\(\int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx\) [403]

Optimal result

Integrand size = 19, antiderivative size = 80 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {(a+b)^3 \log (1-\sin (c+d x))}{2 d}+\frac {(a-b)^3 \log (1+\sin (c+d x))}{2 d}-\frac {3 a b^2 \sin (c+d x)}{d}-\frac {b^3 \sin ^2(c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2747, 716, 647, 31} \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 a b^2 \sin (c+d x)}{d}+\frac {(a-b)^3 \log (\sin (c+d x)+1)}{2 d}-\frac {(a+b)^3 \log (1-\sin (c+d x))}{2 d}-\frac {b^3 \sin ^2(c+d x)}{2 d} \]

[In]

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

[Out]

-1/2*((a + b)^3*Log[1 - Sin[c + d*x]])/d + ((a - b)^3*Log[1 + Sin[c + d*x]])/(2*d) - (3*a*b^2*Sin[c + d*x])/d
- (b^3*Sin[c + d*x]^2)/(2*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 716

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b \text {Subst}\left (\int \frac {(a+x)^3}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b \text {Subst}\left (\int \left (-3 a-x+\frac {a^3+3 a b^2+\left (3 a^2+b^2\right ) x}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {3 a b^2 \sin (c+d x)}{d}-\frac {b^3 \sin ^2(c+d x)}{2 d}+\frac {b \text {Subst}\left (\int \frac {a^3+3 a b^2+\left (3 a^2+b^2\right ) x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {3 a b^2 \sin (c+d x)}{d}-\frac {b^3 \sin ^2(c+d x)}{2 d}-\frac {(a-b)^3 \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{2 d}+\frac {(a+b)^3 \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{2 d} \\ & = -\frac {(a+b)^3 \log (1-\sin (c+d x))}{2 d}+\frac {(a-b)^3 \log (1+\sin (c+d x))}{2 d}-\frac {3 a b^2 \sin (c+d x)}{d}-\frac {b^3 \sin ^2(c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {(a+b)^3 \log (1-\sin (c+d x))-(a-b)^3 \log (1+\sin (c+d x))+6 a b^2 \sin (c+d x)+b^3 \sin ^2(c+d x)}{2 d} \]

[In]

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

[Out]

-1/2*((a + b)^3*Log[1 - Sin[c + d*x]] - (a - b)^3*Log[1 + Sin[c + d*x]] + 6*a*b^2*Sin[c + d*x] + b^3*Sin[c + d
*x]^2)/d

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-3 a^{2} b \ln \left (\cos \left (d x +c \right )\right )+3 a \,b^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(90\)
default	\(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-3 a^{2} b \ln \left (\cos \left (d x +c \right )\right )+3 a \,b^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(90\)
parallelrisch	\(\frac {4 \left (3 a^{2} b +b^{3}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (a +b \right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \left (a -b \right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-12 a \,b^{2} \sin \left (d x +c \right )+\cos \left (2 d x +2 c \right ) b^{3}-b^{3}}{4 d}\)	\(101\)
norman	\(\frac {-\frac {2 b^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {12 a \,b^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a \,b^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {b \left (3 a^{2}+b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(216\)
risch	\(\frac {3 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {2 i b^{3} c}{d}+\frac {6 i a^{2} b c}{d}+3 i a^{2} b x -\frac {3 i a \,b^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+i b^{3} x +\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{d}-\frac {a^{3} \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{d}-\frac {3 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) a^{2} b}{d}-\frac {3 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) a \,b^{2}}{d}-\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) b^{3}}{d}+\frac {b^{3} \cos \left (2 d x +2 c \right )}{4 d}\)	\(264\)

[In]

int(sec(d*x+c)*(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.16 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {b^{3} \cos \left (d x + c\right )^{2} - 6 \, a b^{2} \sin \left (d x + c\right ) + {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, d} \]

[In]

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

[Out]

1/2*(b^3*cos(d*x + c)^2 - 6*a*b^2*sin(d*x + c) + (a^3 - 3*a^2*b + 3*a*b^2 - b^3)*log(sin(d*x + c) + 1) - (a^3
+ 3*a^2*b + 3*a*b^2 + b^3)*log(-sin(d*x + c) + 1))/d

Sympy [F]

\[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \sec {\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((a + b*sin(c + d*x))**3*sec(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.14 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {b^{3} \sin \left (d x + c\right )^{2} + 6 \, a b^{2} \sin \left (d x + c\right ) - {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{2 \, d} \]

[In]

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

[Out]

-1/2*(b^3*sin(d*x + c)^2 + 6*a*b^2*sin(d*x + c) - (a^3 - 3*a^2*b + 3*a*b^2 - b^3)*log(sin(d*x + c) + 1) + (a^3
+ 3*a^2*b + 3*a*b^2 + b^3)*log(sin(d*x + c) - 1))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.16 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {b^{3} \sin \left (d x + c\right )^{2} + 6 \, a b^{2} \sin \left (d x + c\right ) - {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} \]

[In]

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

[Out]

-1/2*(b^3*sin(d*x + c)^2 + 6*a*b^2*sin(d*x + c) - (a^3 - 3*a^2*b + 3*a*b^2 - b^3)*log(abs(sin(d*x + c) + 1)) +
(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(abs(sin(d*x + c) - 1)))/d

Mupad [B] (verification not implemented)

Time = 4.99 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.81 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^3}{2}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,{\left (a-b\right )}^3}{2}+\frac {b^3\,{\sin \left (c+d\,x\right )}^2}{2}+3\,a\,b^2\,\sin \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

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