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

Optimal. Leaf size=195 \[ -\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac {b^5 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}+\frac {\sec ^2(c+d x) \left (a \left (3 a^2-7 b^2\right ) \sin (c+d x)+4 b^3\right )}{8 d \left (a^2-b^2\right )^2} \]

[Out]

-1/16*(3*a^2+9*a*b+8*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d+1/16*(3*a^2-9*a*b+8*b^2)*ln(1+sin(d*x+c))/(a-b)^3/d-b^5*l
n(a+b*sin(d*x+c))/(a^2-b^2)^3/d-1/4*sec(d*x+c)^4*(b-a*sin(d*x+c))/d/(a^2-b^2)+1/8*sec(d*x+c)^2*(4*b^3+a*(3*a^2
-7*b^2)*sin(d*x+c))/(a^2-b^2)^2/d

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Rubi [A]  time = 0.25, antiderivative size = 195, 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} \[ -\frac {b^5 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {\sec ^2(c+d x) \left (a \left (3 a^2-7 b^2\right ) \sin (c+d x)+4 b^3\right )}{8 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a^2 + 9*a*b + 8*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) + ((3*a^2 - 9*a*b + 8*b^2)*Log[1 + Sin[c + d
*x]])/(16*(a - b)^3*d) - (b^5*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^3*d) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x]
))/(4*(a^2 - b^2)*d) + (Sec[c + d*x]^2*(4*b^3 + a*(3*a^2 - 7*b^2)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d)

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)} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {1}{(a+x) \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}+\frac {b^3 \operatorname {Subst}\left (\int \frac {3 a^2-4 b^2+3 a x}{(a+x) \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}+\frac {\sec ^2(c+d x) \left (4 b^3+a \left (3 a^2-7 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac {b \operatorname {Subst}\left (\int \frac {-3 a^4+7 a^2 b^2-8 b^4-a \left (3 a^2-7 b^2\right ) x}{(a+x) \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}+\frac {\sec ^2(c+d x) \left (4 b^3+a \left (3 a^2-7 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac {b \operatorname {Subst}\left (\int \left (-\frac {(a-b)^2 \left (3 a^2+9 a b+8 b^2\right )}{2 b (a+b) (b-x)}+\frac {8 b^4}{(a-b) (a+b) (a+x)}-\frac {(a+b)^2 \left (3 a^2-9 a b+8 b^2\right )}{2 (a-b) b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {b^5 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) \left (4 b^3+a \left (3 a^2-7 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.95, size = 266, normalized size = 1.36 \[ \frac {-\frac {2 \left (3 a^2+9 a b+8 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{(a+b)^3}+\frac {2 \left (3 a^2-9 a b+8 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{(a-b)^3}+\frac {16 b^5 \log (a+b \sin (c+d x))}{\left (b^2-a^2\right )^3}+\frac {3 a+5 b}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {5 b-3 a}{(a-b)^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {1}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}-\frac {1}{(a-b) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*(3*a^2 + 9*a*b + 8*b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(a + b)^3 + (2*(3*a^2 - 9*a*b + 8*b^2)*
Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(a - b)^3 + (16*b^5*Log[a + b*Sin[c + d*x]])/(-a^2 + b^2)^3 + 1/((a
+ b)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + (3*a + 5*b)/((a + b)^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2
) - 1/((a - b)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4) + (-3*a + 5*b)/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c +
d*x)/2])^2))/(16*d)

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fricas [A]  time = 0.64, size = 253, normalized size = 1.30 \[ -\frac {16 \, b^{5} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 15 \, a b^{4} + 8 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 15 \, a b^{4} - 8 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{4} b - 8 \, a^{2} b^{3} + 4 \, b^{5} - 8 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} - 4 \, a^{3} b^{2} + 2 \, a b^{4} + {\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 7 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(16*b^5*cos(d*x + c)^4*log(b*sin(d*x + c) + a) - (3*a^5 - 10*a^3*b^2 + 15*a*b^4 + 8*b^5)*cos(d*x + c)^4*
log(sin(d*x + c) + 1) + (3*a^5 - 10*a^3*b^2 + 15*a*b^4 - 8*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a^4*
b - 8*a^2*b^3 + 4*b^5 - 8*(a^2*b^3 - b^5)*cos(d*x + c)^2 - 2*(2*a^5 - 4*a^3*b^2 + 2*a*b^4 + (3*a^5 - 10*a^3*b^
2 + 7*a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^4)

