3.1010 \(\int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=205 \[ -\frac {a^3 (A-B)}{64 d (a \sin (c+d x)+a)^4}+\frac {a^2 (A+B)}{96 d (a-a \sin (c+d x))^3}-\frac {a^2 (2 A-B)}{48 d (a \sin (c+d x)+a)^3}+\frac {a (5 A+3 B)}{128 d (a-a \sin (c+d x))^2}-\frac {a (5 A-B)}{64 d (a \sin (c+d x)+a)^2}+\frac {5 (3 A+B)}{128 d (a-a \sin (c+d x))}+\frac {5 (7 A+B) \tanh ^{-1}(\sin (c+d x))}{128 a d}-\frac {5 A}{32 d (a \sin (c+d x)+a)} \]

[Out]

5/128*(7*A+B)*arctanh(sin(d*x+c))/a/d+1/96*a^2*(A+B)/d/(a-a*sin(d*x+c))^3+1/128*a*(5*A+3*B)/d/(a-a*sin(d*x+c))
^2+5/128*(3*A+B)/d/(a-a*sin(d*x+c))-1/64*a^3*(A-B)/d/(a+a*sin(d*x+c))^4-1/48*a^2*(2*A-B)/d/(a+a*sin(d*x+c))^3-
1/64*a*(5*A-B)/d/(a+a*sin(d*x+c))^2-5/32*A/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.25, antiderivative size = 205, 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} \[ -\frac {a^3 (A-B)}{64 d (a \sin (c+d x)+a)^4}+\frac {a^2 (A+B)}{96 d (a-a \sin (c+d x))^3}-\frac {a^2 (2 A-B)}{48 d (a \sin (c+d x)+a)^3}+\frac {a (5 A+3 B)}{128 d (a-a \sin (c+d x))^2}-\frac {a (5 A-B)}{64 d (a \sin (c+d x)+a)^2}+\frac {5 (3 A+B)}{128 d (a-a \sin (c+d x))}+\frac {5 (7 A+B) \tanh ^{-1}(\sin (c+d x))}{128 a d}-\frac {5 A}{32 d (a \sin (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(7*A + B)*ArcTanh[Sin[c + d*x]])/(128*a*d) + (a^2*(A + B))/(96*d*(a - a*Sin[c + d*x])^3) + (a*(5*A + 3*B))/
(128*d*(a - a*Sin[c + d*x])^2) + (5*(3*A + B))/(128*d*(a - a*Sin[c + d*x])) - (a^3*(A - B))/(64*d*(a + a*Sin[c
+ d*x])^4) - (a^2*(2*A - B))/(48*d*(a + a*Sin[c + d*x])^3) - (a*(5*A - B))/(64*d*(a + a*Sin[c + d*x])^2) - (5
*A)/(32*d*(a + a*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)} \, dx &=\frac {a^7 \operatorname {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^4 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^7 \operatorname {Subst}\left (\int \left (\frac {A+B}{32 a^5 (a-x)^4}+\frac {5 A+3 B}{64 a^6 (a-x)^3}+\frac {5 (3 A+B)}{128 a^7 (a-x)^2}+\frac {A-B}{16 a^4 (a+x)^5}+\frac {2 A-B}{16 a^5 (a+x)^4}+\frac {5 A-B}{32 a^6 (a+x)^3}+\frac {5 A}{32 a^7 (a+x)^2}+\frac {5 (7 A+B)}{128 a^7 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^2 (A+B)}{96 d (a-a \sin (c+d x))^3}+\frac {a (5 A+3 B)}{128 d (a-a \sin (c+d x))^2}+\frac {5 (3 A+B)}{128 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{64 d (a+a \sin (c+d x))^4}-\frac {a^2 (2 A-B)}{48 d (a+a \sin (c+d x))^3}-\frac {a (5 A-B)}{64 d (a+a \sin (c+d x))^2}-\frac {5 A}{32 d (a+a \sin (c+d x))}+\frac {(5 (7 A+B)) \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{128 d}\\ &=\frac {5 (7 A+B) \tanh ^{-1}(\sin (c+d x))}{128 a d}+\frac {a^2 (A+B)}{96 d (a-a \sin (c+d x))^3}+\frac {a (5 A+3 B)}{128 d (a-a \sin (c+d x))^2}+\frac {5 (3 A+B)}{128 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{64 d (a+a \sin (c+d x))^4}-\frac {a^2 (2 A-B)}{48 d (a+a \sin (c+d x))^3}-\frac {a (5 A-B)}{64 d (a+a \sin (c+d x))^2}-\frac {5 A}{32 d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.88, size = 142, normalized size = 0.69 \[ \frac {\frac {-15 (7 A+B) \sin ^6(c+d x)-15 (7 A+B) \sin ^5(c+d x)+40 (7 A+B) \sin ^4(c+d x)+40 (7 A+B) \sin ^3(c+d x)-33 (7 A+B) \sin ^2(c+d x)-33 (7 A+B) \sin (c+d x)+48 (A-B)}{(\sin (c+d x)-1)^3 (\sin (c+d x)+1)^4}+15 (7 A+B) \tanh ^{-1}(\sin (c+d x))}{384 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(7*A + B)*ArcTanh[Sin[c + d*x]] + (48*(A - B) - 33*(7*A + B)*Sin[c + d*x] - 33*(7*A + B)*Sin[c + d*x]^2 +
40*(7*A + B)*Sin[c + d*x]^3 + 40*(7*A + B)*Sin[c + d*x]^4 - 15*(7*A + B)*Sin[c + d*x]^5 - 15*(7*A + B)*Sin[c +
d*x]^6)/((-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^4))/(384*a*d)

