3.455 \(\int \frac{\sec ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=328 \[ -\frac{3 a b \left (-6 a^2 b^2+a^4-27 b^4\right )}{8 d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}-\frac{3 b \left (-5 a^2 b^2+a^4-4 b^4\right )}{8 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}-\frac{3 b^5 \left (7 a^2+b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^5}-\frac{3 \left (a^2+5 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^5}+\frac{3 \left (a^2-5 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^5}+\frac{\sec ^2(c+d x) \left (a \left (3 a^2-11 b^2\right ) \sin (c+d x)+2 b \left (a^2+3 b^2\right )\right )}{8 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2} \]

[Out]

(-3*(a^2 + 5*a*b + 8*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^5*d) + (3*(a^2 - 5*a*b + 8*b^2)*Log[1 + Sin[c + d
*x]])/(16*(a - b)^5*d) - (3*b^5*(7*a^2 + b^2)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^5*d) - (3*b*(a^4 - 5*a^2*b
^2 - 4*b^4))/(8*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])^2) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x]))/(4*(a^2 - b^2)
*d*(a + b*Sin[c + d*x])^2) - (3*a*b*(a^4 - 6*a^2*b^2 - 27*b^4))/(8*(a^2 - b^2)^4*d*(a + b*Sin[c + d*x])) + (Se
c[c + d*x]^2*(2*b*(a^2 + 3*b^2) + a*(3*a^2 - 11*b^2)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])^2)

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Rubi [A]  time = 0.420186, antiderivative size = 328, 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} \[ -\frac{3 a b \left (-6 a^2 b^2+a^4-27 b^4\right )}{8 d \left (a^2-b^2\right )^4 (a+b \sin (c+d x))}-\frac{3 b \left (-5 a^2 b^2+a^4-4 b^4\right )}{8 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}-\frac{3 b^5 \left (7 a^2+b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^5}-\frac{3 \left (a^2+5 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^5}+\frac{3 \left (a^2-5 a b+8 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^5}+\frac{\sec ^2(c+d x) \left (a \left (3 a^2-11 b^2\right ) \sin (c+d x)+2 b \left (a^2+3 b^2\right )\right )}{8 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + b*Sin[c + d*x])^3,x]

[Out]

(-3*(a^2 + 5*a*b + 8*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^5*d) + (3*(a^2 - 5*a*b + 8*b^2)*Log[1 + Sin[c + d
*x]])/(16*(a - b)^5*d) - (3*b^5*(7*a^2 + b^2)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^5*d) - (3*b*(a^4 - 5*a^2*b
^2 - 4*b^4))/(8*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])^2) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x]))/(4*(a^2 - b^2)
*d*(a + b*Sin[c + d*x])^2) - (3*a*b*(a^4 - 6*a^2*b^2 - 27*b^4))/(8*(a^2 - b^2)^4*d*(a + b*Sin[c + d*x])) + (Se
c[c + d*x]^2*(2*b*(a^2 + 3*b^2) + a*(3*a^2 - 11*b^2)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])^2)

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))^3} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{1}{(a+x)^3 \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))^2}+\frac{b^3 \operatorname{Subst}\left (\int \frac{3 \left (a^2-2 b^2\right )+5 a x}{(a+x)^3 \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))^2}+\frac{\sec ^2(c+d x) \left (2 b \left (a^2+3 b^2\right )+a \left (3 a^2-11 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac{b \operatorname{Subst}\left (\int \frac{-3 \left (a^4-a^2 b^2+8 b^4\right )-3 a \left (3 a^2-11 b^2\right ) x}{(a+x)^3 \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))^2}+\frac{\sec ^2(c+d x) \left (2 b \left (a^2+3 b^2\right )+a \left (3 a^2-11 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac{b \operatorname{Subst}\left (\int \left (-\frac{3 (a-b)^2 \left (a^2+5 a b+8 b^2\right )}{2 b (a+b)^3 (b-x)}-\frac{6 \left (a^4-5 a^2 b^2-4 b^4\right )}{\left (a^2-b^2\right ) (a+x)^3}-\frac{3 \left (a^5-6 a^3 b^2-27 a b^4\right )}{\left (a^2-b^2\right )^2 (a+x)^2}+\frac{24 \left (7 a^2 b^4+b^6\right )}{\left (a^2-b^2\right )^3 (a+x)}-\frac{3 (a+b)^2 \left (a^2-5 a b+8 b^2\right )}{2 (a-b)^3 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+5 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^5 d}+\frac{3 \left (a^2-5 a b+8 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^5 d}-\frac{3 b^5 \left (7 a^2+b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^5 d}-\frac{3 b \left (a^4-5 a^2 b^2-4 b^4\right )}{8 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}-\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))^2}-\frac{3 a b \left (a^4-6 a^2 b^2-27 b^4\right )}{8 \left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac{\sec ^2(c+d x) \left (2 b \left (a^2+3 b^2\right )+a \left (3 a^2-11 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 2.6111, size = 388, normalized size = 1.18 \[ \frac{-\frac{b \left (3 \left (-5 a^2 b^2+a^4-4 b^4\right ) \left (\frac{1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{2 \left (3 a^2+b^2\right ) \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac{4 a}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}-\frac{\log (1-\sin (c+d x))}{b (a+b)^3}+\frac{\log (\sin (c+d x)+1)}{b (a-b)^3}\right )-3 a \left (3 a^2-11 b^2\right ) \left (\frac{1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac{\log (\sin (c+d x)+1)}{2 b (a-b)^2}-\frac{2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}\right )\right )}{b^2-a^2}+\frac{\sec ^2(c+d x) \left (a \left (3 a^2-11 b^2\right ) \sin (c+d x)+2 b \left (a^2+3 b^2\right )\right )}{\left (b^2-a^2\right ) (a+b \sin (c+d x))^2}+\frac{2 \sec ^4(c+d x) (b-a \sin (c+d x))}{(a+b \sin (c+d x))^2}}{8 d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + b*Sin[c + d*x])^3,x]

