3.443 \(\int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=269 \[ -\frac {3 \left (a^2+4 a b+5 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^4}+\frac {3 \left (a^2-4 a b+5 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^4}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {\sec ^2(c+d x) \left (3 a \left (a^2-3 b^2\right ) \sin (c+d x)+b \left (a^2+5 b^2\right )\right )}{8 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {6 a b^5 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}-\frac {3 b \left (a^4-4 a^2 b^2-5 b^4\right )}{8 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))} \]

[Out]

-3/16*(a^2+4*a*b+5*b^2)*ln(1-sin(d*x+c))/(a+b)^4/d+3/16*(a^2-4*a*b+5*b^2)*ln(1+sin(d*x+c))/(a-b)^4/d-6*a*b^5*l
n(a+b*sin(d*x+c))/(a^2-b^2)^4/d-3/8*b*(a^4-4*a^2*b^2-5*b^4)/(a^2-b^2)^3/d/(a+b*sin(d*x+c))-1/4*sec(d*x+c)^4*(b
-a*sin(d*x+c))/(a^2-b^2)/d/(a+b*sin(d*x+c))+1/8*sec(d*x+c)^2*(b*(a^2+5*b^2)+3*a*(a^2-3*b^2)*sin(d*x+c))/(a^2-b
^2)^2/d/(a+b*sin(d*x+c))

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Rubi [A]  time = 0.32, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2668, 741, 823, 801} \[ -\frac {3 b \left (-4 a^2 b^2+a^4-5 b^4\right )}{8 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {6 a b^5 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}-\frac {3 \left (a^2+4 a b+5 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^4}+\frac {3 \left (a^2-4 a b+5 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^4}+\frac {\sec ^2(c+d x) \left (3 a \left (a^2-3 b^2\right ) \sin (c+d x)+b \left (a^2+5 b^2\right )\right )}{8 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + b*Sin[c + d*x])^2,x]

[Out]

(-3*(a^2 + 4*a*b + 5*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^4*d) + (3*(a^2 - 4*a*b + 5*b^2)*Log[1 + Sin[c + d
*x]])/(16*(a - b)^4*d) - (6*a*b^5*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^4*d) - (3*b*(a^4 - 4*a^2*b^2 - 5*b^4))
/(8*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x]))/(4*(a^2 - b^2)*d*(a + b*Sin[
c + d*x])) + (Sec[c + d*x]^2*(b*(a^2 + 5*b^2) + 3*a*(a^2 - 3*b^2)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d*(a + b*Sin
[c + d*x]))

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]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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]

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {1}{(a+x)^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {b^3 \operatorname {Subst}\left (\int \frac {3 a^2-5 b^2+4 a x}{(a+x)^2 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\sec ^2(c+d x) \left (b \left (a^2+5 b^2\right )+3 a \left (a^2-3 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {b \operatorname {Subst}\left (\int \frac {-3 \left (a^4-2 a^2 b^2+5 b^4\right )-6 a \left (a^2-3 b^2\right ) x}{(a+x)^2 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\sec ^2(c+d x) \left (b \left (a^2+5 b^2\right )+3 a \left (a^2-3 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {b \operatorname {Subst}\left (\int \left (-\frac {3 (a-b)^2 \left (a^2+4 a b+5 b^2\right )}{2 b (a+b)^2 (b-x)}-\frac {3 \left (a^4-4 a^2 b^2-5 b^4\right )}{\left (a^2-b^2\right ) (a+x)^2}+\frac {48 a b^4}{\left (a^2-b^2\right )^2 (a+x)}-\frac {3 (a+b)^2 \left (a^2-4 a b+5 b^2\right )}{2 (a-b)^2 b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac {3 \left (a^2+4 a b+5 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^4 d}+\frac {3 \left (a^2-4 a b+5 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^4 d}-\frac {6 a b^5 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4 d}-\frac {3 b \left (a^4-4 a^2 b^2-5 b^4\right )}{8 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\sec ^2(c+d x) \left (b \left (a^2+5 b^2\right )+3 a \left (a^2-3 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 6.11, size = 406, normalized size = 1.51 \[ \frac {b^5 \left (\frac {\sec ^4(c+d x) \left (b^2-a b \sin (c+d x)\right )}{4 b^6 \left (b^2-a^2\right ) (a+b \sin (c+d x))}-\frac {\frac {\left (6 a^2 \left (a^2-3 b^2\right )-3 \left (a^4-2 a^2 b^2+5 b^4\right )\right ) \left (\frac {1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac {\log (\sin (c+d x)+1)}{2 b (a-b)^2}-\frac {2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}\right )-6 a \left (a^2-3 b^2\right ) \left (-\frac {\log (a+b \sin (c+d x))}{a^2-b^2}-\frac {\log (1-\sin (c+d x))}{2 b (a+b)}+\frac {\log (\sin (c+d x)+1)}{2 b (a-b)}\right )}{2 b^2 \left (b^2-a^2\right )}-\frac {\sec ^2(c+d x) \left (-b \left (4 a b^2-a \left (3 a^2-5 b^2\right )\right ) \sin (c+d x)+4 a^2 b^2-b^2 \left (3 a^2-5 b^2\right )\right )}{2 b^4 \left (b^2-a^2\right ) (a+b \sin (c+d x))}}{4 b^2 \left (b^2-a^2\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + b*Sin[c + d*x])^2,x]