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giac [A]  time = 0.47, size = 332, normalized size = 1.70 \[ -\frac {\frac {16 \, b^{6} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (3 \, a^{2} - 9 \, a b + 8 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a^{2} + 9 \, a b + 8 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (6 \, b^{5} \sin \left (d x + c\right )^{4} + 3 \, a^{5} \sin \left (d x + c\right )^{3} - 10 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 7 \, a b^{4} \sin \left (d x + c\right )^{3} + 4 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} - 16 \, b^{5} \sin \left (d x + c\right )^{2} - 5 \, a^{5} \sin \left (d x + c\right ) + 14 \, a^{3} b^{2} \sin \left (d x + c\right ) - 9 \, a b^{4} \sin \left (d x + c\right ) + 2 \, a^{4} b - 8 \, a^{2} b^{3} + 12 \, b^{5}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(16*b^6*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - (3*a^2 - 9*a*b + 8*b^2)*log
(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a^2 + 9*a*b + 8*b^2)*log(abs(sin(d*x + c) - 1))/(
a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2*(6*b^5*sin(d*x + c)^4 + 3*a^5*sin(d*x + c)^3 - 10*a^3*b^2*sin(d*x + c)^3 +
7*a*b^4*sin(d*x + c)^3 + 4*a^2*b^3*sin(d*x + c)^2 - 16*b^5*sin(d*x + c)^2 - 5*a^5*sin(d*x + c) + 14*a^3*b^2*si
n(d*x + c) - 9*a*b^4*sin(d*x + c) + 2*a^4*b - 8*a^2*b^3 + 12*b^5)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(sin(d*
x + c)^2 - 1)^2))/d

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maple [A]  time = 0.18, size = 305, normalized size = 1.56 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {3 a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {5 b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) a^{2}}{16 d \left (a +b \right )^{3}}-\frac {9 \ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) b^{2}}{2 d \left (a +b \right )^{3}}-\frac {b^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {5 b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) a^{2}}{16 d \left (a -b \right )^{3}}-\frac {9 \ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) b^{2}}{2 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/(8*a+8*b)/(sin(d*x+c)-1)^2-3/16/d/(a+b)^2/(sin(d*x+c)-1)*a-5/16/d/(a+b)^2/(sin(d*x+c)-1)*b-3/16/d/(a+b)^
3*ln(sin(d*x+c)-1)*a^2-9/16/d/(a+b)^3*ln(sin(d*x+c)-1)*a*b-1/2/d/(a+b)^3*ln(sin(d*x+c)-1)*b^2-1/d*b^5/(a+b)^3/
(a-b)^3*ln(a+b*sin(d*x+c))-1/2/d/(8*a-8*b)/(1+sin(d*x+c))^2-3/16/d/(a-b)^2/(1+sin(d*x+c))*a+5/16/d/(a-b)^2/(1+
sin(d*x+c))*b+3/16/d/(a-b)^3*ln(1+sin(d*x+c))*a^2-9/16/d/(a-b)^3*ln(1+sin(d*x+c))*a*b+1/2/d/(a-b)^3*ln(1+sin(d
*x+c))*b^2

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maxima [A]  time = 0.34, size = 278, normalized size = 1.43 \[ -\frac {\frac {16 \, b^{5} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (3 \, a^{2} - 9 \, a b + 8 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a^{2} + 9 \, a b + 8 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (4 \, b^{3} \sin \left (d x + c\right )^{2} + {\left (3 \, a^{3} - 7 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} + 2 \, a^{2} b - 6 \, b^{3} - {\left (5 \, a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.59, size = 322, normalized size = 1.65 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {b^2}{8\,{\left (a-b\right )}^3}-\frac {3\,b}{16\,{\left (a-b\right )}^2}+\frac {3}{16\,\left (a-b\right )}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {3\,b}{16\,{\left (a+b\right )}^2}+\frac {3}{16\,\left (a+b\right )}+\frac {b^2}{8\,{\left (a+b\right )}^3}\right )}{d}-\frac {\frac {a^2\,b-3\,b^3}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b^3\,{\sin \left (c+d\,x\right )}^2}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (7\,a\,b^2-3\,a^3\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\sin \left (c+d\,x\right )\,\left (9\,a\,b^2-5\,a^3\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\cos \left (c+d\,x\right )}^2+{\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^2\right )}-\frac {b^5\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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