3.871 \(\int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=96 \[ \frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {4 a^4}{d (a-a \sin (c+d x))}-\frac {a^3 \sin ^2(c+d x)}{2 d}-\frac {3 a^3 \sin (c+d x)}{d}-\frac {6 a^3 \log (1-\sin (c+d x))}{d} \]

[Out]

-6*a^3*ln(1-sin(d*x+c))/d-3*a^3*sin(d*x+c)/d-1/2*a^3*sin(d*x+c)^2/d+1/2*a^5/d/(a-a*sin(d*x+c))^2-4*a^4/d/(a-a*
sin(d*x+c))

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Rubi [A]  time = 0.11, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2836, 12, 43} \[ -\frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {4 a^4}{d (a-a \sin (c+d x))}-\frac {3 a^3 \sin (c+d x)}{d}-\frac {6 a^3 \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a^3*Log[1 - Sin[c + d*x]])/d - (3*a^3*Sin[c + d*x])/d - (a^3*Sin[c + d*x]^2)/(2*d) + a^5/(2*d*(a - a*Sin[c
+ d*x])^2) - (4*a^4)/(d*(a - a*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps

\begin {align*} \int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {x^4}{a^4 (a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a \operatorname {Subst}\left (\int \frac {x^4}{(a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a \operatorname {Subst}\left (\int \left (-3 a+\frac {a^4}{(a-x)^3}-\frac {4 a^3}{(a-x)^2}+\frac {6 a^2}{a-x}-x\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {6 a^3 \log (1-\sin (c+d x))}{d}-\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {4 a^4}{d (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 61, normalized size = 0.64 \[ -\frac {a^3 \left (\sin ^2(c+d x)+6 \sin (c+d x)+\frac {7-8 \sin (c+d x)}{(\sin (c+d x)-1)^2}+12 \log (1-\sin (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a^3*(12*Log[1 - Sin[c + d*x]] + (7 - 8*Sin[c + d*x])/(-1 + Sin[c + d*x])^2 + 6*Sin[c + d*x] + Sin[c + d*
x]^2))/d

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fricas [A]  time = 0.46, size = 128, normalized size = 1.33 \[ \frac {2 \, a^{3} \cos \left (d x + c\right )^{4} + 19 \, a^{3} \cos \left (d x + c\right )^{2} - 8 \, a^{3} - 24 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, a^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(2*a^3*cos(d*x + c)^4 + 19*a^3*cos(d*x + c)^2 - 8*a^3 - 24*(a^3*cos(d*x + c)^2 + 2*a^3*sin(d*x + c) - 2*a^
3)*log(-sin(d*x + c) + 1) - 2*(4*a^3*cos(d*x + c)^2 - 3*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c
) - 2*d)

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giac [B]  time = 0.30, size = 209, normalized size = 2.18 \[ \frac {6 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} + \frac {25 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 106 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 164 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 106 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

(6*a^3*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 12*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - (9*a^3*tan(1/2*d*x + 1/2*
c)^4 + 6*a^3*tan(1/2*d*x + 1/2*c)^3 + 20*a^3*tan(1/2*d*x + 1/2*c)^2 + 6*a^3*tan(1/2*d*x + 1/2*c) + 9*a^3)/(tan
(1/2*d*x + 1/2*c)^2 + 1)^2 + (25*a^3*tan(1/2*d*x + 1/2*c)^4 - 106*a^3*tan(1/2*d*x + 1/2*c)^3 + 164*a^3*tan(1/2
*d*x + 1/2*c)^2 - 106*a^3*tan(1/2*d*x + 1/2*c) + 25*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^4)/d

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maple [B]  time = 0.30, size = 309, normalized size = 3.22 \[ \frac {a^{3} \left (\sin ^{8}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {a^{3} \left (\sin ^{8}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}-\frac {a^{3} \left (\sin ^{6}\left (d x +c \right )\right )}{2 d}-\frac {3 a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d}-\frac {3 a^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {6 a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {3 a^{3} \left (\sin ^{7}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {9 a^{3} \left (\sin ^{7}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {9 a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}-\frac {2 a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{d}-\frac {6 a^{3} \sin \left (d x +c \right )}{d}+\frac {6 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 a^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {3 a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x)

[Out]

1/4/d*a^3*sin(d*x+c)^8/cos(d*x+c)^4-1/2/d*a^3*sin(d*x+c)^8/cos(d*x+c)^2-1/2*a^3*sin(d*x+c)^6/d-3/4*a^3*sin(d*x
+c)^4/d-3/2*a^3*sin(d*x+c)^2/d-6/d*a^3*ln(cos(d*x+c))+3/4/d*a^3*sin(d*x+c)^7/cos(d*x+c)^4-9/8/d*a^3*sin(d*x+c)
^7/cos(d*x+c)^2-9/8*a^3*sin(d*x+c)^5/d-2*a^3*sin(d*x+c)^3/d-6*a^3*sin(d*x+c)/d+6/d*a^3*ln(sec(d*x+c)+tan(d*x+c
))+3/4/d*a^3*tan(d*x+c)^4-3/2/d*a^3*tan(d*x+c)^2+1/4/d*a^3*sin(d*x+c)^5/cos(d*x+c)^4-1/8/d*a^3*sin(d*x+c)^5/co
s(d*x+c)^2

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maxima [A]  time = 0.31, size = 82, normalized size = 0.85 \[ -\frac {a^{3} \sin \left (d x + c\right )^{2} + 12 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, a^{3} \sin \left (d x + c\right ) - \frac {8 \, a^{3} \sin \left (d x + c\right ) - 7 \, a^{3}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(a^3*sin(d*x + c)^2 + 12*a^3*log(sin(d*x + c) - 1) + 6*a^3*sin(d*x + c) - (8*a^3*sin(d*x + c) - 7*a^3)/(s
in(d*x + c)^2 - 2*sin(d*x + c) + 1))/d

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mupad [B]  time = 10.93, size = 263, normalized size = 2.74 \[ \frac {6\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-36\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+52\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-64\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+52\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-36\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+12\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {12\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^4*(a + a*sin(c + d*x))^3)/cos(c + d*x)^5,x)

[Out]

(6*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (52*a^3*tan(c/2 + (d*x)/2)^3 - 36*a^3*tan(c/2 + (d*x)/2)^2 - 64*a^3*
tan(c/2 + (d*x)/2)^4 + 52*a^3*tan(c/2 + (d*x)/2)^5 - 36*a^3*tan(c/2 + (d*x)/2)^6 + 12*a^3*tan(c/2 + (d*x)/2)^7
+ 12*a^3*tan(c/2 + (d*x)/2))/(d*(8*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 12*tan(c/2 + (d*x)/2)^3 + 14
*tan(c/2 + (d*x)/2)^4 - 12*tan(c/2 + (d*x)/2)^5 + 8*tan(c/2 + (d*x)/2)^6 - 4*tan(c/2 + (d*x)/2)^7 + tan(c/2 +
(d*x)/2)^8 + 1)) - (12*a^3*log(tan(c/2 + (d*x)/2) - 1))/d

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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)**5*sin(d*x+c)**4*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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