin(d*x + c)^2 - 1)) - 2*(30*a^2*b^5*sin(d*x + c)^2 + 6*b^7*sin(d*x + c)^2 +
68*a^3*b^4*sin(d*x + c) + 4*a*b^6*sin(d*x + c) + 39*a^4*b^3 - 4*a^2*b^5 + b^7)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 -
4*a^2*b^6 + b^8)*(b*sin(d*x + c) + a)^2))/d