3.638 \(\int (b \sec (c+d x)+a \sin (c+d x))^3 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx\)

Optimal. Leaf size=26 \[ \frac{(a \sin (c+d x)+b \sec (c+d x))^4}{4 d} \]

[Out]

(b*Sec[c + d*x] + a*Sin[c + d*x])^4/(4*d)

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Rubi [A]  time = 0.0441515, antiderivative size = 26, normalized size of antiderivative = 1., 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} \[ \frac{(a \sin (c+d x)+b \sec (c+d x))^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x] + a*Sin[c + d*x])^3*(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])^4/(4*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))^3 (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))^4}{4 d}\\ \end{align*}

Mathematica [B]  time = 6.55879, size = 938, normalized size = 36.08 \[ \frac{a^4 \cos (4 c) \cos (4 d x) (b \sec (c+d x)+a \sin (c+d x))^3 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \cos ^5(c+d x)}{d (3 a \cos (c+d x)+a \cos (3 c+3 d x)+4 b \sin (c+d x)) (2 b+a \sin (2 c+2 d x))^3}-\frac{4 a^3 \cos (2 d x) (a \cos (2 c)+4 b \sin (2 c)) (b \sec (c+d x)+a \sin (c+d x))^3 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \cos ^5(c+d x)}{d (3 a \cos (c+d x)+a \cos (3 c+3 d x)+4 b \sin (c+d x)) (2 b+a \sin (2 c+2 d x))^3}+\frac{4 a^3 (a \sin (2 c)-4 b \cos (2 c)) \sin (2 d x) (b \sec (c+d x)+a \sin (c+d x))^3 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \cos ^5(c+d x)}{d (3 a \cos (c+d x)+a \cos (3 c+3 d x)+4 b \sin (c+d x)) (2 b+a \sin (2 c+2 d x))^3}-\frac{a^4 \sin (4 c) \sin (4 d x) (b \sec (c+d x)+a \sin (c+d x))^3 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \cos ^5(c+d x)}{d (3 a \cos (c+d x)+a \cos (3 c+3 d x)+4 b \sin (c+d x)) (2 b+a \sin (2 c+2 d x))^3}+\frac{32 a^3 b \sec (c) \sin (d x) (b \sec (c+d x)+a \sin (c+d x))^3 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \cos ^4(c+d x)}{d (3 a \cos (c+d x)+a \cos (3 c+3 d x)+4 b \sin (c+d x)) (2 b+a \sin (2 c+2 d x))^3}+\frac{16 a b^2 \sec (c) (3 a \cos (c)+2 b \sin (c)) (b \sec (c+d x)+a \sin (c+d x))^3 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \cos ^3(c+d x)}{d (3 a \cos (c+d x)+a \cos (3 c+3 d x)+4 b \sin (c+d x)) (2 b+a \sin (2 c+2 d x))^3}+\frac{32 a b^3 \sec (c) \sin (d x) (b \sec (c+d x)+a \sin (c+d x))^3 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \cos ^2(c+d x)}{d (3 a \cos (c+d x)+a \cos (3 c+3 d x)+4 b \sin (c+d x)) (2 b+a \sin (2 c+2 d x))^3}+\frac{8 b^4 (b \sec (c+d x)+a \sin (c+d x))^3 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \cos (c+d x)}{d (3 a \cos (c+d x)+a \cos (3 c+3 d x)+4 b \sin (c+d x)) (2 b+a \sin (2 c+2 d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x] + a*Sin[c + d*x])^3*(a*Cos[c + d*x] + b*Sec[c + d*x]*Tan[c + d*x]),x]

[Out]