[Out]

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

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Maple [A]  time = 0.142, size = 398, normalized size = 1.2 \begin{align*}{\frac{{b}^{5}}{2\,d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}+6\,{\frac{{b}^{5}a}{d \left ( a+b \right ) ^{4} \left ( a-b \right ) ^{4} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-21\,{\frac{{b}^{5}\ln \left ( a+b\sin \left ( dx+c \right ) \right ){a}^{2}}{d \left ( a+b \right ) ^{5} \left ( a-b \right ) ^{5}}}-3\,{\frac{{b}^{7}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{5} \left ( a-b \right ) ^{5}}}+{\frac{1}{16\,d \left ( a+b \right ) ^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{3\,a}{16\,d \left ( a+b \right ) ^{4} \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{9\,b}{16\,d \left ( a+b \right ) ^{4} \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 ) ^{5}}}-{\frac{15\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) ab}{16\,d \left ( a+b \right ) ^{5}}}-{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ){b}^{2}}{2\,d \left ( a+b \right ) ^{5}}}-{\frac{1}{16\,d \left ( a-b \right ) ^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,a}{16\,d \left ( a-b \right ) ^{4} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{9\,b}{16\,d \left ( a-b \right ) ^{4} \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 ) ^{5}}}-{\frac{15\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) ab}{16\,d \left ( a-b \right ) ^{5}}}+{\frac{3\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ){b}^{2}}{2\,d \left ( a-b \right ) ^{5}}} \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))^3,x)

[Out]

1/2/d*b^5/(a+b)^3/(a-b)^3/(a+b*sin(d*x+c))^2+6/d*b^5*a/(a+b)^4/(a-b)^4/(a+b*sin(d*x+c))-21/d*b^5/(a+b)^5/(a-b)
^5*ln(a+b*sin(d*x+c))*a^2-3/d*b^7/(a+b)^5/(a-b)^5*ln(a+b*sin(d*x+c))+1/16/d/(a+b)^3/(sin(d*x+c)-1)^2-3/16/d/(a
+b)^4/(sin(d*x+c)-1)*a-9/16/d/(a+b)^4/(sin(d*x+c)-1)*b-3/16/d/(a+b)^5*ln(sin(d*x+c)-1)*a^2-15/16/d/(a+b)^5*ln(
sin(d*x+c)-1)*a*b-3/2/d/(a+b)^5*ln(sin(d*x+c)-1)*b^2-1/16/d/(a-b)^3/(1+sin(d*x+c))^2-3/16/d/(a-b)^4/(1+sin(d*x
+c))*a+9/16/d/(a-b)^4/(1+sin(d*x+c))*b+3/16/d/(a-b)^5*ln(1+sin(d*x+c))*a^2-15/16/d/(a-b)^5*ln(1+sin(d*x+c))*a*
b+3/2/d/(a-b)^5*ln(1+sin(d*x+c))*b^2

