3.1535 \(\int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=118 \[ \frac {(5 a A-b B) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\sec ^6(c+d x) ((a A+b B) \sin (c+d x)+a B+A b)}{6 d}+\frac {(5 a A-b B) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {(5 a A-b B) \tan (c+d x) \sec (c+d x)}{16 d} \]

[Out]

1/16*(5*A*a-B*b)*arctanh(sin(d*x+c))/d+1/6*sec(d*x+c)^6*(A*b+a*B+(A*a+B*b)*sin(d*x+c))/d+1/16*(5*A*a-B*b)*sec(
d*x+c)*tan(d*x+c)/d+1/24*(5*A*a-B*b)*sec(d*x+c)^3*tan(d*x+c)/d

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Rubi [A]  time = 0.11, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2837, 778, 199, 206} \[ \frac {(5 a A-b B) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\sec ^6(c+d x) ((a A+b B) \sin (c+d x)+a B+A b)}{6 d}+\frac {(5 a A-b B) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {(5 a A-b B) \tan (c+d x) \sec (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a*A - b*B)*ArcTanh[Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^6*(A*b + a*B + (a*A + b*B)*Sin[c + d*x]))/(6*d) +
((5*a*A - b*B)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + ((5*a*A - b*B)*Sec[c + d*x]^3*Tan[c + d*x])/(24*d)

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

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

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Mathematica [A]  time = 0.88, size = 104, normalized size = 0.88 \[ -\frac {\sec ^6(c+d x) \left ((3 b B-15 a A) \sin ^5(c+d x)+8 (5 a A-b B) \sin ^3(c+d x)-3 (11 a A+b B) \sin (c+d x)-3 (5 a A-b B) \cos ^6(c+d x) \tanh ^{-1}(\sin (c+d x))-8 (a B+A b)\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/48*(Sec[c + d*x]^6*(-8*(A*b + a*B) - 3*(5*a*A - b*B)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^6 - 3*(11*a*A + b*B
)*Sin[c + d*x] + 8*(5*a*A - b*B)*Sin[c + d*x]^3 + (-15*a*A + 3*b*B)*Sin[c + d*x]^5))/d

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fricas [A]  time = 0.47, size = 135, normalized size = 1.14 \[ \frac {3 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 16 \, B a + 16 \, A b + 2 \, {\left (3 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{2} + 8 \, A a + 8 \, B b\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(5*A*a - B*b)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(5*A*a - B*b)*cos(d*x + c)^6*log(-sin(d*x + c)
+ 1) + 16*B*a + 16*A*b + 2*(3*(5*A*a - B*b)*cos(d*x + c)^4 + 2*(5*A*a - B*b)*cos(d*x + c)^2 + 8*A*a + 8*B*b)*s
in(d*x + c))/(d*cos(d*x + c)^6)

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giac [A]  time = 0.26, size = 139, normalized size = 1.18 \[ \frac {3 \, {\left (5 \, A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a \sin \left (d x + c\right )^{5} - 3 \, B b \sin \left (d x + c\right )^{5} - 40 \, A a \sin \left (d x + c\right )^{3} + 8 \, B b \sin \left (d x + c\right )^{3} + 33 \, A a \sin \left (d x + c\right ) + 3 \, B b \sin \left (d x + c\right ) + 8 \, B a + 8 \, A b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(5*A*a - B*b)*log(abs(sin(d*x + c) + 1)) - 3*(5*A*a - B*b)*log(abs(sin(d*x + c) - 1)) - 2*(15*A*a*sin(
d*x + c)^5 - 3*B*b*sin(d*x + c)^5 - 40*A*a*sin(d*x + c)^3 + 8*B*b*sin(d*x + c)^3 + 33*A*a*sin(d*x + c) + 3*B*b
*sin(d*x + c) + 8*B*a + 8*A*b)/(sin(d*x + c)^2 - 1)^3)/d

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maple [A]  time = 0.56, size = 217, normalized size = 1.84 \[ \frac {a A \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {5 a A \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {5 a A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {5 a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {a B}{6 d \cos \left (d x +c \right )^{6}}+\frac {A b}{6 d \cos \left (d x +c \right )^{6}}+\frac {B b \left (\sin ^{3}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}+\frac {B b \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{4}}+\frac {B b \left (\sin ^{3}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{2}}+\frac {b B \sin \left (d x +c \right )}{16 d}-\frac {B b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d*a*A*tan(d*x+c)*sec(d*x+c)^5+5/24/d*a*A*tan(d*x+c)*sec(d*x+c)^3+5/16/d*a*A*sec(d*x+c)*tan(d*x+c)+5/16/d*a
*A*ln(sec(d*x+c)+tan(d*x+c))+1/6/d*a*B/cos(d*x+c)^6+1/6/d*A*b/cos(d*x+c)^6+1/6/d*B*b*sin(d*x+c)^3/cos(d*x+c)^6
+1/8/d*B*b*sin(d*x+c)^3/cos(d*x+c)^4+1/16/d*B*b*sin(d*x+c)^3/cos(d*x+c)^2+1/16*b*B*sin(d*x+c)/d-1/16/d*B*b*ln(
sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.31, size = 143, normalized size = 1.21 \[ \frac {3 \, {\left (5 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, A a - B b\right )} \sin \left (d x + c\right )^{5} - 8 \, {\left (5 \, A a - B b\right )} \sin \left (d x + c\right )^{3} + 8 \, B a + 8 \, A b + 3 \, {\left (11 \, A a + B b\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(5*A*a - B*b)*log(sin(d*x + c) + 1) - 3*(5*A*a - B*b)*log(sin(d*x + c) - 1) - 2*(3*(5*A*a - B*b)*sin(d
*x + c)^5 - 8*(5*A*a - B*b)*sin(d*x + c)^3 + 8*B*a + 8*A*b + 3*(11*A*a + B*b)*sin(d*x + c))/(sin(d*x + c)^6 -
3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1))/d

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mupad [B]  time = 12.44, size = 120, normalized size = 1.02 \[ \frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {5\,A\,a}{16}-\frac {B\,b}{16}\right )}{d}-\frac {\left (\frac {5\,A\,a}{16}-\frac {B\,b}{16}\right )\,{\sin \left (c+d\,x\right )}^5+\left (\frac {B\,b}{6}-\frac {5\,A\,a}{6}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {11\,A\,a}{16}+\frac {B\,b}{16}\right )\,\sin \left (c+d\,x\right )+\frac {A\,b}{6}+\frac {B\,a}{6}}{d\,\left ({\sin \left (c+d\,x\right )}^6-3\,{\sin \left (c+d\,x\right )}^4+3\,{\sin \left (c+d\,x\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh(sin(c + d*x))*((5*A*a)/16 - (B*b)/16))/d - ((A*b)/6 + (B*a)/6 + sin(c + d*x)*((11*A*a)/16 + (B*b)/16) -
sin(c + d*x)^3*((5*A*a)/6 - (B*b)/6) + sin(c + d*x)^5*((5*A*a)/16 - (B*b)/16))/(d*(3*sin(c + d*x)^2 - 3*sin(c
+ d*x)^4 + sin(c + d*x)^6 - 1))

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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+B*sin(d*x+c)),x)

[Out]

Timed out

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