3.1.39 \(\int (a+a \sin (c+d x))^4 \tan ^2(c+d x) \, dx\) [39]

Optimal. Leaf size=113 \[ -\frac {95 a^4 x}{8}+\frac {12 a^4 \cos (c+d x)}{d}-\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {31 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d} \]

[Out]

-95/8*a^4*x+12*a^4*cos(d*x+c)/d-4/3*a^4*cos(d*x+c)^3/d+8*a^4*cos(d*x+c)/d/(1-sin(d*x+c))+31/8*a^4*cos(d*x+c)*s
in(d*x+c)/d+1/4*a^4*cos(d*x+c)*sin(d*x+c)^3/d

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Rubi [A]
time = 0.12, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2788, 2727, 2718, 2715, 8, 2713} \begin {gather*} -\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {12 a^4 \cos (c+d x)}{d}+\frac {a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {31 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {95 a^4 x}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^4*Tan[c + d*x]^2,x]

[Out]

(-95*a^4*x)/8 + (12*a^4*Cos[c + d*x])/d - (4*a^4*Cos[c + d*x]^3)/(3*d) + (8*a^4*Cos[c + d*x])/(d*(1 - Sin[c +
d*x])) + (31*a^4*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^4*Cos[c + d*x]*Sin[c + d*x]^3)/(4*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 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))^4 \tan ^2(c+d x) \, dx &=a^2 \int \left (-8 a^2-\frac {8 a^2}{-1+\sin (c+d x)}-8 a^2 \sin (c+d x)-7 a^2 \sin ^2(c+d x)-4 a^2 \sin ^3(c+d x)-a^2 \sin ^4(c+d x)\right ) \, dx\\ &=-8 a^4 x-a^4 \int \sin ^4(c+d x) \, dx-\left (4 a^4\right ) \int \sin ^3(c+d x) \, dx-\left (7 a^4\right ) \int \sin ^2(c+d x) \, dx-\left (8 a^4\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx-\left (8 a^4\right ) \int \sin (c+d x) \, dx\\ &=-8 a^4 x+\frac {8 a^4 \cos (c+d x)}{d}+\frac {8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {7 a^4 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac {1}{4} \left (3 a^4\right ) \int \sin ^2(c+d x) \, dx-\frac {1}{2} \left (7 a^4\right ) \int 1 \, dx+\frac {\left (4 a^4\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {23 a^4 x}{2}+\frac {12 a^4 \cos (c+d x)}{d}-\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {31 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac {1}{8} \left (3 a^4\right ) \int 1 \, dx\\ &=-\frac {95 a^4 x}{8}+\frac {12 a^4 \cos (c+d x)}{d}-\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {31 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]
time = 0.71, size = 125, normalized size = 1.11 \begin {gather*} \frac {(a+a \sin (c+d x))^4 \left (-1140 (c+d x)+1056 \cos (c+d x)-32 \cos (3 (c+d x))+\frac {1536 \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 )}+192 \sin (2 (c+d x))-3 \sin (4 (c+d x))\right )}{96 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^4*Tan[c + d*x]^2,x]

[Out]

((a + a*Sin[c + d*x])^4*(-1140*(c + d*x) + 1056*Cos[c + d*x] - 32*Cos[3*(c + d*x)] + (1536*Sin[(c + d*x)/2])/(
Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + 192*Sin[2*(c + d*x)] - 3*Sin[4*(c + d*x)]))/(96*d*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2])^8)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(230\) vs. \(2(105)=210\).
time = 0.27, size = 231, normalized size = 2.04

method	result	size
risch	\(-\frac {95 a^{4} x}{8}+\frac {11 a^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {11 a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {16 a^{4}}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}-\frac {a^{4} \sin \left (4 d x +4 c \right )}{32 d}-\frac {a^{4} \cos \left (3 d x +3 c \right )}{3 d}+\frac {2 a^{4} \sin \left (2 d x +2 c \right )}{d}\)	\(115\)
derivativedivides	\(\frac {a^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+4 a^{4} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+6 a^{4} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+4 a^{4} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a^{4} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(231\)
default	\(\frac {a^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+4 a^{4} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+6 a^{4} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+4 a^{4} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a^{4} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(231\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^4*tan(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.53, size = 181, normalized size = 1.60 \begin {gather*} -\frac {32 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{4} + 3 \, {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{4} + 72 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{4} + 24 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{4} - 96 \, a^{4} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{24 \, 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)^2,x, algorithm="maxima")

[Out]

-1/24*(32*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*a^4 + 3*(15*d*x + 15*c - (9*tan(d*x + c)^3 + 7*ta
n(d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1) - 8*tan(d*x + c))*a^4 + 72*(3*d*x + 3*c - tan(d*x + c)/(ta
n(d*x + c)^2 + 1) - 2*tan(d*x + c))*a^4 + 24*(d*x + c - tan(d*x + c))*a^4 - 96*a^4*(1/cos(d*x + c) + cos(d*x +
c)))/d

