3.1.34 \(\int (a+a \sin (c+d x))^4 \tan ^5(c+d x) \, dx\) [34]

Optimal. Leaf size=129 \[ -\frac {25 a^4 \log (1-\sin (c+d x))}{d}-\frac {16 a^4 \sin (c+d x)}{d}-\frac {9 a^4 \sin ^2(c+d x)}{2 d}-\frac {4 a^4 \sin ^3(c+d x)}{3 d}-\frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {a^6}{d (a-a \sin (c+d x))^2}-\frac {11 a^5}{d (a-a \sin (c+d x))} \]

[Out]

-25*a^4*ln(1-sin(d*x+c))/d-16*a^4*sin(d*x+c)/d-9/2*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+a^6/d/(a-a*sin(d*x+c))^2-11*a^5/d/(a-a*sin(d*x+c))

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Rubi [A]
time = 0.07, antiderivative size = 129, 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, 78} \begin {gather*} \frac {a^6}{d (a-a \sin (c+d x))^2}-\frac {11 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 {9 a^4 \sin ^2(c+d x)}{2 d}-\frac {16 a^4 \sin (c+d x)}{d}-\frac {25 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]^5,x]

[Out]

(-25*a^4*Log[1 - Sin[c + d*x]])/d - (16*a^4*Sin[c + d*x])/d - (9*a^4*Sin[c + d*x]^2)/(2*d) - (4*a^4*Sin[c + d*
x]^3)/(3*d) - (a^4*Sin[c + d*x]^4)/(4*d) + a^6/(d*(a - a*Sin[c + d*x])^2) - (11*a^5)/(d*(a - a*Sin[c + d*x]))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 ^5(c+d x) \, dx &=\frac {\text {Subst}\left (\int \frac {x^5 (a+x)}{(a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (-16 a^3+\frac {2 a^6}{(a-x)^3}-\frac {11 a^5}{(a-x)^2}+\frac {25 a^4}{a-x}-9 a^2 x-4 a x^2-x^3\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {25 a^4 \log (1-\sin (c+d x))}{d}-\frac {16 a^4 \sin (c+d x)}{d}-\frac {9 a^4 \sin ^2(c+d x)}{2 d}-\frac {4 a^4 \sin ^3(c+d x)}{3 d}-\frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {a^6}{d (a-a \sin (c+d x))^2}-\frac {11 a^5}{d (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 0.30, size = 83, normalized size = 0.64 \begin {gather*} -\frac {a^4 \left (300 \log (1-\sin (c+d x))+\frac {120-132 \sin (c+d x)}{(-1+\sin (c+d x))^2}+192 \sin (c+d x)+54 \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]^5,x]

[Out]

-1/12*(a^4*(300*Log[1 - Sin[c + d*x]] + (120 - 132*Sin[c + d*x])/(-1 + Sin[c + d*x])^2 + 192*Sin[c + d*x] + 54
*Sin[c + d*x]^2 + 16*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4))/d

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(392\) vs. \(2(123)=246\).
time = 0.21, size = 393, normalized size = 3.05

method	result	size
risch	\(25 i a^{4} x -\frac {i a^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{6 d}+\frac {19 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {17 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {17 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {19 a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}+\frac {i a^{4} {\mathrm e}^{-3 i \left (d x +c \right )}}{6 d}+\frac {50 i a^{4} c}{d}+\frac {2 i \left (-11 \,{\mathrm e}^{i \left (d x +c \right )} a^{4}-20 i a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+11 a^{4} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {50 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a^{4} \cos \left (4 d x +4 c \right )}{32 d}\)	\(227\)
derivativedivides	\(\frac {a^{4} \left (\frac {\sin ^{10}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{10}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{8}\left (d x +c \right )\right )}{4}-\left (\sin ^{6}\left (d x +c \right )\right )-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{2}-3 \left (\sin ^{2}\left (d x +c \right )\right )-6 \ln \left (\cos \left (d x +c \right )\right )\right )+4 a^{4} \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+6 a^{4} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+4 a^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{4} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(393\)
default	\(\frac {a^{4} \left (\frac {\sin ^{10}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{10}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{8}\left (d x +c \right )\right )}{4}-\left (\sin ^{6}\left (d x +c \right )\right )-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{2}-3 \left (\sin ^{2}\left (d x +c \right )\right )-6 \ln \left (\cos \left (d x +c \right )\right )\right )+4 a^{4} \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+6 a^{4} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+4 a^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{4} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(393\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^4*tan(d*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^4*(1/4*sin(d*x+c)^10/cos(d*x+c)^4-3/4*sin(d*x+c)^10/cos(d*x+c)^2-3/4*sin(d*x+c)^8-sin(d*x+c)^6-3/2*sin(
d*x+c)^4-3*sin(d*x+c)^2-6*ln(cos(d*x+c)))+4*a^4*(1/4*sin(d*x+c)^9/cos(d*x+c)^4-5/8*sin(d*x+c)^9/cos(d*x+c)^2-5
/8*sin(d*x+c)^7-7/8*sin(d*x+c)^5-35/24*sin(d*x+c)^3-35/8*sin(d*x+c)+35/8*ln(sec(d*x+c)+tan(d*x+c)))+6*a^4*(1/4
*sin(d*x+c)^8/cos(d*x+c)^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*(1/4*sin(d*x+c)^7/cos(d*x+c)^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8*sin(d*x+c)^5-5/8*sin(d*
x+c)^3-15/8*sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c)))+a^4*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c))))

