3.1366 \(\int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=177 \[ \frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \left (4 a b^2-b \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac {a b^4 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {b (a+3 b) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {b (a-3 b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2837, 12, 823, 801} \[ \frac {a b^4 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \left (4 a b^2-b \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}-\frac {b (a+3 b) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {b (a-3 b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^4*Tan[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {x}{b (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^4 \operatorname {Subst}\left (\int \frac {x}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {b^2 \operatorname {Subst}\left (\int \frac {-a b^2+3 b^2 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) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a b^2-b \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac {\operatorname {Subst}\left (\int \frac {a b^2 \left (a^2-5 b^2\right )+b^2 \left (a^2+3 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) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a b^2-b \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac {\operatorname {Subst}\left (\int \left (\frac {(a-b)^2 b (a+3 b)}{2 (a+b) (b-x)}+\frac {8 a b^4}{(a-b) (a+b) (a+x)}+\frac {(a-3 b) b (a+b)^2}{2 (a-b) (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac {b (a+3 b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {(a-3 b) b \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a b^4 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a b^2-b \left (a^2+3 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 = 0.94, size = 244, normalized size = 1.38 \[ \frac {\frac {16 a b^4 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3}+\frac {a+3 b}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {a-3 b}{(a-b)^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {1}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {1}{(a-b) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}-\frac {2 b (a+3 b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{(a+b)^3}+\frac {2 b (a-3 b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{(a-b)^3}}{16 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^4*Tan[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

((-2*b*(a + 3*b)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(a + b)^3 + (2*(a - 3*b)*b*Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]])/(a - b)^3 + (16*a*b^4*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) + (a + 3*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) + (a - 3*b)/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2))/(16*d)

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fricas [A]  time = 1.00, size = 255, normalized size = 1.44 \[ \frac {16 \, a b^{4} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{4} b - 6 \, a^{2} b^{3} - 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{4} b - 6 \, a^{2} b^{3} + 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{5} - 8 \, a^{3} b^{2} + 4 \, a b^{4} - 8 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - {\left (a^{4} b + 2 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*a*b^4*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + (a^4*b - 6*a^2*b^3 - 8*a*b^4 - 3*b^5)*cos(d*x + c)^4*l
og(sin(d*x + c) + 1) - (a^4*b - 6*a^2*b^3 + 8*a*b^4 - 3*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a^5 - 8
*a^3*b^2 + 4*a*b^4 - 8*(a^3*b^2 - a*b^4)*cos(d*x + c)^2 - 2*(2*a^4*b - 4*a^2*b^3 + 2*b^5 - (a^4*b + 2*a^2*b^3
- 3*b^5)*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 = 0.27, size = 323, normalized size = 1.82 \[ \frac {\frac {16 \, a b^{5} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} + \frac {{\left (a b - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (6 \, a b^{4} \sin \left (d x + c\right )^{4} - a^{4} b \sin \left (d x + c\right )^{3} - 2 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} + 3 \, b^{5} \sin \left (d x + c\right )^{3} + 4 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} - 16 \, a b^{4} \sin \left (d x + c\right )^{2} - a^{4} b \sin \left (d x + c\right ) + 6 \, a^{2} b^{3} \sin \left (d x + c\right ) - 5 \, b^{5} \sin \left (d x + c\right ) + 2 \, a^{5} - 8 \, a^{3} b^{2} + 12 \, a b^{4}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*a*b^5*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) + (a*b - 3*b^2)*log(abs(sin(
d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (a*b + 3*b^2)*log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b + 3*a
*b^2 + b^3) + 2*(6*a*b^4*sin(d*x + c)^4 - a^4*b*sin(d*x + c)^3 - 2*a^2*b^3*sin(d*x + c)^3 + 3*b^5*sin(d*x + c)
^3 + 4*a^3*b^2*sin(d*x + c)^2 - 16*a*b^4*sin(d*x + c)^2 - a^4*b*sin(d*x + c) + 6*a^2*b^3*sin(d*x + c) - 5*b^5*
sin(d*x + c) + 2*a^5 - 8*a^3*b^2 + 12*a*b^4)/((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 = 0.35, size = 259, normalized size = 1.46 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {3 b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) b^{2}}{16 d \left (a +b \right )^{3}}+\frac {a \,b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {3 b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) b^{2}}{16 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 267, normalized size = 1.51 \[ \frac {\frac {16 \, a b^{4} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (a b - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (4 \, a b^{2} \sin \left (d x + c\right )^{2} - {\left (a^{2} b + 3 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 2 \, a^{3} - 6 \, a b^{2} - {\left (a^{2} b - 5 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 12.40, size = 483, normalized size = 2.73 \[ \frac {a\,b^4\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (a^2\,b-5\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,a^2\,b-3\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (7\,a^2\,b-3\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {4\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-5\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}+\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-3\,b\right )}{8\,d\,{\left (a-b\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*b^4*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - ((2
*tan(c/2 + (d*x)/2)^2*(2*a*b^2 - a^3))/(a^4 + b^4 - 2*a^2*b^2) + (2*tan(c/2 + (d*x)/2)^6*(2*a*b^2 - a^3))/(a^4
+ b^4 - 2*a^2*b^2) + (tan(c/2 + (d*x)/2)^7*(a^2*b - 5*b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)
^3*(7*a^2*b - 3*b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(7*a^2*b - 3*b^3))/(4*(a^4 + b^4 - 2
*a^2*b^2)) - (4*a*b^2*tan(c/2 + (d*x)/2)^4)/(a^4 + b^4 - 2*a^2*b^2) + (b*tan(c/2 + (d*x)/2)*(a^2 - 5*b^2))/(4*
(a^4 + b^4 - 2*a^2*b^2)))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c
/2 + (d*x)/2)^8 + 1)) - (log(tan(c/2 + (d*x)/2) - 1)*(b/(8*(a + b)^2) + b^2/(4*(a + b)^3)))/d + (b*log(tan(c/2
+ (d*x)/2) + 1)*(a - 3*b))/(8*d*(a - b)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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