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Fricas [A]
time = 0.34, size = 179, normalized size = 1.58 \begin {gather*} -\frac {6 \, a^{4} \cos \left (d x + c\right )^{5} + 32 \, a^{4} \cos \left (d x + c\right )^{4} - 73 \, a^{4} \cos \left (d x + c\right )^{3} + 285 \, a^{4} d x - 288 \, a^{4} \cos \left (d x + c\right )^{2} - 192 \, a^{4} + 3 \, {\left (95 \, a^{4} d x - 127 \, a^{4}\right )} \cos \left (d x + c\right ) + {\left (6 \, a^{4} \cos \left (d x + c\right )^{4} - 26 \, a^{4} \cos \left (d x + c\right )^{3} - 285 \, a^{4} d x - 99 \, a^{4} \cos \left (d x + c\right )^{2} + 189 \, a^{4} \cos \left (d x + c\right ) - 192 \, a^{4}\right )} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right ) - 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)^2,x, algorithm="fricas")

[Out]

-1/24*(6*a^4*cos(d*x + c)^5 + 32*a^4*cos(d*x + c)^4 - 73*a^4*cos(d*x + c)^3 + 285*a^4*d*x - 288*a^4*cos(d*x +
c)^2 - 192*a^4 + 3*(95*a^4*d*x - 127*a^4)*cos(d*x + c) + (6*a^4*cos(d*x + c)^4 - 26*a^4*cos(d*x + c)^3 - 285*a
^4*d*x - 99*a^4*cos(d*x + c)^2 + 189*a^4*cos(d*x + c) - 192*a^4)*sin(d*x + c))/(d*cos(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 ^{2}{\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{2}{\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)**2,x)

[Out]

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

[Out]

Timed out

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Mupad [B]
time = 10.26, size = 363, normalized size = 3.21 \begin {gather*} -\frac {95\,a^4\,x}{8}-\frac {\frac {95\,a^4\,\left (c+d\,x\right )}{8}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{8}-\frac {a^4\,\left (285\,c+285\,d\,x-326\right )}{24}\right )-\frac {a^4\,\left (285\,c+285\,d\,x-896\right )}{24}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{8}-\frac {a^4\,\left (285\,c+285\,d\,x-570\right )}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{2}-\frac {a^4\,\left (1140\,c+1140\,d\,x-570\right )}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{2}-\frac {a^4\,\left (1140\,c+1140\,d\,x-1430\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{2}-\frac {a^4\,\left (1140\,c+1140\,d\,x-2154\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{2}-\frac {a^4\,\left (1140\,c+1140\,d\,x-3014\right )}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {285\,a^4\,\left (c+d\,x\right )}{4}-\frac {a^4\,\left (1710\,c+1710\,d\,x-1770\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {285\,a^4\,\left (c+d\,x\right )}{4}-\frac {a^4\,\left (1710\,c+1710\,d\,x-3606\right )}{24}\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2*(a + a*sin(c + d*x))^4,x)

[Out]

- (95*a^4*x)/8 - ((95*a^4*(c + d*x))/8 - tan(c/2 + (d*x)/2)*((95*a^4*(c + d*x))/8 - (a^4*(285*c + 285*d*x - 32
6))/24) - (a^4*(285*c + 285*d*x - 896))/24 + tan(c/2 + (d*x)/2)^8*((95*a^4*(c + d*x))/8 - (a^4*(285*c + 285*d*
x - 570))/24) - tan(c/2 + (d*x)/2)^7*((95*a^4*(c + d*x))/2 - (a^4*(1140*c + 1140*d*x - 570))/24) - tan(c/2 + (
d*x)/2)^3*((95*a^4*(c + d*x))/2 - (a^4*(1140*c + 1140*d*x - 1430))/24) + tan(c/2 + (d*x)/2)^6*((95*a^4*(c + d*
x))/2 - (a^4*(1140*c + 1140*d*x - 2154))/24) + tan(c/2 + (d*x)/2)^2*((95*a^4*(c + d*x))/2 - (a^4*(1140*c + 114
0*d*x - 3014))/24) - tan(c/2 + (d*x)/2)^5*((285*a^4*(c + d*x))/4 - (a^4*(1710*c + 1710*d*x - 1770))/24) + tan(
c/2 + (d*x)/2)^4*((285*a^4*(c + d*x))/4 - (a^4*(1710*c + 1710*d*x - 3606))/24))/(d*(tan(c/2 + (d*x)/2) - 1)*(t
an(c/2 + (d*x)/2)^2 + 1)^4)

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