3.1018 \(\int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=236 \[ -\frac {a^3 (A-B)}{80 d (a \sin (c+d x)+a)^5}-\frac {a^2 (2 A-B)}{64 d (a \sin (c+d x)+a)^4}+\frac {3 (7 A+3 B)}{256 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 (7 A+B)}{256 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {7 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{128 a^2 d}+\frac {a (A+B)}{192 d (a-a \sin (c+d x))^3}-\frac {a (5 A-B)}{96 d (a \sin (c+d x)+a)^3}+\frac {3 A+2 B}{128 d (a-a \sin (c+d x))^2}-\frac {5 A}{64 d (a \sin (c+d x)+a)^2} \]

[Out]

7/128*(4*A+B)*arctanh(sin(d*x+c))/a^2/d+1/192*a*(A+B)/d/(a-a*sin(d*x+c))^3+1/128*(3*A+2*B)/d/(a-a*sin(d*x+c))^
2-1/80*a^3*(A-B)/d/(a+a*sin(d*x+c))^5-1/64*a^2*(2*A-B)/d/(a+a*sin(d*x+c))^4-1/96*a*(5*A-B)/d/(a+a*sin(d*x+c))^
3-5/64*A/d/(a+a*sin(d*x+c))^2+3/256*(7*A+3*B)/d/(a^2-a^2*sin(d*x+c))-5/256*(7*A+B)/d/(a^2+a^2*sin(d*x+c))

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Rubi [A]  time = 0.28, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2836, 77, 206} \[ -\frac {a^3 (A-B)}{80 d (a \sin (c+d x)+a)^5}-\frac {a^2 (2 A-B)}{64 d (a \sin (c+d x)+a)^4}+\frac {3 (7 A+3 B)}{256 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 (7 A+B)}{256 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {7 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{128 a^2 d}+\frac {a (A+B)}{192 d (a-a \sin (c+d x))^3}-\frac {a (5 A-B)}{96 d (a \sin (c+d x)+a)^3}+\frac {3 A+2 B}{128 d (a-a \sin (c+d x))^2}-\frac {5 A}{64 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

(7*(4*A + B)*ArcTanh[Sin[c + d*x]])/(128*a^2*d) + (a*(A + B))/(192*d*(a - a*Sin[c + d*x])^3) + (3*A + 2*B)/(12
8*d*(a - a*Sin[c + d*x])^2) - (a^3*(A - B))/(80*d*(a + a*Sin[c + d*x])^5) - (a^2*(2*A - B))/(64*d*(a + a*Sin[c
+ d*x])^4) - (a*(5*A - B))/(96*d*(a + a*Sin[c + d*x])^3) - (5*A)/(64*d*(a + a*Sin[c + d*x])^2) + (3*(7*A + 3*
B))/(256*d*(a^2 - a^2*Sin[c + d*x])) - (5*(7*A + B))/(256*d*(a^2 + a^2*Sin[c + d*x]))

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2836

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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps

\begin {align*} \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx &=\frac {a^7 \operatorname {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^4 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^7 \operatorname {Subst}\left (\int \left (\frac {A+B}{64 a^6 (a-x)^4}+\frac {3 A+2 B}{64 a^7 (a-x)^3}+\frac {3 (7 A+3 B)}{256 a^8 (a-x)^2}+\frac {A-B}{16 a^4 (a+x)^6}+\frac {2 A-B}{16 a^5 (a+x)^5}+\frac {5 A-B}{32 a^6 (a+x)^4}+\frac {5 A}{32 a^7 (a+x)^3}+\frac {5 (7 A+B)}{256 a^8 (a+x)^2}+\frac {7 (4 A+B)}{128 a^8 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a (A+B)}{192 d (a-a \sin (c+d x))^3}+\frac {3 A+2 B}{128 d (a-a \sin (c+d x))^2}-\frac {a^3 (A-B)}{80 d (a+a \sin (c+d x))^5}-\frac {a^2 (2 A-B)}{64 d (a+a \sin (c+d x))^4}-\frac {a (5 A-B)}{96 d (a+a \sin (c+d x))^3}-\frac {5 A}{64 d (a+a \sin (c+d x))^2}+\frac {3 (7 A+3 B)}{256 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 (7 A+B)}{256 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {(7 (4 A+B)) \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{128 a d}\\ &=\frac {7 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{128 a^2 d}+\frac {a (A+B)}{192 d (a-a \sin (c+d x))^3}+\frac {3 A+2 B}{128 d (a-a \sin (c+d x))^2}-\frac {a^3 (A-B)}{80 d (a+a \sin (c+d x))^5}-\frac {a^2 (2 A-B)}{64 d (a+a \sin (c+d x))^4}-\frac {a (5 A-B)}{96 d (a+a \sin (c+d x))^3}-\frac {5 A}{64 d (a+a \sin (c+d x))^2}+\frac {3 (7 A+3 B)}{256 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 (7 A+B)}{256 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.48, size = 160, normalized size = 0.68 \[ \frac {210 (4 A+B) \tanh ^{-1}(\sin (c+d x))-\frac {2 \left (105 (4 A+B) \sin ^7(c+d x)+210 (4 A+B) \sin ^6(c+d x)-175 (4 A+B) \sin ^5(c+d x)-560 (4 A+B) \sin ^4(c+d x)-49 (4 A+B) \sin ^3(c+d x)+462 (4 A+B) \sin ^2(c+d x)+183 (4 A+B) \sin (c+d x)+48 (3 B-8 A)\right )}{(\sin (c+d x)-1)^3 (\sin (c+d x)+1)^5}}{3840 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