(8*b^4*Cos[c + d*x]*(b*Sec[c + d*x] + a*Sin[c + d*x])^3*(a*Cos[c + d*x] + b*Sec[c + d*x]*Tan[c + d*x]))/(d*(3*
a*Cos[c + d*x] + a*Cos[3*c + 3*d*x] + 4*b*Sin[c + d*x])*(2*b + a*Sin[2*c + 2*d*x])^3) + (a^4*Cos[4*c]*Cos[4*d*
x]*Cos[c + d*x]^5*(b*Sec[c + d*x] + a*Sin[c + d*x])^3*(a*Cos[c + d*x] + b*Sec[c + d*x]*Tan[c + d*x]))/(d*(3*a*
Cos[c + d*x] + a*Cos[3*c + 3*d*x] + 4*b*Sin[c + d*x])*(2*b + a*Sin[2*c + 2*d*x])^3) + (16*a*b^2*Cos[c + d*x]^3
*Sec[c]*(3*a*Cos[c] + 2*b*Sin[c])*(b*Sec[c + d*x] + a*Sin[c + d*x])^3*(a*Cos[c + d*x] + b*Sec[c + d*x]*Tan[c +
d*x]))/(d*(3*a*Cos[c + d*x] + a*Cos[3*c + 3*d*x] + 4*b*Sin[c + d*x])*(2*b + a*Sin[2*c + 2*d*x])^3) - (4*a^3*C
os[2*d*x]*Cos[c + d*x]^5*(a*Cos[2*c] + 4*b*Sin[2*c])*(b*Sec[c + d*x] + a*Sin[c + d*x])^3*(a*Cos[c + d*x] + b*S
ec[c + d*x]*Tan[c + d*x]))/(d*(3*a*Cos[c + d*x] + a*Cos[3*c + 3*d*x] + 4*b*Sin[c + d*x])*(2*b + a*Sin[2*c + 2*
d*x])^3) + (32*a*b^3*Cos[c + d*x]^2*Sec[c]*Sin[d*x]*(b*Sec[c + d*x] + a*Sin[c + d*x])^3*(a*Cos[c + d*x] + b*Se
c[c + d*x]*Tan[c + d*x]))/(d*(3*a*Cos[c + d*x] + a*Cos[3*c + 3*d*x] + 4*b*Sin[c + d*x])*(2*b + a*Sin[2*c + 2*d
*x])^3) + (32*a^3*b*Cos[c + d*x]^4*Sec[c]*Sin[d*x]*(b*Sec[c + d*x] + a*Sin[c + d*x])^3*(a*Cos[c + d*x] + b*Sec
[c + d*x]*Tan[c + d*x]))/(d*(3*a*Cos[c + d*x] + a*Cos[3*c + 3*d*x] + 4*b*Sin[c + d*x])*(2*b + a*Sin[2*c + 2*d*
x])^3) + (4*a^3*Cos[c + d*x]^5*(-4*b*Cos[2*c] + a*Sin[2*c])*Sin[2*d*x]*(b*Sec[c + d*x] + a*Sin[c + d*x])^3*(a*
Cos[c + d*x] + b*Sec[c + d*x]*Tan[c + d*x]))/(d*(3*a*Cos[c + d*x] + a*Cos[3*c + 3*d*x] + 4*b*Sin[c + d*x])*(2*
b + a*Sin[2*c + 2*d*x])^3) - (a^4*Cos[c + d*x]^5*Sin[4*c]*Sin[4*d*x]*(b*Sec[c + d*x] + a*Sin[c + d*x])^3*(a*Co
s[c + d*x] + b*Sec[c + d*x]*Tan[c + d*x]))/(d*(3*a*Cos[c + d*x] + a*Cos[3*c + 3*d*x] + 4*b*Sin[c + d*x])*(2*b
+ a*Sin[2*c + 2*d*x])^3)

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Maple [B]  time = 0.2, size = 137, normalized size = 5.3 \begin{align*}{\frac{{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{a}^{3}b \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{d\cos \left ( dx+c \right ) }}+{\frac{{a}^{3}b\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{3\,{a}^{2}{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{a{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{a{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{4}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c)+a*sin(d*x+c))^3*(a*cos(d*x+c)+b*sec(d*x+c)*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.963945, size = 32, normalized size = 1.23 \begin{align*} \frac{{\left (b \sec \left (d x + c\right ) + a \sin \left (d x + c\right )\right )}^{4}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c)+a*sin(d*x+c))^3*(a*cos(d*x+c)+b*sec(d*x+c)*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/4*(b*sec(d*x + c) + a*sin(d*x + c))^4/d