[Out]

(b^5*((Sec[c + d*x]^4*(b^2 - a*b*Sin[c + d*x]))/(4*b^6*(-a^2 + b^2)*(a + b*Sin[c + d*x])) - (-1/2*(Sec[c + d*x
]^2*(4*a^2*b^2 - b^2*(3*a^2 - 5*b^2) - b*(4*a*b^2 - a*(3*a^2 - 5*b^2))*Sin[c + d*x]))/(b^4*(-a^2 + b^2)*(a + b
*Sin[c + d*x])) + (-6*a*(a^2 - 3*b^2)*(-1/2*Log[1 - Sin[c + d*x]]/(b*(a + b)) + Log[1 + Sin[c + d*x]]/(2*(a -
b)*b) - Log[a + b*Sin[c + d*x]]/(a^2 - b^2)) + (6*a^2*(a^2 - 3*b^2) - 3*(a^4 - 2*a^2*b^2 + 5*b^4))*(-1/2*Log[1
- Sin[c + d*x]]/(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*b^2*(-a^2 + b^2)))/(4*b^2*(-a^2 + b^2))))/d

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fricas [B]  time = 0.77, size = 527, normalized size = 1.96 \[ -\frac {4 \, a^{6} b - 12 \, a^{4} b^{3} + 12 \, a^{2} b^{5} - 4 \, b^{7} + 6 \, {\left (a^{6} b - 5 \, a^{4} b^{3} - a^{2} b^{5} + 5 \, b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} - 9 \, a^{2} b^{5} + 5 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 96 \, {\left (a b^{6} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a^{2} b^{5} \cos \left (d x + c\right )^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 3 \, {\left ({\left (a^{6} b - 5 \, a^{4} b^{3} + 15 \, a^{2} b^{5} + 16 \, a b^{6} + 5 \, b^{7}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (a^{7} - 5 \, a^{5} b^{2} + 15 \, a^{3} b^{4} + 16 \, a^{2} b^{5} + 5 \, a b^{6}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (a^{6} b - 5 \, a^{4} b^{3} + 15 \, a^{2} b^{5} - 16 \, a b^{6} + 5 \, b^{7}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (a^{7} - 5 \, a^{5} b^{2} + 15 \, a^{3} b^{4} - 16 \, a^{2} b^{5} + 5 \, a b^{6}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, a^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} + 3 \, {\left (a^{7} - 5 \, a^{5} b^{2} + 7 \, a^{3} b^{4} - 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left ({\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/16*(4*a^6*b - 12*a^4*b^3 + 12*a^2*b^5 - 4*b^7 + 6*(a^6*b - 5*a^4*b^3 - a^2*b^5 + 5*b^7)*cos(d*x + c)^4 - 2*
(a^6*b + 3*a^4*b^3 - 9*a^2*b^5 + 5*b^7)*cos(d*x + c)^2 + 96*(a*b^6*cos(d*x + c)^4*sin(d*x + c) + a^2*b^5*cos(d
*x + c)^4)*log(b*sin(d*x + c) + a) - 3*((a^6*b - 5*a^4*b^3 + 15*a^2*b^5 + 16*a*b^6 + 5*b^7)*cos(d*x + c)^4*sin
(d*x + c) + (a^7 - 5*a^5*b^2 + 15*a^3*b^4 + 16*a^2*b^5 + 5*a*b^6)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) + 3*((
a^6*b - 5*a^4*b^3 + 15*a^2*b^5 - 16*a*b^6 + 5*b^7)*cos(d*x + c)^4*sin(d*x + c) + (a^7 - 5*a^5*b^2 + 15*a^3*b^4
- 16*a^2*b^5 + 5*a*b^6)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 2*(2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 +
3*(a^7 - 5*a^5*b^2 + 7*a^3*b^4 - 3*a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a
^2*b^7 + b^9)*d*cos(d*x + c)^4*sin(d*x + c) + (a^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)
^4)

