3.1543 \(\int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=160 \[ \frac {\left (5 a^2 A-2 a b B-A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\sec ^4(c+d x) \left (\left (5 a^2 A-2 a b B+3 A b^2\right ) \sin (c+d x)+2 b (4 a A-b B)\right )}{24 d}+\frac {\left (5 a^2 A-2 a b B-A b^2\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {\sec ^6(c+d x) (a+b \sin (c+d x))^2 (A \sin (c+d x)+B)}{6 d} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2837, 821, 778, 199, 206} \[ \frac {\left (5 a^2 A-2 a b B-A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\sec ^4(c+d x) \left (\left (5 a^2 A-2 a b B+3 A b^2\right ) \sin (c+d x)+2 b (4 a A-b B)\right )}{24 d}+\frac {\left (5 a^2 A-2 a b B-A b^2\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {\sec ^6(c+d x) (a+b \sin (c+d x))^2 (A \sin (c+d x)+B)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a^2*A - A*b^2 - 2*a*b*B)*ArcTanh[Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^6*(B + A*Sin[c + d*x])*(a + b*Sin[c
+ d*x])^2)/(6*d) + (Sec[c + d*x]^4*(2*b*(4*a*A - b*B) + (5*a^2*A + 3*A*b^2 - 2*a*b*B)*Sin[c + d*x]))/(24*d) +
((5*a^2*A - A*b^2 - 2*a*b*B)*Sec[c + d*x]*Tan[c + d*x])/(16*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 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*
x^2)^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (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 \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac {b^7 \operatorname {Subst}\left (\int \frac {(a+x)^2 \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) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}-\frac {b^5 \operatorname {Subst}\left (\int \frac {(a+x) (-5 a A+2 b B-3 A x)}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{6 d}\\ &=\frac {\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}+\frac {\sec ^4(c+d x) \left (2 b (4 a A-b B)+\left (5 a^2 A+3 A b^2-2 a b B\right ) \sin (c+d x)\right )}{24 d}+\frac {\left (b^3 \left (5 a^2 A-A b^2-2 a b B\right )\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) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}+\frac {\sec ^4(c+d x) \left (2 b (4 a A-b B)+\left (5 a^2 A+3 A b^2-2 a b B\right ) \sin (c+d x)\right )}{24 d}+\frac {\left (5 a^2 A-A b^2-2 a b B\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {\left (b \left (5 a^2 A-A b^2-2 a b B\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=\frac {\left (5 a^2 A-A b^2-2 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}+\frac {\sec ^4(c+d x) \left (2 b (4 a A-b B)+\left (5 a^2 A+3 A b^2-2 a b B\right ) \sin (c+d x)\right )}{24 d}+\frac {\left (5 a^2 A-A b^2-2 a b B\right ) \sec (c+d x) \tan (c+d x)}{16 d}\\ \end {align*}

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Mathematica [A]  time = 1.54, size = 242, normalized size = 1.51 \[ \frac {-\frac {3 b \left (-5 a^2 A+2 a b B+A b^2\right ) \left (-2 \left (a^4-b^4\right ) \tan (c+d x) \sec (c+d x)+\left (4 a b^3-6 a^3 b\right ) \tan ^2(c+d x)+2 a^3 b \sec ^2(c+d x)+\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (\sin (c+d x)+1))\right )}{16 (a-b) (a+b)}+b \sec ^6(c+d x) (a+b \sin (c+d x))^3 ((b B-a A) \sin (c+d x)-a B+A b)+\frac {1}{4} b \sec ^4(c+d x) (a+b \sin (c+d x))^3 ((2 b B-5 a A) \sin (c+d x)+3 A b)}{6 b d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sec[c + d*x]^6*(a + b*Sin[c + d*x])^3*(A*b - a*B + (-(a*A) + b*B)*Sin[c + d*x]) + (b*Sec[c + d*x]^4*(a + b*
Sin[c + d*x])^3*(3*A*b + (-5*a*A + 2*b*B)*Sin[c + d*x]))/4 - (3*b*(-5*a^2*A + A*b^2 + 2*a*b*B)*((a^2 - b^2)^2*
(Log[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]]) + 2*a^3*b*Sec[c + d*x]^2 - 2*(a^4 - b^4)*Sec[c + d*x]*Tan[c +
d*x] + (-6*a^3*b + 4*a*b^3)*Tan[c + d*x]^2))/(16*(a - b)*(a + b)))/(6*b*(-a^2 + b^2)*d)

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fricas [A]  time = 0.49, size = 203, normalized size = 1.27 \[ \frac {3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 24 \, B b^{2} \cos \left (d x + c\right )^{2} + 16 \, B a^{2} + 32 \, A a b + 16 \, B b^{2} + 2 \, {\left (3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, A a^{2} + 16 \, B a b + 8 \, A b^{2} + 2 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d
*x + c)^6*log(-sin(d*x + c) + 1) - 24*B*b^2*cos(d*x + c)^2 + 16*B*a^2 + 32*A*a*b + 16*B*b^2 + 2*(3*(5*A*a^2 -
2*B*a*b - A*b^2)*cos(d*x + c)^4 + 8*A*a^2 + 16*B*a*b + 8*A*b^2 + 2*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^2)
*sin(d*x + c))/(d*cos(d*x + c)^6)

