3.639 \(\int (b \sec (c+d x)+a \sin (c+d x))^2 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx\)

Optimal. Leaf size=26 \[ \frac {(a \sin (c+d x)+b \sec (c+d x))^3}{3 d} \]

[Out]

1/3*(b*sec(d*x+c)+a*sin(d*x+c))^3/d

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Rubi [A]  time = 0.04, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {4385} \[ \frac {(a \sin (c+d x)+b \sec (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x] + a*Sin[c + d*x])^2*(a*Cos[c + d*x] + b*Sec[c + d*x]*Tan[c + d*x]),x]

[Out]

(b*Sec[c + d*x] + a*Sin[c + d*x])^3/(3*d)

Rule 4385

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[ActivateTrig[y], ActivateTrig[u], x]}, Simp[(q*A
ctivateTrig[y^(m + 1)])/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1] &&  !InertTrigFreeQ[u]

Rubi steps

\begin {align*} \int (b \sec (c+d x)+a \sin (c+d x))^2 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx &=\frac {(b \sec (c+d x)+a \sin (c+d x))^3}{3 d}\\ \end {align*}

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Mathematica [A]  time = 1.22, size = 31, normalized size = 1.19 \[ \frac {\sec ^3(c+d x) (a \sin (2 (c+d x))+2 b)^3}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x] + a*Sin[c + d*x])^2*(a*Cos[c + d*x] + b*Sec[c + d*x]*Tan[c + d*x]),x]

[Out]

(Sec[c + d*x]^3*(2*b + a*Sin[2*(c + d*x)])^3)/(24*d)

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fricas [B]  time = 0.98, size = 92, normalized size = 3.54 \[ -\frac {3 \, a^{2} b \cos \left (d x + c\right )^{4} - 3 \, a^{2} b \cos \left (d x + c\right )^{2} - b^{3} + {\left (a^{3} \cos \left (d x + c\right )^{5} - a^{3} \cos \left (d x + c\right )^{3} - 3 \, a b^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c)+a*sin(d*x+c))^2*(a*cos(d*x+c)+b*sec(d*x+c)*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/3*(3*a^2*b*cos(d*x + c)^4 - 3*a^2*b*cos(d*x + c)^2 - b^3 + (a^3*cos(d*x + c)^5 - a^3*cos(d*x + c)^3 - 3*a*b
^2*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [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((b*sec(d*x+c)+a*sin(d*x+c))^2*(a*cos(d*x+c)+b*sec(d*x+c)*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.56, size = 118, normalized size = 4.54 \[ \frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} b \left (\sin ^{4}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {a^{2} b \left (\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{d}+\frac {a \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {a \,b^{2} \sin \left (d x +c \right )}{d}+\frac {b^{3}}{3 d \cos \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c)+a*sin(d*x+c))^2*(a*cos(d*x+c)+b*sec(d*x+c)*tan(d*x+c)),x)

[Out]

1/3/d*a^3*sin(d*x+c)^3+1/d*a^2*b*sin(d*x+c)^4/cos(d*x+c)+1/d*a^2*b*sin(d*x+c)^2*cos(d*x+c)+1/d*a*b^2*sin(d*x+c
)^3/cos(d*x+c)^2+1/d*a*b^2*sin(d*x+c)+1/3/d*b^3/cos(d*x+c)^3

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maxima [A]  time = 0.31, size = 24, normalized size = 0.92 \[ \frac {{\left (b \sec \left (d x + c\right ) + a \sin \left (d x + c\right )\right )}^{3}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c)+a*sin(d*x+c))^2*(a*cos(d*x+c)+b*sec(d*x+c)*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/3*(b*sec(d*x + c) + a*sin(d*x + c))^3/d

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mupad [B]  time = 3.22, size = 100, normalized size = 3.85 \[ \frac {a^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a^2\,b\,{\cos \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )\,a\,b^2\,\cos \left (c+d\,x\right )+\frac {b^3}{3}}{d\,{\cos \left (c+d\,x\right )}^3}-\frac {a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}-\frac {a^2\,b\,\cos \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*sin(c + d*x))/(3*d) + (b^3/3 + a^2*b*cos(c + d*x)^2 + a*b^2*cos(c + d*x)*sin(c + d*x))/(d*cos(c + d*x)^3)
- (a^3*cos(c + d*x)^2*sin(c + d*x))/(3*d) - (a^2*b*cos(c + d*x))/d

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sympy [A]  time = 11.16, size = 100, normalized size = 3.85 \[ \begin {cases} \frac {a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{2} b \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{d} + \frac {a b^{2} \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{d} + \frac {b^{3} \sec ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + b \sec {\relax (c )}\right )^{2} \left (a \cos {\relax (c )} + b \tan {\relax (c )} \sec {\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c)+a*sin(d*x+c))**2*(a*cos(d*x+c)+b*sec(d*x+c)*tan(d*x+c)),x)

[Out]

Piecewise((a**3*sin(c + d*x)**3/(3*d) + a**2*b*sin(c + d*x)**2*sec(c + d*x)/d + a*b**2*sin(c + d*x)*sec(c + d*
x)**2/d + b**3*sec(c + d*x)**3/(3*d), Ne(d, 0)), (x*(a*sin(c) + b*sec(c))**2*(a*cos(c) + b*tan(c)*sec(c)), Tru
e))

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