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

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

[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)

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Rubi [A]  time = 0.254862, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, 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 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 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 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 ^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*}

Mathematica [A]  time = 0.953295, size = 266, normalized size = 1.36 \[ \frac{\frac{16 b^5 \log (a+b \sin (c+d x))}{\left (b^2-a^2\right )^3}-\frac{2 \left (3 a^2+9 a b+8 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{(a+b)^3}+\frac{2 \left (3 a^2-9 a b+8 b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{(a-b)^3}+\frac{3 a+5 b}{(a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{5 b-3 a}{(a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{1}{(a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{1}{(a-b) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}}{16 d} \]

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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Maple [A]  time = 0.061, size = 305, normalized size = 1.6 \begin{align*} -{\frac{{b}^{5}\ln \left ( a+b\sin \left ( dx+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 ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{3\,a}{16\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{5\,b}{16\,d \left ( a+b \right ) ^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ){a}^{2}}{16\,d \left ( a+b \right ) ^{3}}}-{\frac{9\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) ab}{16\,d \left ( a+b \right ) ^{3}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ){b}^{2}}{2\,d \left ( a+b \right ) ^{3}}}-{\frac{1}{2\,d \left ( 8\,a-8\,b \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,a}{16\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{5\,b}{16\,d \left ( a-b \right ) ^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{3\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ){a}^{2}}{16\,d \left ( a-b \right ) ^{3}}}-{\frac{9\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) ab}{16\,d \left ( a-b \right ) ^{3}}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ){b}^{2}}{2\,d \left ( a-b \right ) ^{3}}} \end{align*}

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/d*b^5/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))+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^2-1/2/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 = 0.985268, size = 375, normalized size = 1.92 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 4.30634, size = 579, normalized size = 2.97 \begin{align*} -\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}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\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)**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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Giac [A]  time = 1.16194, size = 448, normalized size = 2.3 \begin{align*} -\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} \end{align*}

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