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fricas [A]  time = 0.61, size = 224, normalized size = 1.09 \[ -\frac {30 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6} - 10 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left ({\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left ({\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{4} + 10 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{2} + 56 \, A + 8 \, B\right )} \sin \left (d x + c\right ) - 16 \, A - 112 \, B}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \]

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)),x, algorithm="fricas")

[Out]

-1/768*(30*(7*A + B)*cos(d*x + c)^6 - 10*(7*A + B)*cos(d*x + c)^4 - 4*(7*A + B)*cos(d*x + c)^2 - 15*((7*A + B)
*cos(d*x + c)^6*sin(d*x + c) + (7*A + B)*cos(d*x + c)^6)*log(sin(d*x + c) + 1) + 15*((7*A + B)*cos(d*x + c)^6*
sin(d*x + c) + (7*A + B)*cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 2*(15*(7*A + B)*cos(d*x + c)^4 + 10*(7*A + B
)*cos(d*x + c)^2 + 56*A + 8*B)*sin(d*x + c) - 16*A - 112*B)/(a*d*cos(d*x + c)^6*sin(d*x + c) + a*d*cos(d*x + c
)^6)

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giac [A]  time = 0.31, size = 236, normalized size = 1.15 \[ \frac {\frac {60 \, {\left (7 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {60 \, {\left (7 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (385 \, A \sin \left (d x + c\right )^{3} + 55 \, B \sin \left (d x + c\right )^{3} - 1335 \, A \sin \left (d x + c\right )^{2} - 225 \, B \sin \left (d x + c\right )^{2} + 1575 \, A \sin \left (d x + c\right ) + 321 \, B \sin \left (d x + c\right ) - 641 \, A - 167 \, B\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {875 \, A \sin \left (d x + c\right )^{4} + 125 \, B \sin \left (d x + c\right )^{4} + 3980 \, A \sin \left (d x + c\right )^{3} + 500 \, B \sin \left (d x + c\right )^{3} + 6930 \, A \sin \left (d x + c\right )^{2} + 702 \, B \sin \left (d x + c\right )^{2} + 5548 \, A \sin \left (d x + c\right ) + 340 \, B \sin \left (d x + c\right ) + 1771 \, A - 35 \, B}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \]

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)),x, algorithm="giac")

[Out]

1/3072*(60*(7*A + B)*log(abs(sin(d*x + c) + 1))/a - 60*(7*A + B)*log(abs(sin(d*x + c) - 1))/a + 2*(385*A*sin(d
*x + c)^3 + 55*B*sin(d*x + c)^3 - 1335*A*sin(d*x + c)^2 - 225*B*sin(d*x + c)^2 + 1575*A*sin(d*x + c) + 321*B*s
in(d*x + c) - 641*A - 167*B)/(a*(sin(d*x + c) - 1)^3) - (875*A*sin(d*x + c)^4 + 125*B*sin(d*x + c)^4 + 3980*A*
sin(d*x + c)^3 + 500*B*sin(d*x + c)^3 + 6930*A*sin(d*x + c)^2 + 702*B*sin(d*x + c)^2 + 5548*A*sin(d*x + c) + 3
40*B*sin(d*x + c) + 1771*A - 35*B)/(a*(sin(d*x + c) + 1)^4))/d

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maple [A]  time = 0.50, size = 321, normalized size = 1.57 \[ -\frac {35 \ln \left (\sin \left (d x +c \right )-1\right ) A}{256 a d}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right ) B}{256 a d}+\frac {5 A}{128 a d \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3 B}{128 a d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {A}{96 a d \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {B}{96 a d \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {15 A}{128 a d \left (\sin \left (d x +c \right )-1\right )}-\frac {5 B}{128 a d \left (\sin \left (d x +c \right )-1\right )}-\frac {5 A}{32 a d \left (1+\sin \left (d x +c \right )\right )}-\frac {A}{64 a d \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {B}{64 a d \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {A}{24 a d \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {B}{48 a d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5 A}{64 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {B}{64 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {35 \ln \left (1+\sin \left (d x +c \right )\right ) A}{256 d a}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right ) B}{256 d a} \]