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giac [A]  time = 0.27, size = 229, normalized size = 1.43 \[ \frac {3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a^{2} \sin \left (d x + c\right )^{5} - 6 \, B a b \sin \left (d x + c\right )^{5} - 3 \, A b^{2} \sin \left (d x + c\right )^{5} - 40 \, A a^{2} \sin \left (d x + c\right )^{3} + 16 \, B a b \sin \left (d x + c\right )^{3} + 8 \, A b^{2} \sin \left (d x + c\right )^{3} + 12 \, B b^{2} \sin \left (d x + c\right )^{2} + 33 \, A a^{2} \sin \left (d x + c\right ) + 6 \, B a b \sin \left (d x + c\right ) + 3 \, A b^{2} \sin \left (d x + c\right ) + 8 \, B a^{2} + 16 \, A a b - 4 \, B b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x + c) + 1)) - 3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x
+ c) - 1)) - 2*(15*A*a^2*sin(d*x + c)^5 - 6*B*a*b*sin(d*x + c)^5 - 3*A*b^2*sin(d*x + c)^5 - 40*A*a^2*sin(d*x
+ c)^3 + 16*B*a*b*sin(d*x + c)^3 + 8*A*b^2*sin(d*x + c)^3 + 12*B*b^2*sin(d*x + c)^2 + 33*A*a^2*sin(d*x + c) +
6*B*a*b*sin(d*x + c) + 3*A*b^2*sin(d*x + c) + 8*B*a^2 + 16*A*a*b - 4*B*b^2)/(sin(d*x + c)^2 - 1)^3)/d

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maple [B]  time = 0.58, size = 396, normalized size = 2.48 \[ \frac {a^{2} A \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {5 a^{2} A \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {5 a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {5 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {B \,a^{2}}{6 d \cos \left (d x +c \right )^{6}}+\frac {A a b}{3 d \cos \left (d x +c \right )^{6}}+\frac {B a b \left (\sin ^{3}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{6}}+\frac {B a b \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {B a b \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {B a b \sin \left (d x +c \right )}{8 d}-\frac {B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {A \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}+\frac {A \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{4}}+\frac {A \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{2}}+\frac {A \,b^{2} \sin \left (d x +c \right )}{16 d}-\frac {A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {B \,b^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}+\frac {B \,b^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{12 d \cos \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d*a^2*A*tan(d*x+c)*sec(d*x+c)^5+5/24/d*a^2*A*tan(d*x+c)*sec(d*x+c)^3+5/16/d*a^2*A*sec(d*x+c)*tan(d*x+c)+5/
16/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c))+1/6/d*B*a^2/cos(d*x+c)^6+1/3/d*A*a*b/cos(d*x+c)^6+1/3/d*B*a*b*sin(d*x+c)^
3/cos(d*x+c)^6+1/4/d*B*a*b*sin(d*x+c)^3/cos(d*x+c)^4+1/8/d*B*a*b*sin(d*x+c)^3/cos(d*x+c)^2+1/8/d*B*a*b*sin(d*x
+c)-1/8/d*B*a*b*ln(sec(d*x+c)+tan(d*x+c))+1/6/d*A*b^2*sin(d*x+c)^3/cos(d*x+c)^6+1/8/d*A*b^2*sin(d*x+c)^3/cos(d
*x+c)^4+1/16/d*A*b^2*sin(d*x+c)^3/cos(d*x+c)^2+1/16/d*A*b^2*sin(d*x+c)-1/16/d*A*b^2*ln(sec(d*x+c)+tan(d*x+c))+
1/6/d*B*b^2*sin(d*x+c)^4/cos(d*x+c)^6+1/12/d*B*b^2*sin(d*x+c)^4/cos(d*x+c)^4

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maxima [A]  time = 0.45, size = 211, normalized size = 1.32 \[ \frac {3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{5} + 12 \, B b^{2} \sin \left (d x + c\right )^{2} - 8 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} + 8 \, B a^{2} + 16 \, A a b - 4 \, B b^{2} + 3 \, {\left (11 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(sin(d*x + c) + 1) - 3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(sin(d*x + c) - 1
) - 2*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^5 + 12*B*b^2*sin(d*x + c)^2 - 8*(5*A*a^2 - 2*B*a*b - A*b^2)*
sin(d*x + c)^3 + 8*B*a^2 + 16*A*a*b - 4*B*b^2 + 3*(11*A*a^2 + 2*B*a*b + A*b^2)*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 = 12.45, size = 220, normalized size = 1.38 \[ -\frac {\mathrm {atanh}\left (\frac {4\,\sin \left (c+d\,x\right )\,\left (-\frac {5\,A\,a^2}{32}+\frac {B\,a\,b}{16}+\frac {A\,b^2}{32}\right )}{-\frac {5\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}}\right )\,\left (-\frac {5\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )}{d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {11\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )+\frac {B\,a^2}{6}-\frac {B\,b^2}{12}+{\sin \left (c+d\,x\right )}^3\,\left (-\frac {5\,A\,a^2}{6}+\frac {B\,a\,b}{3}+\frac {A\,b^2}{6}\right )-{\sin \left (c+d\,x\right )}^5\,\left (-\frac {5\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )+\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^2}{4}+\frac {A\,a\,b}{3}}{d\,\left ({\sin \left (c+d\,x\right )}^6-3\,{\sin \left (c+d\,x\right )}^4+3\,{\sin \left (c+d\,x\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (atanh((4*sin(c + d*x)*((A*b^2)/32 - (5*A*a^2)/32 + (B*a*b)/16))/((A*b^2)/8 - (5*A*a^2)/8 + (B*a*b)/4))*((A*
b^2)/16 - (5*A*a^2)/16 + (B*a*b)/8))/d - (sin(c + d*x)*((11*A*a^2)/16 + (A*b^2)/16 + (B*a*b)/8) + (B*a^2)/6 -
(B*b^2)/12 + sin(c + d*x)^3*((A*b^2)/6 - (5*A*a^2)/6 + (B*a*b)/3) - sin(c + d*x)^5*((A*b^2)/16 - (5*A*a^2)/16
+ (B*a*b)/8) + (B*b^2*sin(c + d*x)^2)/4 + (A*a*b)/3)/(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))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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