(210*(4*A + B)*ArcTanh[Sin[c + d*x]] - (2*(48*(-8*A + 3*B) + 183*(4*A + B)*Sin[c + d*x] + 462*(4*A + B)*Sin[c
+ d*x]^2 - 49*(4*A + B)*Sin[c + d*x]^3 - 560*(4*A + B)*Sin[c + d*x]^4 - 175*(4*A + B)*Sin[c + d*x]^5 + 210*(4*
A + B)*Sin[c + d*x]^6 + 105*(4*A + B)*Sin[c + d*x]^7))/((-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^5))/(3840*a^2
*d)

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fricas [A]  time = 0.77, size = 290, normalized size = 1.23 \[ \frac {420 \, {\left (4 \, A + B\right )} \cos \left (d x + c\right )^{6} - 140 \, {\left (4 \, A + B\right )} \cos \left (d x + c\right )^{4} - 56 \, {\left (4 \, A + B\right )} \cos \left (d x + c\right )^{2} + 105 \, {\left ({\left (4 \, A + B\right )} \cos \left (d x + c\right )^{8} - 2 \, {\left (4 \, A + B\right )} \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, {\left (4 \, A + B\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (4 \, A + B\right )} \cos \left (d x + c\right )^{8} - 2 \, {\left (4 \, A + B\right )} \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, {\left (4 \, A + B\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (105 \, {\left (4 \, A + B\right )} \cos \left (d x + c\right )^{6} - 140 \, {\left (4 \, A + B\right )} \cos \left (d x + c\right )^{4} - 84 \, {\left (4 \, A + B\right )} \cos \left (d x + c\right )^{2} - 256 \, A - 64 \, B\right )} \sin \left (d x + c\right ) - 128 \, A - 512 \, B}{3840 \, {\left (a^{2} d \cos \left (d x + c\right )^{8} - 2 \, a^{2} d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3840*(420*(4*A + B)*cos(d*x + c)^6 - 140*(4*A + B)*cos(d*x + c)^4 - 56*(4*A + B)*cos(d*x + c)^2 + 105*((4*A
+ B)*cos(d*x + c)^8 - 2*(4*A + B)*cos(d*x + c)^6*sin(d*x + c) - 2*(4*A + B)*cos(d*x + c)^6)*log(sin(d*x + c) +
1) - 105*((4*A + B)*cos(d*x + c)^8 - 2*(4*A + B)*cos(d*x + c)^6*sin(d*x + c) - 2*(4*A + B)*cos(d*x + c)^6)*lo
g(-sin(d*x + c) + 1) + 2*(105*(4*A + B)*cos(d*x + c)^6 - 140*(4*A + B)*cos(d*x + c)^4 - 84*(4*A + B)*cos(d*x +
c)^2 - 256*A - 64*B)*sin(d*x + c) - 128*A - 512*B)/(a^2*d*cos(d*x + c)^8 - 2*a^2*d*cos(d*x + c)^6*sin(d*x + c
) - 2*a^2*d*cos(d*x + c)^6)