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Maxima [A]
time = 0.29, size = 109, normalized size = 0.84 \begin {gather*} -\frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 54 \, a^{4} \sin \left (d x + c\right )^{2} + 300 \, a^{4} \log \left (\sin \left (d x + c\right ) - 1\right ) + 192 \, a^{4} \sin \left (d x + c\right ) - \frac {12 \, {\left (11 \, a^{4} \sin \left (d x + c\right ) - 10 \, a^{4}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \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)^5,x, algorithm="maxima")

[Out]

-1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 54*a^4*sin(d*x + c)^2 + 300*a^4*log(sin(d*x + c) - 1) +
192*a^4*sin(d*x + c) - 12*(11*a^4*sin(d*x + c) - 10*a^4)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d

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Fricas [A]
time = 0.37, size = 154, normalized size = 1.19 \begin {gather*} -\frac {24 \, a^{4} \cos \left (d x + c\right )^{6} - 272 \, a^{4} \cos \left (d x + c\right )^{4} - 2393 \, a^{4} \cos \left (d x + c\right )^{2} + 1906 \, a^{4} + 2400 \, {\left (a^{4} \cos \left (d x + c\right )^{2} + 2 \, a^{4} \sin \left (d x + c\right ) - 2 \, a^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \, {\left (8 \, a^{4} \cos \left (d x + c\right )^{4} - 96 \, a^{4} \cos \left (d x + c\right )^{2} + 181 \, a^{4}\right )} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, 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)^5,x, algorithm="fricas")

[Out]

-1/96*(24*a^4*cos(d*x + c)^6 - 272*a^4*cos(d*x + c)^4 - 2393*a^4*cos(d*x + c)^2 + 1906*a^4 + 2400*(a^4*cos(d*x
+ c)^2 + 2*a^4*sin(d*x + c) - 2*a^4)*log(-sin(d*x + c) + 1) - 10*(8*a^4*cos(d*x + c)^4 - 96*a^4*cos(d*x + c)^
2 + 181*a^4)*sin(d*x + c))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*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 ^{5}{\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int \tan ^{5}{\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)**5,x)

[Out]

a**4*(Integral(4*sin(c + d*x)*tan(c + d*x)**5, x) + Integral(6*sin(c + d*x)**2*tan(c + d*x)**5, x) + Integral(
4*sin(c + d*x)**3*tan(c + d*x)**5, x) + Integral(sin(c + d*x)**4*tan(c + d*x)**5, x) + Integral(tan(c + d*x)**
5, 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)^5,x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 7.88, size = 379, normalized size = 2.94 \begin {gather*} \frac {25\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {50\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d}-\frac {50\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-150\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {950\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{3}-\frac {1700\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {2180\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {2452\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {2180\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {1700\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {950\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-150\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+50\,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 )}^{12}-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+31\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+44\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+31\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+10\,{\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(25*a^4*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (50*a^4*log(tan(c/2 + (d*x)/2) - 1))/d - ((950*a^4*tan(c/2 + (d*x)/
2)^3)/3 - 150*a^4*tan(c/2 + (d*x)/2)^2 - (1700*a^4*tan(c/2 + (d*x)/2)^4)/3 + (2180*a^4*tan(c/2 + (d*x)/2)^5)/3
- (2452*a^4*tan(c/2 + (d*x)/2)^6)/3 + (2180*a^4*tan(c/2 + (d*x)/2)^7)/3 - (1700*a^4*tan(c/2 + (d*x)/2)^8)/3 +
(950*a^4*tan(c/2 + (d*x)/2)^9)/3 - 150*a^4*tan(c/2 + (d*x)/2)^10 + 50*a^4*tan(c/2 + (d*x)/2)^11 + 50*a^4*tan(
c/2 + (d*x)/2))/(d*(10*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 20*tan(c/2 + (d*x)/2)^3 + 31*tan(c/2 + (d
*x)/2)^4 - 40*tan(c/2 + (d*x)/2)^5 + 44*tan(c/2 + (d*x)/2)^6 - 40*tan(c/2 + (d*x)/2)^7 + 31*tan(c/2 + (d*x)/2)
^8 - 20*tan(c/2 + (d*x)/2)^9 + 10*tan(c/2 + (d*x)/2)^10 - 4*tan(c/2 + (d*x)/2)^11 + tan(c/2 + (d*x)/2)^12 + 1)
)

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