3.1541 \(\int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=112 \[ -\frac {(a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))}{4 d}+\frac {(a-b) (a A+b (A-2 B)) \log (\sin (c+d x)+1)}{4 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) ((a A+b B) \sin (c+d x)+a B+A b)}{2 d} \]

[Out]

-1/4*(a+b)*(a*A-b*(A+2*B))*ln(1-sin(d*x+c))/d+1/4*(a-b)*(a*A+b*(A-2*B))*ln(1+sin(d*x+c))/d+1/2*sec(d*x+c)^2*(a
+b*sin(d*x+c))*(A*b+a*B+(A*a+B*b)*sin(d*x+c))/d

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Rubi [A]  time = 0.18, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2837, 819, 633, 31} \[ -\frac {(a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))}{4 d}+\frac {(a-b) (a A+b (A-2 B)) \log (\sin (c+d x)+1)}{4 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) ((a A+b B) \sin (c+d x)+a B+A b)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*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, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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 ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac {b^3 \operatorname {Subst}\left (\int \frac {(a+x)^2 \left (A+\frac {B x}{b}\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (A b+a B+(a A+b B) \sin (c+d x))}{2 d}-\frac {b \operatorname {Subst}\left (\int \frac {-a^2 A+A b^2+2 a b B+2 b B x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 d}\\ &=\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (A b+a B+(a A+b B) \sin (c+d x))}{2 d}-\frac {((a-b) (a A+b (A-2 B))) \operatorname {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{4 d}+\frac {((a+b) (a A-b (A+2 B))) \operatorname {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=-\frac {(a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))}{4 d}+\frac {(a-b) (a A+b (A-2 B)) \log (1+\sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (A b+a B+(a A+b B) \sin (c+d x))}{2 d}\\ \end {align*}

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Mathematica [A]  time = 1.53, size = 174, normalized size = 1.55 \[ \frac {\left (-6 a^3 A b+4 a A b^3+2 b^4 B\right ) \tan ^2(c+d x)-2 a^3 (a B-A b) \sec ^2(c+d x)+\left (a^2-b^2\right ) ((a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))-(a-b) (a A+b (A-2 B)) \log (\sin (c+d x)+1))-2 \left (a^2-b^2\right ) \left (a^2 A+2 a b B+A b^2\right ) \tan (c+d x) \sec (c+d x)}{4 d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

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

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fricas [A]  time = 0.48, size = 136, normalized size = 1.21 \[ \frac {{\left (A a^{2} - 2 \, B a b - {\left (A - 2 \, B\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} - 2 \, B a b - {\left (A + 2 \, B\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a^{2} + 4 \, A a b + 2 \, B b^{2} + 2 \, {\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/4*((A*a^2 - 2*B*a*b - (A - 2*B)*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (A*a^2 - 2*B*a*b - (A + 2*B)*b^2
)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*B*a^2 + 4*A*a*b + 2*B*b^2 + 2*(A*a^2 + 2*B*a*b + A*b^2)*sin(d*x +
c))/(d*cos(d*x + c)^2)

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giac [A]  time = 0.26, size = 146, normalized size = 1.30 \[ \frac {{\left (A a^{2} - 2 \, B a b - A b^{2} + 2 \, B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (A a^{2} - 2 \, B a b - A b^{2} - 2 \, B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (B b^{2} \sin \left (d x + c\right )^{2} + A a^{2} \sin \left (d x + c\right ) + 2 \, B a b \sin \left (d x + c\right ) + A b^{2} \sin \left (d x + c\right ) + B a^{2} + 2 \, A a b\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/4*((A*a^2 - 2*B*a*b - A*b^2 + 2*B*b^2)*log(abs(sin(d*x + c) + 1)) - (A*a^2 - 2*B*a*b - A*b^2 - 2*B*b^2)*log(
abs(sin(d*x + c) - 1)) - 2*(B*b^2*sin(d*x + c)^2 + A*a^2*sin(d*x + c) + 2*B*a*b*sin(d*x + c) + A*b^2*sin(d*x +
c) + B*a^2 + 2*A*a*b)/(sin(d*x + c)^2 - 1))/d

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maple [B]  time = 0.58, size = 231, normalized size = 2.06 \[ \frac {a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {B \,a^{2}}{2 d \cos \left (d x +c \right )^{2}}+\frac {A a b}{d \cos \left (d x +c \right )^{2}}+\frac {B a b \left (\sin ^{3}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {B a b \sin \left (d x +c \right )}{d}-\frac {B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {A \,b^{2} \sin \left (d x +c \right )}{2 d}-\frac {A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {B \,b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {B \,b^{2} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/2/d*a^2*A*sec(d*x+c)*tan(d*x+c)+1/2/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*B*a^2/cos(d*x+c)^2+1/d*A*a*b/cos
(d*x+c)^2+1/d*B*a*b*sin(d*x+c)^3/cos(d*x+c)^2+1/d*B*a*b*sin(d*x+c)-1/d*B*a*b*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*A
*b^2*sin(d*x+c)^3/cos(d*x+c)^2+1/2/d*A*b^2*sin(d*x+c)-1/2/d*A*b^2*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*B*b^2*tan(d*
x+c)^2+1/d*B*b^2*ln(cos(d*x+c))

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maxima [A]  time = 0.34, size = 122, normalized size = 1.09 \[ \frac {{\left (A a^{2} - 2 \, B a b - {\left (A - 2 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} - 2 \, B a b - {\left (A + 2 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (B a^{2} + 2 \, A a b + B b^{2} + {\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/4*((A*a^2 - 2*B*a*b - (A - 2*B)*b^2)*log(sin(d*x + c) + 1) - (A*a^2 - 2*B*a*b - (A + 2*B)*b^2)*log(sin(d*x +
c) - 1) - 2*(B*a^2 + 2*A*a*b + B*b^2 + (A*a^2 + 2*B*a*b + A*b^2)*sin(d*x + c))/(sin(d*x + c)^2 - 1))/d

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mupad [B]  time = 12.36, size = 118, normalized size = 1.05 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-b\right )\,\left (A\,a+A\,b-2\,B\,b\right )}{4\,d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {A\,a^2}{2}+B\,a\,b+\frac {A\,b^2}{2}\right )+\frac {B\,a^2}{2}+\frac {B\,b^2}{2}+A\,a\,b}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (A\,b-A\,a+2\,B\,b\right )}{4\,d} \]

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)^3,x)

[Out]

(log(sin(c + d*x) + 1)*(a - b)*(A*a + A*b - 2*B*b))/(4*d) - (sin(c + d*x)*((A*a^2)/2 + (A*b^2)/2 + B*a*b) + (B
*a^2)/2 + (B*b^2)/2 + A*a*b)/(d*(sin(c + d*x)^2 - 1)) + (log(sin(c + d*x) - 1)*(a + b)*(A*b - A*a + 2*B*b))/(4
*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \sin {\left (c + d x \right )}\right ) \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Integral((A + B*sin(c + d*x))*(a + b*sin(c + d*x))**2*sec(c + d*x)**3, x)

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