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giac [A]  time = 0.39, size = 258, normalized size = 1.09 \[ \frac {\frac {420 \, {\left (4 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {420 \, {\left (4 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac {10 \, {\left (308 \, A \sin \left (d x + c\right )^{3} + 77 \, B \sin \left (d x + c\right )^{3} - 1050 \, A \sin \left (d x + c\right )^{2} - 285 \, B \sin \left (d x + c\right )^{2} + 1212 \, A \sin \left (d x + c\right ) + 363 \, B \sin \left (d x + c\right ) - 478 \, A - 163 \, B\right )}}{a^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {3836 \, A \sin \left (d x + c\right )^{5} + 959 \, B \sin \left (d x + c\right )^{5} + 21280 \, A \sin \left (d x + c\right )^{4} + 5095 \, B \sin \left (d x + c\right )^{4} + 47960 \, A \sin \left (d x + c\right )^{3} + 10790 \, B \sin \left (d x + c\right )^{3} + 55360 \, A \sin \left (d x + c\right )^{2} + 11230 \, B \sin \left (d x + c\right )^{2} + 33260 \, A \sin \left (d x + c\right ) + 5435 \, B \sin \left (d x + c\right ) + 8608 \, A + 667 \, B}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{15360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/15360*(420*(4*A + B)*log(abs(sin(d*x + c) + 1))/a^2 - 420*(4*A + B)*log(abs(sin(d*x + c) - 1))/a^2 + 10*(308
*A*sin(d*x + c)^3 + 77*B*sin(d*x + c)^3 - 1050*A*sin(d*x + c)^2 - 285*B*sin(d*x + c)^2 + 1212*A*sin(d*x + c) +
363*B*sin(d*x + c) - 478*A - 163*B)/(a^2*(sin(d*x + c) - 1)^3) - (3836*A*sin(d*x + c)^5 + 959*B*sin(d*x + c)^
5 + 21280*A*sin(d*x + c)^4 + 5095*B*sin(d*x + c)^4 + 47960*A*sin(d*x + c)^3 + 10790*B*sin(d*x + c)^3 + 55360*A
*sin(d*x + c)^2 + 11230*B*sin(d*x + c)^2 + 33260*A*sin(d*x + c) + 5435*B*sin(d*x + c) + 8608*A + 667*B)/(a^2*(
sin(d*x + c) + 1)^5))/d

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maple [A]  time = 0.76, size = 359, normalized size = 1.52 \[ -\frac {7 \ln \left (\sin \left (d x +c \right )-1\right ) A}{64 d \,a^{2}}-\frac {7 \ln \left (\sin \left (d x +c \right )-1\right ) B}{256 d \,a^{2}}+\frac {3 A}{128 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {B}{64 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {A}{192 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {B}{192 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {21 A}{256 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {9 B}{256 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {5 A}{64 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {A}{80 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {B}{80 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {A}{32 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {B}{64 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5 A}{96 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {B}{96 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {7 \ln \left (1+\sin \left (d x +c \right )\right ) A}{64 d \,a^{2}}+\frac {7 B \ln \left (1+\sin \left (d x +c \right )\right )}{256 a^{2} d}-\frac {35 A}{256 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {5 B}{256 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x)

[Out]

-7/64/d/a^2*ln(sin(d*x+c)-1)*A-7/256/d/a^2*ln(sin(d*x+c)-1)*B+3/128/d/a^2/(sin(d*x+c)-1)^2*A+1/64/d/a^2/(sin(d
*x+c)-1)^2*B-1/192/d/a^2/(sin(d*x+c)-1)^3*A-1/192/d/a^2/(sin(d*x+c)-1)^3*B-21/256/d/a^2/(sin(d*x+c)-1)*A-9/256
/d/a^2/(sin(d*x+c)-1)*B-5/64/d/a^2/(1+sin(d*x+c))^2*A-1/80/d/a^2/(1+sin(d*x+c))^5*A+1/80/d/a^2/(1+sin(d*x+c))^
5*B-1/32/d/a^2/(1+sin(d*x+c))^4*A+1/64/d/a^2/(1+sin(d*x+c))^4*B-5/96/d/a^2/(1+sin(d*x+c))^3*A+1/96/d/a^2/(1+si
n(d*x+c))^3*B+7/64/d/a^2*ln(1+sin(d*x+c))*A+7/256*B*ln(1+sin(d*x+c))/a^2/d-35/256/d/a^2/(1+sin(d*x+c))*A-5/256
/d/a^2/(1+sin(d*x+c))*B