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giac [A]  time = 0.44, size = 460, normalized size = 1.71 \[ -\frac {\frac {96 \, a b^{6} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}} - \frac {3 \, {\left (a^{2} - 4 \, a b + 5 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {3 \, {\left (a^{2} + 4 \, a b + 5 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {16 \, {\left (6 \, a b^{6} \sin \left (d x + c\right ) + 7 \, a^{2} b^{5} - b^{7}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}} + \frac {2 \, {\left (36 \, a b^{5} \sin \left (d x + c\right )^{4} + 3 \, a^{6} \sin \left (d x + c\right )^{3} - 15 \, a^{4} b^{2} \sin \left (d x + c\right )^{3} + 5 \, a^{2} b^{4} \sin \left (d x + c\right )^{3} + 7 \, b^{6} \sin \left (d x + c\right )^{3} + 16 \, a^{3} b^{3} \sin \left (d x + c\right )^{2} - 88 \, a b^{5} \sin \left (d x + c\right )^{2} - 5 \, a^{6} \sin \left (d x + c\right ) + 17 \, a^{4} b^{2} \sin \left (d x + c\right ) - 3 \, a^{2} b^{4} \sin \left (d x + c\right ) - 9 \, b^{6} \sin \left (d x + c\right ) + 4 \, a^{5} b - 24 \, a^{3} b^{3} + 56 \, a b^{5}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(96*a*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) - 3*(a^2 - 4*a*
b + 5*b^2)*log(abs(sin(d*x + c) + 1))/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) + 3*(a^2 + 4*a*b + 5*b^2)*lo
g(abs(sin(d*x + c) - 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 16*(6*a*b^6*sin(d*x + c) + 7*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*(36*a*b^5*sin(d*x + c)^4 + 3*a
^6*sin(d*x + c)^3 - 15*a^4*b^2*sin(d*x + c)^3 + 5*a^2*b^4*sin(d*x + c)^3 + 7*b^6*sin(d*x + c)^3 + 16*a^3*b^3*s
in(d*x + c)^2 - 88*a*b^5*sin(d*x + c)^2 - 5*a^6*sin(d*x + c) + 17*a^4*b^2*sin(d*x + c) - 3*a^2*b^4*sin(d*x + c
) - 9*b^6*sin(d*x + c) + 4*a^5*b - 24*a^3*b^3 + 56*a*b^5)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(si
n(d*x + c)^2 - 1)^2))/d

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maple [A]  time = 0.31, size = 331, normalized size = 1.23 \[ \frac {1}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {7 b}{16 d \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}-\frac {3 a}{16 d \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) a^{2}}{16 d \left (a +b \right )^{4}}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) a b}{4 d \left (a +b \right )^{4}}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right ) b^{2}}{16 d \left (a +b \right )^{4}}+\frac {b^{5}}{d \left (a +b \right )^{3} \left (a -b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )}-\frac {6 b^{5} a \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}-\frac {1}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {7 b}{16 d \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}-\frac {3 a}{16 d \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) a^{2}}{16 d \left (a -b \right )^{4}}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) a b}{4 d \left (a -b \right )^{4}}+\frac {15 \ln \left (1+\sin \left (d x +c \right )\right ) b^{2}}{16 d \left (a -b \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+b*sin(d*x+c))^2,x)

[Out]

1/16/d/(a+b)^2/(sin(d*x+c)-1)^2-7/16/d/(a+b)^3/(sin(d*x+c)-1)*b-3/16/d/(a+b)^3/(sin(d*x+c)-1)*a-3/16/d/(a+b)^4
*ln(sin(d*x+c)-1)*a^2-3/4/d/(a+b)^4*ln(sin(d*x+c)-1)*a*b-15/16/d/(a+b)^4*ln(sin(d*x+c)-1)*b^2+1/d*b^5/(a+b)^3/
(a-b)^3/(a+b*sin(d*x+c))-6/d*b^5*a/(a+b)^4/(a-b)^4*ln(a+b*sin(d*x+c))-1/16/d/(a-b)^2/(1+sin(d*x+c))^2+7/16/d/(
a-b)^3/(1+sin(d*x+c))*b-3/16/d/(a-b)^3/(1+sin(d*x+c))*a+3/16/d/(a-b)^4*ln(1+sin(d*x+c))*a^2-3/4/d/(a-b)^4*ln(1
+sin(d*x+c))*a*b+15/16/d/(a-b)^4*ln(1+sin(d*x+c))*b^2

