3.999 \(\int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=69 \[ a^3 B x-\frac {2 a^5 B \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 69, 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, 2670, 2680, 8} \[ -\frac {2 a^5 B \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )}+a^3 B x+\frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

a^3*B*x + ((A + B)*Sec[c + d*x]^3*(a + a*Sin[c + d*x])^3)/(3*d) - (2*a^5*B*Cos[c + d*x])/(d*(a^2 - a^2*Sin[c +
d*x]))

Rule 8

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

Rule 2670

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

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

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]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-(a B) \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=\frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-\left (a^5 B\right ) \int \frac {\cos ^2(c+d x)}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-\frac {2 a^5 B \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )}+\left (a^3 B\right ) \int 1 \, dx\\ &=a^3 B x+\frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-\frac {2 a^5 B \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.11, size = 121, normalized size = 1.75 \[ -\frac {a^3 \left (-3 \cos \left (\frac {1}{2} (c+d x)\right ) (2 A+3 B (c+d x+2))+\cos \left (\frac {3}{2} (c+d x)\right ) (2 A+B (3 c+3 d x+14))+6 B \sin \left (\frac {1}{2} (c+d x)\right ) (2 (c+d x+2)+(c+d x) \cos (c+d x))\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

-1/6*(a^3*(-3*(2*A + 3*B*(2 + c + d*x))*Cos[(c + d*x)/2] + (2*A + B*(14 + 3*c + 3*d*x))*Cos[(3*(c + d*x))/2] +
6*B*(2*(2 + c + d*x) + (c + d*x)*Cos[c + d*x])*Sin[(c + d*x)/2]))/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)

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fricas [B]  time = 0.70, size = 167, normalized size = 2.42 \[ -\frac {6 \, B a^{3} d x + 2 \, {\left (A + B\right )} a^{3} - {\left (3 \, B a^{3} d x + {\left (A + 7 \, B\right )} a^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, B a^{3} d x + {\left (A - 5 \, B\right )} a^{3}\right )} \cos \left (d x + c\right ) - {\left (6 \, B a^{3} d x - 2 \, {\left (A + B\right )} a^{3} + {\left (3 \, B a^{3} d x - {\left (A + 7 \, B\right )} a^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/3*(6*B*a^3*d*x + 2*(A + B)*a^3 - (3*B*a^3*d*x + (A + 7*B)*a^3)*cos(d*x + c)^2 + (3*B*a^3*d*x + (A - 5*B)*a^
3)*cos(d*x + c) - (6*B*a^3*d*x - 2*(A + B)*a^3 + (3*B*a^3*d*x - (A + 7*B)*a^3)*cos(d*x + c))*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 = 0.25, size = 93, normalized size = 1.35 \[ \frac {3 \, {\left (d x + c\right )} B a^{3} - \frac {2 \, {\left (3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{3} - 5 \, B a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)*B*a^3 - 2*(3*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 3*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*B*a^3*tan(1/2
*d*x + 1/2*c) + A*a^3 - 5*B*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

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maple [B]  time = 0.67, size = 248, normalized size = 3.59 \[ \frac {a^{3} A \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+B \,a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+\frac {a^{3} A \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{3} A}{\cos \left (d x +c \right )^{3}}+\frac {B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}-a^{3} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {B \,a^{3}}{3 \cos \left (d x +c \right )^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

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

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maxima [B]  time = 0.42, size = 164, normalized size = 2.38 \[ \frac {3 \, A a^{3} \tan \left (d x + c\right )^{3} + 3 \, B a^{3} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} B a^{3} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} A a^{3}}{\cos \left (d x + c\right )^{3}} - \frac {3 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} B a^{3}}{\cos \left (d x + c\right )^{3}} + \frac {3 \, A a^{3}}{\cos \left (d x + c\right )^{3}} + \frac {B a^{3}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.81, size = 140, normalized size = 2.03 \[ B\,a^3\,x-\frac {\frac {a^3\,\left (2\,A-10\,B+3\,B\,\left (c+d\,x\right )\right )}{3}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3\,\left (6\,A-6\,B+9\,B\,\left (c+d\,x\right )\right )}{3}-3\,B\,a^3\,\left (c+d\,x\right )\right )+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^3\,\left (24\,B-9\,B\,\left (c+d\,x\right )\right )}{3}+3\,B\,a^3\,\left (c+d\,x\right )\right )-B\,a^3\,\left (c+d\,x\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3)/cos(c + d*x)^4,x)

[Out]

B*a^3*x - ((a^3*(2*A - 10*B + 3*B*(c + d*x)))/3 + tan(c/2 + (d*x)/2)^2*((a^3*(6*A - 6*B + 9*B*(c + d*x)))/3 -
3*B*a^3*(c + d*x)) + tan(c/2 + (d*x)/2)*((a^3*(24*B - 9*B*(c + d*x)))/3 + 3*B*a^3*(c + d*x)) - B*a^3*(c + d*x)
)/(d*(tan(c/2 + (d*x)/2) - 1)^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))**3*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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