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maxima [A]  time = 0.35, size = 252, normalized size = 1.07 \[ -\frac {\frac {2 \, {\left (105 \, {\left (4 \, A + B\right )} \sin \left (d x + c\right )^{7} + 210 \, {\left (4 \, A + B\right )} \sin \left (d x + c\right )^{6} - 175 \, {\left (4 \, A + B\right )} \sin \left (d x + c\right )^{5} - 560 \, {\left (4 \, A + B\right )} \sin \left (d x + c\right )^{4} - 49 \, {\left (4 \, A + B\right )} \sin \left (d x + c\right )^{3} + 462 \, {\left (4 \, A + B\right )} \sin \left (d x + c\right )^{2} + 183 \, {\left (4 \, A + B\right )} \sin \left (d x + c\right ) - 384 \, A + 144 \, B\right )}}{a^{2} \sin \left (d x + c\right )^{8} + 2 \, a^{2} \sin \left (d x + c\right )^{7} - 2 \, a^{2} \sin \left (d x + c\right )^{6} - 6 \, a^{2} \sin \left (d x + c\right )^{5} + 6 \, a^{2} \sin \left (d x + c\right )^{3} + 2 \, a^{2} \sin \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - a^{2}} - \frac {105 \, {\left (4 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {105 \, {\left (4 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{3840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3840*(2*(105*(4*A + B)*sin(d*x + c)^7 + 210*(4*A + B)*sin(d*x + c)^6 - 175*(4*A + B)*sin(d*x + c)^5 - 560*(
4*A + B)*sin(d*x + c)^4 - 49*(4*A + B)*sin(d*x + c)^3 + 462*(4*A + B)*sin(d*x + c)^2 + 183*(4*A + B)*sin(d*x +
c) - 384*A + 144*B)/(a^2*sin(d*x + c)^8 + 2*a^2*sin(d*x + c)^7 - 2*a^2*sin(d*x + c)^6 - 6*a^2*sin(d*x + c)^5
+ 6*a^2*sin(d*x + c)^3 + 2*a^2*sin(d*x + c)^2 - 2*a^2*sin(d*x + c) - a^2) - 105*(4*A + B)*log(sin(d*x + c) + 1
)/a^2 + 105*(4*A + B)*log(sin(d*x + c) - 1)/a^2)/d

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mupad [B]  time = 9.60, size = 240, normalized size = 1.02 \[ \frac {\left (\frac {7\,A}{32}+\frac {7\,B}{128}\right )\,{\sin \left (c+d\,x\right )}^7+\left (\frac {7\,A}{16}+\frac {7\,B}{64}\right )\,{\sin \left (c+d\,x\right )}^6+\left (-\frac {35\,A}{96}-\frac {35\,B}{384}\right )\,{\sin \left (c+d\,x\right )}^5+\left (-\frac {7\,A}{6}-\frac {7\,B}{24}\right )\,{\sin \left (c+d\,x\right )}^4+\left (-\frac {49\,A}{480}-\frac {49\,B}{1920}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {77\,A}{80}+\frac {77\,B}{320}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {61\,A}{160}+\frac {61\,B}{640}\right )\,\sin \left (c+d\,x\right )-\frac {A}{5}+\frac {3\,B}{40}}{d\,\left (-a^2\,{\sin \left (c+d\,x\right )}^8-2\,a^2\,{\sin \left (c+d\,x\right )}^7+2\,a^2\,{\sin \left (c+d\,x\right )}^6+6\,a^2\,{\sin \left (c+d\,x\right )}^5-6\,a^2\,{\sin \left (c+d\,x\right )}^3-2\,a^2\,{\sin \left (c+d\,x\right )}^2+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )}+\frac {7\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (4\,A+B\right )}{128\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(c + d*x))/(cos(c + d*x)^7*(a + a*sin(c + d*x))^2),x)

[Out]

((3*B)/40 - A/5 + sin(c + d*x)*((61*A)/160 + (61*B)/640) - sin(c + d*x)^4*((7*A)/6 + (7*B)/24) + sin(c + d*x)^
6*((7*A)/16 + (7*B)/64) + sin(c + d*x)^7*((7*A)/32 + (7*B)/128) - sin(c + d*x)^5*((35*A)/96 + (35*B)/384) + si
n(c + d*x)^2*((77*A)/80 + (77*B)/320) - sin(c + d*x)^3*((49*A)/480 + (49*B)/1920))/(d*(2*a^2*sin(c + d*x) + a^
2 - 2*a^2*sin(c + d*x)^2 - 6*a^2*sin(c + d*x)^3 + 6*a^2*sin(c + d*x)^5 + 2*a^2*sin(c + d*x)^6 - 2*a^2*sin(c +
d*x)^7 - a^2*sin(c + d*x)^8)) + (7*atanh(sin(c + d*x))*(4*A + B))/(128*a^2*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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