\(\int (b \sec (c+d x)+a \sin (c+d x))^2 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx\) [639]

Optimal result

Integrand size = 43, antiderivative size = 26 \[ \int (b \sec (c+d x)+a \sin (c+d x))^2 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\frac {(b \sec (c+d x)+a \sin (c+d x))^3}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {4470} \[ \int (b \sec (c+d x)+a \sin (c+d x))^2 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\frac {(a \sin (c+d x)+b \sec (c+d x))^3}{3 d} \]

[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 4470

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*} \text {integral}& = \frac {(b \sec (c+d x)+a \sin (c+d x))^3}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.95 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int (b \sec (c+d x)+a \sin (c+d x))^2 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\frac {(b \sec (c+d x)+a \sin (c+d x))^3}{3 d} \]

[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]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(151\) vs. \(2(24)=48\).

Time = 48.38 (sec) , antiderivative size = 152, normalized size of antiderivative = 5.85

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (d x +c \right )^{3} a^{3}}{3}+a^{2} b \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )-2 \cos \left (d x +c \right ) a^{2} b +2 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {b^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(152\)
default	\(\frac {\frac {\sin \left (d x +c \right )^{3} a^{3}}{3}+a^{2} b \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )-2 \cos \left (d x +c \right ) a^{2} b +2 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\frac {b^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(152\)
parts	\(\frac {\frac {\sin \left (d x +c \right )^{3} a^{3}}{3}-2 \cos \left (d x +c \right ) a^{2} b +a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{3} \sec \left (d x +c \right )^{3}}{3 d}+\frac {a^{2} b \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}+\frac {2 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(162\)
risch	\(\frac {-3 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}-12 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2} b +24 \,{\mathrm e}^{3 i \left (d x +c \right )} a^{2} b -i a^{3} {\mathrm e}^{-3 i \left (d x +c \right )}+64 \,{\mathrm e}^{3 i \left (d x +c \right )} b^{3}+i a^{3} {\mathrm e}^{9 i \left (d x +c \right )}-12 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+48 i {\mathrm e}^{i \left (d x +c \right )} a \,b^{2}-48 i {\mathrm e}^{5 i \left (d x +c \right )} a \,b^{2}}{24 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(171\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (24) = 48\).

Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.54 \[ \int (b \sec (c+d x)+a \sin (c+d x))^2 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=-\frac {3 \, a^{2} b \cos \left (d x + c\right )^{4} - 3 \, a^{2} b \cos \left (d x + c\right )^{2} - b^{3} + {\left (a^{3} \cos \left (d x + c\right )^{5} - a^{3} \cos \left (d x + c\right )^{3} - 3 \, a b^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (20) = 40\).

Time = 1.00 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.85 \[ \int (b \sec (c+d x)+a \sin (c+d x))^2 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\begin {cases} \frac {a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{2} b \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{d} + \frac {a b^{2} \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{d} + \frac {b^{3} \sec ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + b \sec {\left (c \right )}\right )^{2} \left (a \cos {\left (c \right )} + b \tan {\left (c \right )} \sec {\left (c \right )}\right ) & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int (b \sec (c+d x)+a \sin (c+d x))^2 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\frac {{\left (b \sec \left (d x + c\right ) + a \sin \left (d x + c\right )\right )}^{3}}{3 \, d} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (24) = 48\).

Time = 0.70 (sec) , antiderivative size = 321, normalized size of antiderivative = 12.35 \[ \int (b \sec (c+d x)+a \sin (c+d x))^2 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\frac {2 \, {\left (3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 4 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 10 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 4 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{3}\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{3} d} \]

[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]

2/3*(3*a*b^2*tan(1/2*d*x + 1/2*c)^11 - 6*a^2*b*tan(1/2*d*x + 1/2*c)^10 - 3*b^3*tan(1/2*d*x + 1/2*c)^10 + 4*a^3
*tan(1/2*d*x + 1/2*c)^9 + 9*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 9*b^3*tan(1/2*d*x + 1/2*c)^8 - 12*a^3*tan(1/2*d*x +
1/2*c)^7 + 6*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^6 - 10*b^3*tan(1/2*d*x + 1/2*c)^6 +
12*a^3*tan(1/2*d*x + 1/2*c)^5 - 6*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*b^3*tan(1/2*d*x + 1/2*c)^4 - 4*a^3*tan(1/2
*d*x + 1/2*c)^3 - 9*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 6*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 3*b^3*tan(1/2*d*x + 1/2*c)
^2 - 3*a*b^2*tan(1/2*d*x + 1/2*c) - b^3)/((tan(1/2*d*x + 1/2*c)^4 - 1)^3*d)

Mupad [B] (verification not implemented)

Time = 26.84 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.85 \[ \int (b \sec (c+d x)+a \sin (c+d x))^2 (a \cos (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx=\frac {a^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a^2\,b\,{\cos \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )\,a\,b^2\,\cos \left (c+d\,x\right )+\frac {b^3}{3}}{d\,{\cos \left (c+d\,x\right )}^3}-\frac {a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}-\frac {a^2\,b\,\cos \left (c+d\,x\right )}{d} \]

[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