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Fricas [B]  time = 2.62149, size = 292, normalized size = 11.23 \begin{align*} \frac{8 \, a^{4} \cos \left (d x + c\right )^{8} - 16 \, a^{4} \cos \left (d x + c\right )^{6} + 5 \, a^{4} \cos \left (d x + c\right )^{4} + 48 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} + 8 \, b^{4} - 32 \,{\left (a^{3} b \cos \left (d x + c\right )^{5} - a^{3} b \cos \left (d x + c\right )^{3} - a b^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{32 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c)+a*sin(d*x+c))^3*(a*cos(d*x+c)+b*sec(d*x+c)*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/32*(8*a^4*cos(d*x + c)^8 - 16*a^4*cos(d*x + c)^6 + 5*a^4*cos(d*x + c)^4 + 48*a^2*b^2*cos(d*x + c)^2 + 8*b^4
- 32*(a^3*b*cos(d*x + c)^5 - a^3*b*cos(d*x + c)^3 - a*b^3*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

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Sympy [A]  time = 65.903, size = 129, normalized size = 4.96 \begin{align*} \begin{cases} \frac{a^{4} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac{a^{3} b \sin ^{3}{\left (c + d x \right )} \sec{\left (c + d x \right )}}{d} + \frac{3 a^{2} b^{2} \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{2 d} + \frac{a b^{3} \sin{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{d} + \frac{b^{4} \sec ^{4}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + b \sec{\left (c \right )}\right )^{3} \left (a \cos{\left (c \right )} + b \tan{\left (c \right )} \sec{\left (c \right )}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c)+a*sin(d*x+c))**3*(a*cos(d*x+c)+b*sec(d*x+c)*tan(d*x+c)),x)

[Out]

Piecewise((a**4*sin(c + d*x)**4/(4*d) + a**3*b*sin(c + d*x)**3*sec(c + d*x)/d + 3*a**2*b**2*sin(c + d*x)**2*se
c(c + d*x)**2/(2*d) + a*b**3*sin(c + d*x)*sec(c + d*x)**3/d + b**4*sec(c + d*x)**4/(4*d), Ne(d, 0)), (x*(a*sin
(c) + b*sec(c))**3*(a*cos(c) + b*tan(c)*sec(c)), True))

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Giac [B]  time = 1.29462, size = 192, normalized size = 7.38 \begin{align*} \frac{b^{4} \tan \left (d x + c\right )^{4} + 4 \, a b^{3} \tan \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 2 \, b^{4} \tan \left (d x + c\right )^{2} + 4 \, a^{3} b \tan \left (d x + c\right ) + 4 \, a b^{3} \tan \left (d x + c\right ) - \frac{4 \, a^{3} b \tan \left (d x + c\right )^{3} + 2 \, a^{4} \tan \left (d x + c\right )^{2} + 4 \, a^{3} b \tan \left (d x + c\right ) + a^{4}}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c)+a*sin(d*x+c))^3*(a*cos(d*x+c)+b*sec(d*x+c)*tan(d*x+c)),x, algorithm="giac")

[Out]

1/4*(b^4*tan(d*x + c)^4 + 4*a*b^3*tan(d*x + c)^3 + 6*a^2*b^2*tan(d*x + c)^2 + 2*b^4*tan(d*x + c)^2 + 4*a^3*b*t
an(d*x + c) + 4*a*b^3*tan(d*x + c) - (4*a^3*b*tan(d*x + c)^3 + 2*a^4*tan(d*x + c)^2 + 4*a^3*b*tan(d*x + c) + a
^4)/(tan(d*x + c)^2 + 1)^2)/d