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Maxima [B]  time = 1.04404, size = 979, normalized size = 2.98 \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)^5/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/16*(48*(7*a^2*b^5 + b^7)*log(b*sin(d*x + c) + a)/(a^10 - 5*a^8*b^2 + 10*a^6*b^4 - 10*a^4*b^6 + 5*a^2*b^8 -
b^10) - 3*(a^2 - 5*a*b + 8*b^2)*log(sin(d*x + c) + 1)/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5
) + 3*(a^2 + 5*a*b + 8*b^2)*log(sin(d*x + c) - 1)/(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5) +
2*(6*a^6*b - 44*a^4*b^3 - 62*a^2*b^5 + 4*b^7 + 3*(a^5*b^2 - 6*a^3*b^4 - 27*a*b^6)*sin(d*x + c)^5 + 6*(a^6*b -
6*a^4*b^3 - 13*a^2*b^5 + 2*b^7)*sin(d*x + c)^4 + (3*a^7 - 23*a^5*b^2 + 61*a^3*b^4 + 151*a*b^6)*sin(d*x + c)^3
- 2*(5*a^6*b - 37*a^4*b^3 - 73*a^2*b^5 + 9*b^7)*sin(d*x + c)^2 - (5*a^7 - 26*a^5*b^2 + 49*a^3*b^4 + 68*a*b^6)*
sin(d*x + c))/(a^10 - 4*a^8*b^2 + 6*a^6*b^4 - 4*a^4*b^6 + a^2*b^8 + (a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b
^8 + b^10)*sin(d*x + c)^6 + 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*sin(d*x + c)^5 + (a^10 - 6*a
^8*b^2 + 14*a^6*b^4 - 16*a^4*b^6 + 9*a^2*b^8 - 2*b^10)*sin(d*x + c)^4 - 4*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a
^3*b^7 + a*b^9)*sin(d*x + c)^3 - (2*a^10 - 9*a^8*b^2 + 16*a^6*b^4 - 14*a^4*b^6 + 6*a^2*b^8 - b^10)*sin(d*x + c
)^2 + 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*sin(d*x + c)))/d

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Fricas [B]  time = 9.12643, size = 1995, normalized size = 6.08 \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)^5/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/16*(4*a^8*b - 16*a^6*b^3 + 24*a^4*b^5 - 16*a^2*b^7 + 4*b^9 + 12*(a^8*b - 7*a^6*b^3 - 7*a^4*b^5 + 15*a^2*b^7
- 2*b^9)*cos(d*x + c)^4 - 4*(a^8*b - 6*a^4*b^5 + 8*a^2*b^7 - 3*b^9)*cos(d*x + c)^2 - 48*((7*a^2*b^7 + b^9)*cos
(d*x + c)^6 - 2*(7*a^3*b^6 + a*b^8)*cos(d*x + c)^4*sin(d*x + c) - (7*a^4*b^5 + 8*a^2*b^7 + b^9)*cos(d*x + c)^4
)*log(b*sin(d*x + c) + a) + 3*((a^7*b^2 - 7*a^5*b^4 + 35*a^3*b^6 + 56*a^2*b^7 + 35*a*b^8 + 8*b^9)*cos(d*x + c)
^6 - 2*(a^8*b - 7*a^6*b^3 + 35*a^4*b^5 + 56*a^3*b^6 + 35*a^2*b^7 + 8*a*b^8)*cos(d*x + c)^4*sin(d*x + c) - (a^9
- 6*a^7*b^2 + 28*a^5*b^4 + 56*a^4*b^5 + 70*a^3*b^6 + 64*a^2*b^7 + 35*a*b^8 + 8*b^9)*cos(d*x + c)^4)*log(sin(d
*x + c) + 1) - 3*((a^7*b^2 - 7*a^5*b^4 + 35*a^3*b^6 - 56*a^2*b^7 + 35*a*b^8 - 8*b^9)*cos(d*x + c)^6 - 2*(a^8*b
- 7*a^6*b^3 + 35*a^4*b^5 - 56*a^3*b^6 + 35*a^2*b^7 - 8*a*b^8)*cos(d*x + c)^4*sin(d*x + c) - (a^9 - 6*a^7*b^2
+ 28*a^5*b^4 - 56*a^4*b^5 + 70*a^3*b^6 - 64*a^2*b^7 + 35*a*b^8 - 8*b^9)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1)
- 2*(2*a^9 - 8*a^7*b^2 + 12*a^5*b^4 - 8*a^3*b^6 + 2*a*b^8 - 3*(a^7*b^2 - 7*a^5*b^4 - 21*a^3*b^6 + 27*a*b^8)*c
os(d*x + c)^4 + (3*a^9 - 20*a^7*b^2 + 42*a^5*b^4 - 36*a^3*b^6 + 11*a*b^8)*cos(d*x + c)^2)*sin(d*x + c))/((a^10
*b^2 - 5*a^8*b^4 + 10*a^6*b^6 - 10*a^4*b^8 + 5*a^2*b^10 - b^12)*d*cos(d*x + c)^6 - 2*(a^11*b - 5*a^9*b^3 + 10*
a^7*b^5 - 10*a^5*b^7 + 5*a^3*b^9 - a*b^11)*d*cos(d*x + c)^4*sin(d*x + c) - (a^12 - 4*a^10*b^2 + 5*a^8*b^4 - 5*
a^4*b^8 + 4*a^2*b^10 - b^12)*d*cos(d*x + c)^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.20583, size = 776, normalized size = 2.37 \begin{align*} -\frac{\frac{48 \,{\left (7 \, a^{2} b^{6} + b^{8}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{10} b - 5 \, a^{8} b^{3} + 10 \, a^{6} b^{5} - 10 \, a^{4} b^{7} + 5 \, a^{2} b^{9} - b^{11}} - \frac{3 \,{\left (a^{2} - 5 \, a b + 8 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}} + \frac{3 \,{\left (a^{2} + 5 \, a b + 8 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}} + \frac{2 \,{\left (3 \, a^{5} b^{2} \sin \left (d x + c\right )^{5} - 18 \, a^{3} b^{4} \sin \left (d x + c\right )^{5} - 81 \, a b^{6} \sin \left (d x + c\right )^{5} + 6 \, a^{6} b \sin \left (d x + c\right )^{4} - 36 \, a^{4} b^{3} \sin \left (d x + c\right )^{4} - 78 \, a^{2} b^{5} \sin \left (d x + c\right )^{4} + 12 \, b^{7} \sin \left (d x + c\right )^{4} + 3 \, a^{7} \sin \left (d x + c\right )^{3} - 23 \, a^{5} b^{2} \sin \left (d x + c\right )^{3} + 61 \, a^{3} b^{4} \sin \left (d x + c\right )^{3} + 151 \, a b^{6} \sin \left (d x + c\right )^{3} - 10 \, a^{6} b \sin \left (d x + c\right )^{2} + 74 \, a^{4} b^{3} \sin \left (d x + c\right )^{2} + 146 \, a^{2} b^{5} \sin \left (d x + c\right )^{2} - 18 \, b^{7} \sin \left (d x + c\right )^{2} - 5 \, a^{7} \sin \left (d x + c\right ) + 26 \, a^{5} b^{2} \sin \left (d x + c\right ) - 49 \, a^{3} b^{4} \sin \left (d x + c\right ) - 68 \, a b^{6} \sin \left (d x + c\right ) + 6 \, a^{6} b - 44 \, a^{4} b^{3} - 62 \, a^{2} b^{5} + 4 \, 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 )^{3} + a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - a\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))^3,x, algorithm="giac")