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maxima [A]  time = 0.34, size = 505, normalized size = 1.88 \[ -\frac {\frac {96 \, a b^{5} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac {3 \, {\left (a^{2} - 4 \, a b + 5 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {3 \, {\left (a^{2} + 4 \, a b + 5 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {2 \, {\left (4 \, a^{4} b - 20 \, a^{2} b^{3} - 8 \, b^{5} + 3 \, {\left (a^{4} b - 4 \, a^{2} b^{3} - 5 \, b^{5}\right )} \sin \left (d x + c\right )^{4} + 3 \, {\left (a^{5} - 4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \sin \left (d x + c\right )^{3} - {\left (5 \, a^{4} b - 28 \, a^{2} b^{3} - 25 \, b^{5}\right )} \sin \left (d x + c\right )^{2} - {\left (5 \, a^{5} - 16 \, a^{3} b^{2} + 11 \, a b^{4}\right )} \sin \left (d x + c\right )\right )}}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6} + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )^{5} + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \sin \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )^{3} - 2 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/16*(96*a*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) - 3*(a^2 - 4*a*b + 5*b
^2)*log(sin(d*x + c) + 1)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) + 3*(a^2 + 4*a*b + 5*b^2)*log(sin(d*x +
c) - 1)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + 2*(4*a^4*b - 20*a^2*b^3 - 8*b^5 + 3*(a^4*b - 4*a^2*b^3 -
5*b^5)*sin(d*x + c)^4 + 3*(a^5 - 4*a^3*b^2 + 3*a*b^4)*sin(d*x + c)^3 - (5*a^4*b - 28*a^2*b^3 - 25*b^5)*sin(d*
x + c)^2 - (5*a^5 - 16*a^3*b^2 + 11*a*b^4)*sin(d*x + c))/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 + (a^6*b - 3*a^4
*b^3 + 3*a^2*b^5 - b^7)*sin(d*x + c)^5 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*sin(d*x + c)^4 - 2*(a^6*b - 3*a
^4*b^3 + 3*a^2*b^5 - b^7)*sin(d*x + c)^3 - 2*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*sin(d*x + c)^2 + (a^6*b - 3
*a^4*b^3 + 3*a^2*b^5 - b^7)*sin(d*x + c)))/d

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mupad [B]  time = 5.94, size = 449, normalized size = 1.67 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {3\,b^2}{8\,{\left (a-b\right )}^4}-\frac {3\,b}{8\,{\left (a-b\right )}^3}+\frac {3}{16\,{\left (a-b\right )}^2}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {3\,b}{8\,{\left (a+b\right )}^3}+\frac {3}{16\,{\left (a+b\right )}^2}+\frac {3\,b^2}{8\,{\left (a+b\right )}^4}\right )}{d}+\frac {\frac {-a^4\,b+5\,a^2\,b^3+2\,b^5}{2\,\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {3\,{\sin \left (c+d\,x\right )}^3\,\left (3\,a\,b^2-a^3\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {3\,{\sin \left (c+d\,x\right )}^4\,\left (-a^4\,b+4\,a^2\,b^3+5\,b^5\right )}{8\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {\sin \left (c+d\,x\right )\,\left (11\,a\,b^2-5\,a^3\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (-5\,a^4\,b+28\,a^2\,b^3+25\,b^5\right )}{8\,\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left (b\,{\sin \left (c+d\,x\right )}^5+a\,{\sin \left (c+d\,x\right )}^4-2\,b\,{\sin \left (c+d\,x\right )}^3-2\,a\,{\sin \left (c+d\,x\right )}^2+b\,\sin \left (c+d\,x\right )+a\right )}-\frac {6\,a\,b^5\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + b*sin(c + d*x))^2),x)

[Out]

(log(sin(c + d*x) + 1)*((3*b^2)/(8*(a - b)^4) - (3*b)/(8*(a - b)^3) + 3/(16*(a - b)^2)))/d - (log(sin(c + d*x)
- 1)*((3*b)/(8*(a + b)^3) + 3/(16*(a + b)^2) + (3*b^2)/(8*(a + b)^4)))/d + ((2*b^5 - a^4*b + 5*a^2*b^3)/(2*(a
^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (3*sin(c + d*x)^3*(3*a*b^2 - a^3))/(8*(a^4 + b^4 - 2*a^2*b^2)) + (3*sin(c
+ d*x)^4*(5*b^5 - a^4*b + 4*a^2*b^3))/(8*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (sin(c + d*x)*(11*a*b^2 - 5*a
^3))/(8*(a^4 + b^4 - 2*a^2*b^2)) - (sin(c + d*x)^2*(25*b^5 - 5*a^4*b + 28*a^2*b^3))/(8*(a^2 - b^2)*(a^4 + b^4
- 2*a^2*b^2)))/(d*(a + b*sin(c + d*x) - 2*a*sin(c + d*x)^2 + a*sin(c + d*x)^4 - 2*b*sin(c + d*x)^3 + b*sin(c +
d*x)^5)) - (6*a*b^5*log(a + b*sin(c + d*x)))/(d*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+b*sin(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)**5/(a + b*sin(c + d*x))**2, x)

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