3.1.30 \(\int (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx\) [30]

Optimal. Leaf size=119 \[ \frac {17 a^3 x}{2}-\frac {6 a^3 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {25 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}-\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

17/2*a^3*x-6*a^3*cos(d*x+c)/d+1/3*a^3*cos(d*x+c)^3/d+2/3*a^3*cos(d*x+c)/d/(1-sin(d*x+c))^2-25/3*a^3*cos(d*x+c)
/d/(1-sin(d*x+c))-3/2*a^3*cos(d*x+c)*sin(d*x+c)/d

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Rubi [A]
time = 0.14, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2788, 2729, 2727, 2718, 2715, 8, 2713} \begin {gather*} \frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {6 a^3 \cos (c+d x)}{d}-\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {25 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {17 a^3 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(17*a^3*x)/2 - (6*a^3*Cos[c + d*x])/d + (a^3*Cos[c + d*x]^3)/(3*d) + (2*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*
x])^2) - (25*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])) - (3*a^3*Cos[c + d*x]*Sin[c + d*x])/(2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rubi steps

\begin {align*} \int (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx &=a^4 \int \left (\frac {7}{a}+\frac {2}{a (-1+\sin (c+d x))^2}+\frac {9}{a (-1+\sin (c+d x))}+\frac {5 \sin (c+d x)}{a}+\frac {3 \sin ^2(c+d x)}{a}+\frac {\sin ^3(c+d x)}{a}\right ) \, dx\\ &=7 a^3 x+a^3 \int \sin ^3(c+d x) \, dx+\left (2 a^3\right ) \int \frac {1}{(-1+\sin (c+d x))^2} \, dx+\left (3 a^3\right ) \int \sin ^2(c+d x) \, dx+\left (5 a^3\right ) \int \sin (c+d x) \, dx+\left (9 a^3\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx\\ &=7 a^3 x-\frac {5 a^3 \cos (c+d x)}{d}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {9 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{3} \left (2 a^3\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx+\frac {1}{2} \left (3 a^3\right ) \int 1 \, dx-\frac {a^3 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {17 a^3 x}{2}-\frac {6 a^3 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {25 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}-\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]
time = 1.36, size = 177, normalized size = 1.49 \begin {gather*} \frac {(a+a \sin (c+d x))^3 \left (102 (c+d x)-69 \cos (c+d x)+\cos (3 (c+d x))+\frac {8}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {16 \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {200 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-9 \sin (2 (c+d x))\right )}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + a*Sin[c + d*x])^3*(102*(c + d*x) - 69*Cos[c + d*x] + Cos[3*(c + d*x)] + 8/(Cos[(c + d*x)/2] - Sin[(c + d
*x)/2])^2 + (16*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 - (200*Sin[(c + d*x)/2])/(Cos[(c + d
*x)/2] - Sin[(c + d*x)/2]) - 9*Sin[2*(c + d*x)]))/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(265\) vs. \(2(109)=218\).
time = 0.28, size = 266, normalized size = 2.24

method	result	size
risch	\(\frac {17 a^{3} x}{2}+\frac {3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {23 a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {23 a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {3 i a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 \left (-48 i a^{3} {\mathrm e}^{i \left (d x +c \right )}+27 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-25 a^{3}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}+\frac {a^{3} \cos \left (3 d x +3 c \right )}{12 d}\)	\(149\)
derivativedivides	\(\frac {a^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \left (\sin ^{8}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )}{3}\right )+3 a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+3 a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )}{d}\)	\(266\)
default	\(\frac {a^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \left (\sin ^{8}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )}{3}\right )+3 a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+3 a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )}{d}\)	\(266\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 165, normalized size = 1.39 \begin {gather*} \frac {2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {9 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} - 9 \, \cos \left (d x + c\right )\right )} a^{3} + 3 \, {\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac {3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a^{3} + 2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} - 6 \, a^{3} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(cos(d*x + c)^3 - (9*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 - 9*cos(d*x + c))*a^3 + 3*(2*tan(d*x + c)^3 + 1
5*d*x + 15*c - 3*tan(d*x + c)/(tan(d*x + c)^2 + 1) - 12*tan(d*x + c))*a^3 + 2*(tan(d*x + c)^3 + 3*d*x + 3*c -
3*tan(d*x + c))*a^3 - 6*a^3*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c)))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (105) = 210\).
time = 0.37, size = 220, normalized size = 1.85 \begin {gather*} \frac {2 \, a^{3} \cos \left (d x + c\right )^{5} + 7 \, a^{3} \cos \left (d x + c\right )^{4} - 22 \, a^{3} \cos \left (d x + c\right )^{3} - 102 \, a^{3} d x - 4 \, a^{3} + {\left (51 \, a^{3} d x + 77 \, a^{3}\right )} \cos \left (d x + c\right )^{2} - {\left (51 \, a^{3} d x - 100 \, a^{3}\right )} \cos \left (d x + c\right ) + {\left (2 \, a^{3} \cos \left (d x + c\right )^{4} - 5 \, a^{3} \cos \left (d x + c\right )^{3} + 102 \, a^{3} d x - 27 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} + {\left (51 \, a^{3} d x - 104 \, a^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \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))^3*tan(d*x+c)^4,x, algorithm="fricas")

