3.981 \(\int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=73 \[ \frac {a^2 (A-2 B) \tan (c+d x)}{3 d}+\frac {a^2 (A-2 B) \sec (c+d x)}{3 d}+\frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2855, 2669, 3767, 8} \[ \frac {a^2 (A-2 B) \tan (c+d x)}{3 d}+\frac {a^2 (A-2 B) \sec (c+d x)}{3 d}+\frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(A - 2*B)*Sec[c + d*x])/(3*d) + ((A + B)*Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2)/(3*d) + (a^2*(A - 2*B)*Ta
n[c + d*x])/(3*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 121, normalized size = 1.66 \[ -\frac {a^2 A \tan ^3(c+d x)}{3 d}+\frac {2 a^2 A \sec ^3(c+d x)}{3 d}+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{d}+\frac {2 a^2 B \tan ^3(c+d x)}{3 d}-\frac {a^2 B \sec ^3(c+d x)}{3 d}+\frac {a^2 B \tan ^2(c+d x) \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(2*a^2*A*Sec[c + d*x]^3)/(3*d) - (a^2*B*Sec[c + d*x]^3)/(3*d) + (a^2*A*Sec[c + d*x]^2*Tan[c + d*x])/d + (a^2*B
*Sec[c + d*x]*Tan[c + d*x]^2)/d - (a^2*A*Tan[c + d*x]^3)/(3*d) + (2*a^2*B*Tan[c + d*x]^3)/(3*d)

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fricas [A]  time = 0.49, size = 120, normalized size = 1.64 \[ -\frac {{\left (A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right ) + {\left (A + B\right )} a^{2} - {\left ({\left (A - 2 \, B\right )} a^{2} \cos \left (d x + c\right ) - {\left (A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 78, normalized size = 1.07 \[ -\frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A a^{2} - B a^{2}\right )}}{3 \, d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-2/3*(3*A*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*A*a^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2*tan(1/2*d*x + 1/2*c) + 2*A*a^2 -
B*a^2)/(d*(tan(1/2*d*x + 1/2*c) - 1)^3)

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maple [B]  time = 0.66, size = 162, normalized size = 2.22 \[ \frac {\frac {a^{2} A \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {2 a^{2} A}{3 \cos \left (d x +c \right )^{3}}+\frac {2 B \,a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}-a^{2} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {B \,a^{2}}{3 \cos \left (d x +c \right )^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/d*(1/3*a^2*A*sin(d*x+c)^3/cos(d*x+c)^3+B*a^2*(1/3*sin(d*x+c)^4/cos(d*x+c)^3-1/3*sin(d*x+c)^4/cos(d*x+c)-1/3*
(2+sin(d*x+c)^2)*cos(d*x+c))+2/3*a^2*A/cos(d*x+c)^3+2/3*B*a^2*sin(d*x+c)^3/cos(d*x+c)^3-a^2*A*(-2/3-1/3*sec(d*
x+c)^2)*tan(d*x+c)+1/3*B*a^2/cos(d*x+c)^3)

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maxima [A]  time = 0.44, size = 108, normalized size = 1.48 \[ \frac {A a^{2} \tan \left (d x + c\right )^{3} + 2 \, B a^{2} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} B a^{2}}{\cos \left (d x + c\right )^{3}} + \frac {2 \, A a^{2}}{\cos \left (d x + c\right )^{3}} + \frac {B a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.15, size = 77, normalized size = 1.05 \[ -\frac {\sqrt {2}\,a^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B}{2}-\frac {5\,A}{2}+\frac {A\,\cos \left (c+d\,x\right )}{2}+\frac {B\,\cos \left (c+d\,x\right )}{2}+\frac {3\,A\,\sin \left (c+d\,x\right )}{2}-\frac {3\,B\,\sin \left (c+d\,x\right )}{2}\right )}{6\,d\,{\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d\,x}{2}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^4,x)

[Out]

-(2^(1/2)*a^2*cos(c/2 + (d*x)/2)*(B/2 - (5*A)/2 + (A*cos(c + d*x))/2 + (B*cos(c + d*x))/2 + (3*A*sin(c + d*x))
/2 - (3*B*sin(c + d*x))/2))/(6*d*cos(c/2 + pi/4 + (d*x)/2)^3)

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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)**4*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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