3.1.35 \(\int (a+a \sin (c+d x))^4 \tan ^3(c+d x) \, dx\) [35]

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

[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 = 0.06, 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 = {2786, 90} \begin {gather*} \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} \end {gather*}

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 90

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 2786

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 {\text {Subst}\left (\int \frac {x^3 (a+x)^2}{(a-x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\text {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 = 0.11, size = 76, normalized size = 0.71 \begin {gather*} \frac {a^4 \left (192 \log (1-\sin (c+d x))+\frac {48}{1-\sin (c+d x)}+144 \sin (c+d x)+48 \sin ^2(c+d x)+16 \sin ^3(c+d x)+3 \sin ^4(c+d x)\right )}{12 d} \end {gather*}

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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(266\) vs. \(2(103)=206\).
time = 0.17, size = 267, normalized size = 2.50

method	result	size
risch	\(-16 i a^{4} x -\frac {13 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {13 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {32 i a^{4} c}{d}-\frac {8 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d}+\frac {32 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a^{4} \cos \left (4 d x +4 c \right )}{32 d}-\frac {a^{4} \sin \left (3 d x +3 c \right )}{3 d}-\frac {17 a^{4} \cos \left (2 d x +2 c \right )}{8 d}\)	\(159\)
derivativedivides	\(\frac {a^{4} \left (\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}+3 \ln \left (\cos \left (d x +c \right )\right )\right )+4 a^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+4 a^{4} \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(267\)
default	\(\frac {a^{4} \left (\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}+3 \ln \left (\cos \left (d x +c \right )\right )\right )+4 a^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+4 a^{4} \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(267\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 85, normalized size = 0.79 \begin {gather*} \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} \end {gather*}

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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Fricas [A]
time = 0.35, size = 116, normalized size = 1.08 \begin {gather*} \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 )}} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 ) \end {gather*}

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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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Mupad [B]
time = 7.56, size = 320, normalized size = 2.99 \begin {gather*} \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} \end {gather*}

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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