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

Optimal. Leaf size=107 \[ \frac {4 a^5}{d (a-a \sin (c+d x))}+\frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {4 a^4 \sin ^2(c+d x)}{d}+\frac {12 a^4 \sin (c+d x)}{d}+\frac {16 a^4 \log (1-\sin (c+d x))}{d} \]

[Out]

16*a^4*ln(1-sin(d*x+c))/d+12*a^4*sin(d*x+c)/d+4*a^4*sin(d*x+c)^2/d+4/3*a^4*sin(d*x+c)^3/d+1/4*a^4*sin(d*x+c)^4
/d+4*a^5/d/(a-a*sin(d*x+c))

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Rubi [A]  time = 0.08, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2707, 88} \[ \frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {4 a^4 \sin ^2(c+d x)}{d}+\frac {4 a^5}{d (a-a \sin (c+d x))}+\frac {12 a^4 \sin (c+d x)}{d}+\frac {16 a^4 \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a^4*Log[1 - Sin[c + d*x]])/d + (12*a^4*Sin[c + d*x])/d + (4*a^4*Sin[c + d*x]^2)/d + (4*a^4*Sin[c + d*x]^3)
/(3*d) + (a^4*Sin[c + d*x]^4)/(4*d) + (4*a^5)/(d*(a - a*Sin[c + d*x]))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2707

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

Rubi steps

\begin {align*} \int (a+a \sin (c+d x))^4 \tan ^3(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^3 (a+x)^2}{(a-x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (12 a^3+\frac {4 a^5}{(a-x)^2}-\frac {16 a^4}{a-x}+8 a^2 x+4 a x^2+x^3\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {16 a^4 \log (1-\sin (c+d x))}{d}+\frac {12 a^4 \sin (c+d x)}{d}+\frac {4 a^4 \sin ^2(c+d x)}{d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {4 a^5}{d (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 76, normalized size = 0.71 \[ \frac {a^4 \left (3 \sin ^4(c+d x)+16 \sin ^3(c+d x)+48 \sin ^2(c+d x)+144 \sin (c+d x)+\frac {48}{1-\sin (c+d x)}+192 \log (1-\sin (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(192*Log[1 - Sin[c + d*x]] + 48/(1 - Sin[c + d*x]) + 144*Sin[c + d*x] + 48*Sin[c + d*x]^2 + 16*Sin[c + d*
x]^3 + 3*Sin[c + d*x]^4))/(12*d)

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fricas [A]  time = 0.46, size = 116, normalized size = 1.08 \[ \frac {104 \, a^{4} \cos \left (d x + c\right )^{4} - 976 \, a^{4} \cos \left (d x + c\right )^{2} + 689 \, a^{4} + 1536 \, {\left (a^{4} \sin \left (d x + c\right ) - a^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (24 \, a^{4} \cos \left (d x + c\right )^{4} - 304 \, a^{4} \cos \left (d x + c\right )^{2} - 1073 \, a^{4}\right )} \sin \left (d x + c\right )}{96 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(104*a^4*cos(d*x + c)^4 - 976*a^4*cos(d*x + c)^2 + 689*a^4 + 1536*(a^4*sin(d*x + c) - a^4)*log(-sin(d*x +
c) + 1) + (24*a^4*cos(d*x + c)^4 - 304*a^4*cos(d*x + c)^2 - 1073*a^4)*sin(d*x + c))/(d*sin(d*x + c) - d)

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

[Out]

Timed out

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maple [B]  time = 0.20, size = 245, normalized size = 2.29 \[ \frac {a^{4} \left (\sin ^{8}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{4} \left (\sin ^{6}\left (d x +c \right )\right )}{2 d}+\frac {15 a^{4} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d}+\frac {15 a^{4} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}+\frac {16 a^{4} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {2 a^{4} \left (\sin ^{7}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {2 a^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{d}+\frac {16 a^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}+\frac {16 a^{4} \sin \left (d x +c \right )}{d}-\frac {16 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 a^{4} \left (\sin ^{6}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {2 a^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {a^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*a^4*sin(d*x+c)^8/cos(d*x+c)^2+1/2/d*a^4*sin(d*x+c)^6+15/4*a^4*sin(d*x+c)^4/d+15/2*a^4*sin(d*x+c)^2/d+16/
d*a^4*ln(cos(d*x+c))+2/d*a^4*sin(d*x+c)^7/cos(d*x+c)^2+2/d*a^4*sin(d*x+c)^5+16/3*a^4*sin(d*x+c)^3/d+16*a^4*sin
(d*x+c)/d-16/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+3/d*a^4*sin(d*x+c)^6/cos(d*x+c)^2+2/d*a^4*sin(d*x+c)^5/cos(d*x+c)
^2+1/2/d*a^4*tan(d*x+c)^2

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maxima [A]  time = 0.29, size = 85, normalized size = 0.79 \[ \frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 48 \, a^{4} \sin \left (d x + c\right )^{2} + 192 \, a^{4} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, a^{4} \sin \left (d x + c\right ) - \frac {48 \, a^{4}}{\sin \left (d x + c\right ) - 1}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 48*a^4*sin(d*x + c)^2 + 192*a^4*log(sin(d*x + c) - 1) + 1
44*a^4*sin(d*x + c) - 48*a^4/(sin(d*x + c) - 1))/d

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mupad [B]  time = 7.56, size = 320, normalized size = 2.99 \[ \frac {32\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d}+\frac {32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {320\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {340\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {424\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {340\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {320\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+32\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {16\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(32*a^4*log(tan(c/2 + (d*x)/2) - 1))/d + ((320*a^4*tan(c/2 + (d*x)/2)^3)/3 - 32*a^4*tan(c/2 + (d*x)/2)^2 - (34
0*a^4*tan(c/2 + (d*x)/2)^4)/3 + (424*a^4*tan(c/2 + (d*x)/2)^5)/3 - (340*a^4*tan(c/2 + (d*x)/2)^6)/3 + (320*a^4
*tan(c/2 + (d*x)/2)^7)/3 - 32*a^4*tan(c/2 + (d*x)/2)^8 + 32*a^4*tan(c/2 + (d*x)/2)^9 + 32*a^4*tan(c/2 + (d*x)/
2))/(d*(5*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2) - 8*tan(c/2 + (d*x)/2)^3 + 10*tan(c/2 + (d*x)/2)^4 - 12*
tan(c/2 + (d*x)/2)^5 + 10*tan(c/2 + (d*x)/2)^6 - 8*tan(c/2 + (d*x)/2)^7 + 5*tan(c/2 + (d*x)/2)^8 - 2*tan(c/2 +
(d*x)/2)^9 + tan(c/2 + (d*x)/2)^10 + 1)) - (16*a^4*log(tan(c/2 + (d*x)/2)^2 + 1))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{4} \left (\int 4 \sin {\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{3}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*(Integral(4*sin(c + d*x)*tan(c + d*x)**3, x) + Integral(6*sin(c + d*x)**2*tan(c + d*x)**3, x) + Integral(
4*sin(c + d*x)**3*tan(c + d*x)**3, x) + Integral(sin(c + d*x)**4*tan(c + d*x)**3, x) + Integral(tan(c + d*x)**
3, x))

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