[Out]

-1/16*(48*(7*a^2*b^6 + b^8)*log(abs(b*sin(d*x + c) + a))/(a^10*b - 5*a^8*b^3 + 10*a^6*b^5 - 10*a^4*b^7 + 5*a^2
*b^9 - b^11) - 3*(a^2 - 5*a*b + 8*b^2)*log(abs(sin(d*x + c) + 1))/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5
*a*b^4 - b^5) + 3*(a^2 + 5*a*b + 8*b^2)*log(abs(sin(d*x + c) - 1))/(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 +
5*a*b^4 + b^5) + 2*(3*a^5*b^2*sin(d*x + c)^5 - 18*a^3*b^4*sin(d*x + c)^5 - 81*a*b^6*sin(d*x + c)^5 + 6*a^6*b*s
in(d*x + c)^4 - 36*a^4*b^3*sin(d*x + c)^4 - 78*a^2*b^5*sin(d*x + c)^4 + 12*b^7*sin(d*x + c)^4 + 3*a^7*sin(d*x
+ c)^3 - 23*a^5*b^2*sin(d*x + c)^3 + 61*a^3*b^4*sin(d*x + c)^3 + 151*a*b^6*sin(d*x + c)^3 - 10*a^6*b*sin(d*x +
c)^2 + 74*a^4*b^3*sin(d*x + c)^2 + 146*a^2*b^5*sin(d*x + c)^2 - 18*b^7*sin(d*x + c)^2 - 5*a^7*sin(d*x + c) +
26*a^5*b^2*sin(d*x + c) - 49*a^3*b^4*sin(d*x + c) - 68*a*b^6*sin(d*x + c) + 6*a^6*b - 44*a^4*b^3 - 62*a^2*b^5
+ 4*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)^3 + a*sin(d*x + c)^2 - b*sin(d*x + c
) - a)^2))/d