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)),x)

[Out]

-35/256/a/d*ln(sin(d*x+c)-1)*A-5/256/a/d*ln(sin(d*x+c)-1)*B+5/128/a/d/(sin(d*x+c)-1)^2*A+3/128/a/d/(sin(d*x+c)
-1)^2*B-1/96/a/d/(sin(d*x+c)-1)^3*A-1/96/a/d/(sin(d*x+c)-1)^3*B-15/128/a/d/(sin(d*x+c)-1)*A-5/128/a/d/(sin(d*x
+c)-1)*B-5/32/a/d/(1+sin(d*x+c))*A-1/64/a/d/(1+sin(d*x+c))^4*A+1/64/a/d/(1+sin(d*x+c))^4*B-1/24/a/d/(1+sin(d*x
+c))^3*A+1/48/a/d/(1+sin(d*x+c))^3*B-5/64/a/d/(1+sin(d*x+c))^2*A+1/64/a/d/(1+sin(d*x+c))^2*B+35/256/d/a*ln(1+s
in(d*x+c))*A+5/256/d/a*ln(1+sin(d*x+c))*B

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maxima [A]  time = 0.34, size = 220, normalized size = 1.07 \[ \frac {\frac {15 \, {\left (7 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {15 \, {\left (7 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {2 \, {\left (15 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{6} + 15 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{5} - 40 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{4} - 40 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{3} + 33 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{2} + 33 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right ) - 48 \, A + 48 \, B\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a}}{768 \, d} \]

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)),x, algorithm="maxima")

[Out]

1/768*(15*(7*A + B)*log(sin(d*x + c) + 1)/a - 15*(7*A + B)*log(sin(d*x + c) - 1)/a - 2*(15*(7*A + B)*sin(d*x +
c)^6 + 15*(7*A + B)*sin(d*x + c)^5 - 40*(7*A + B)*sin(d*x + c)^4 - 40*(7*A + B)*sin(d*x + c)^3 + 33*(7*A + B)
*sin(d*x + c)^2 + 33*(7*A + B)*sin(d*x + c) - 48*A + 48*B)/(a*sin(d*x + c)^7 + a*sin(d*x + c)^6 - 3*a*sin(d*x
+ c)^5 - 3*a*sin(d*x + c)^4 + 3*a*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - a*sin(d*x + c) - a))/d

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mupad [B]  time = 9.35, size = 206, normalized size = 1.00 \[ \frac {\left (\frac {35\,A}{128}+\frac {5\,B}{128}\right )\,{\sin \left (c+d\,x\right )}^6+\left (\frac {35\,A}{128}+\frac {5\,B}{128}\right )\,{\sin \left (c+d\,x\right )}^5+\left (-\frac {35\,A}{48}-\frac {5\,B}{48}\right )\,{\sin \left (c+d\,x\right )}^4+\left (-\frac {35\,A}{48}-\frac {5\,B}{48}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {77\,A}{128}+\frac {11\,B}{128}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {77\,A}{128}+\frac {11\,B}{128}\right )\,\sin \left (c+d\,x\right )-\frac {A}{8}+\frac {B}{8}}{d\,\left (-a\,{\sin \left (c+d\,x\right )}^7-a\,{\sin \left (c+d\,x\right )}^6+3\,a\,{\sin \left (c+d\,x\right )}^5+3\,a\,{\sin \left (c+d\,x\right )}^4-3\,a\,{\sin \left (c+d\,x\right )}^3-3\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )}+\frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (7\,A+B\right )}{128\,a\,d} \]

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))),x)

[Out]

(B/8 - A/8 + sin(c + d*x)*((77*A)/128 + (11*B)/128) - sin(c + d*x)^3*((35*A)/48 + (5*B)/48) - sin(c + d*x)^4*(
(35*A)/48 + (5*B)/48) + sin(c + d*x)^5*((35*A)/128 + (5*B)/128) + sin(c + d*x)^6*((35*A)/128 + (5*B)/128) + si
n(c + d*x)^2*((77*A)/128 + (11*B)/128))/(d*(a + a*sin(c + d*x) - 3*a*sin(c + d*x)^2 - 3*a*sin(c + d*x)^3 + 3*a
*sin(c + d*x)^4 + 3*a*sin(c + d*x)^5 - a*sin(c + d*x)^6 - a*sin(c + d*x)^7)) + (5*atanh(sin(c + d*x))*(7*A + B
))/(128*a*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)),x)

[Out]

Timed out

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