[Out]

1/6*(2*a^3*cos(d*x + c)^5 + 7*a^3*cos(d*x + c)^4 - 22*a^3*cos(d*x + c)^3 - 102*a^3*d*x - 4*a^3 + (51*a^3*d*x +
77*a^3)*cos(d*x + c)^2 - (51*a^3*d*x - 100*a^3)*cos(d*x + c) + (2*a^3*cos(d*x + c)^4 - 5*a^3*cos(d*x + c)^3 +
102*a^3*d*x - 27*a^3*cos(d*x + c)^2 - 4*a^3 + (51*a^3*d*x - 104*a^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x +
c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 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^{3} \left (\int 3 \sin {\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}\, dx + \int \tan ^{4}{\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))**3*tan(d*x+c)**4,x)

[Out]

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

[Out]

Timed out

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Mupad [B]
time = 10.50, size = 371, normalized size = 3.12 \begin {gather*} \frac {17\,a^3\,x}{2}+\frac {\frac {17\,a^3\,\left (c+d\,x\right )}{2}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (153\,c+153\,d\,x-378\right )}{6}\right )-\frac {a^3\,\left (51\,c+51\,d\,x-160\right )}{6}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (153\,c+153\,d\,x-102\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (51\,a^3\,\left (c+d\,x\right )-\frac {a^3\,\left (306\,c+306\,d\,x-306\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (51\,a^3\,\left (c+d\,x\right )-\frac {a^3\,\left (306\,c+306\,d\,x-654\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (85\,a^3\,\left (c+d\,x\right )-\frac {a^3\,\left (510\,c+510\,d\,x-578\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (85\,a^3\,\left (c+d\,x\right )-\frac {a^3\,\left (510\,c+510\,d\,x-1022\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (102\,a^3\,\left (c+d\,x\right )-\frac {a^3\,\left (612\,c+612\,d\,x-918\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (102\,a^3\,\left (c+d\,x\right )-\frac {a^3\,\left (612\,c+612\,d\,x-1002\right )}{6}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(17*a^3*x)/2 + ((17*a^3*(c + d*x))/2 - tan(c/2 + (d*x)/2)*((51*a^3*(c + d*x))/2 - (a^3*(153*c + 153*d*x - 378)
)/6) - (a^3*(51*c + 51*d*x - 160))/6 + tan(c/2 + (d*x)/2)^8*((51*a^3*(c + d*x))/2 - (a^3*(153*c + 153*d*x - 10
2))/6) - tan(c/2 + (d*x)/2)^7*(51*a^3*(c + d*x) - (a^3*(306*c + 306*d*x - 306))/6) + tan(c/2 + (d*x)/2)^2*(51*
a^3*(c + d*x) - (a^3*(306*c + 306*d*x - 654))/6) + tan(c/2 + (d*x)/2)^6*(85*a^3*(c + d*x) - (a^3*(510*c + 510*
d*x - 578))/6) - tan(c/2 + (d*x)/2)^3*(85*a^3*(c + d*x) - (a^3*(510*c + 510*d*x - 1022))/6) - tan(c/2 + (d*x)/
2)^5*(102*a^3*(c + d*x) - (a^3*(612*c + 612*d*x - 918))/6) + tan(c/2 + (d*x)/2)^4*(102*a^3*(c + d*x) - (a^3*(6
12*c + 612*d*x - 1002))/6))/(d*(tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^3 - 1)^3)

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