3.4.82 \(\int (a+b \tan ^4(c+d x))^3 \, dx\) [382]

Optimal. Leaf size=144 \[ (a+b)^3 x-\frac {b \left (3 a^2+3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b \left (3 a^2+3 a b+b^2\right ) \tan ^3(c+d x)}{3 d}-\frac {b^2 (3 a+b) \tan ^5(c+d x)}{5 d}+\frac {b^2 (3 a+b) \tan ^7(c+d x)}{7 d}-\frac {b^3 \tan ^9(c+d x)}{9 d}+\frac {b^3 \tan ^{11}(c+d x)}{11 d} \]

[Out]

(a+b)^3*x-b*(3*a^2+3*a*b+b^2)*tan(d*x+c)/d+1/3*b*(3*a^2+3*a*b+b^2)*tan(d*x+c)^3/d-1/5*b^2*(3*a+b)*tan(d*x+c)^5
/d+1/7*b^2*(3*a+b)*tan(d*x+c)^7/d-1/9*b^3*tan(d*x+c)^9/d+1/11*b^3*tan(d*x+c)^11/d

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Rubi [A]
time = 0.06, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3742, 1168, 209} \begin {gather*} \frac {b \left (3 a^2+3 a b+b^2\right ) \tan ^3(c+d x)}{3 d}-\frac {b \left (3 a^2+3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b^2 (3 a+b) \tan ^7(c+d x)}{7 d}-\frac {b^2 (3 a+b) \tan ^5(c+d x)}{5 d}+x (a+b)^3+\frac {b^3 \tan ^{11}(c+d x)}{11 d}-\frac {b^3 \tan ^9(c+d x)}{9 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^4)^3,x]

[Out]

(a + b)^3*x - (b*(3*a^2 + 3*a*b + b^2)*Tan[c + d*x])/d + (b*(3*a^2 + 3*a*b + b^2)*Tan[c + d*x]^3)/(3*d) - (b^2
*(3*a + b)*Tan[c + d*x]^5)/(5*d) + (b^2*(3*a + b)*Tan[c + d*x]^7)/(7*d) - (b^3*Tan[c + d*x]^9)/(9*d) + (b^3*Ta
n[c + d*x]^11)/(11*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 1168

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a
+ c*x^4)^p, x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 3742

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[c*(ff/f), Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \left (a+b \tan ^4(c+d x)\right )^3 \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b x^4\right )^3}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (-b \left (3 a^2+3 a b+b^2\right )+b \left (3 a^2+3 a b+b^2\right ) x^2-b^2 (3 a+b) x^4+b^2 (3 a+b) x^6-b^3 x^8+b^3 x^{10}+\frac {(a+b)^3}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {b \left (3 a^2+3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b \left (3 a^2+3 a b+b^2\right ) \tan ^3(c+d x)}{3 d}-\frac {b^2 (3 a+b) \tan ^5(c+d x)}{5 d}+\frac {b^2 (3 a+b) \tan ^7(c+d x)}{7 d}-\frac {b^3 \tan ^9(c+d x)}{9 d}+\frac {b^3 \tan ^{11}(c+d x)}{11 d}+\frac {(a+b)^3 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=(a+b)^3 x-\frac {b \left (3 a^2+3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b \left (3 a^2+3 a b+b^2\right ) \tan ^3(c+d x)}{3 d}-\frac {b^2 (3 a+b) \tan ^5(c+d x)}{5 d}+\frac {b^2 (3 a+b) \tan ^7(c+d x)}{7 d}-\frac {b^3 \tan ^9(c+d x)}{9 d}+\frac {b^3 \tan ^{11}(c+d x)}{11 d}\\ \end {align*}

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Mathematica [A]
time = 0.67, size = 128, normalized size = 0.89 \begin {gather*} \frac {(a+b)^3 \text {ArcTan}(\tan (c+d x))}{d}+\frac {b \tan (c+d x) \left (-3465 \left (3 a^2+3 a b+b^2\right )+1155 \left (3 a^2+3 a b+b^2\right ) \tan ^2(c+d x)-693 b (3 a+b) \tan ^4(c+d x)+495 b (3 a+b) \tan ^6(c+d x)-385 b^2 \tan ^8(c+d x)+315 b^2 \tan ^{10}(c+d x)\right )}{3465 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x]^4)^3,x]

[Out]

((a + b)^3*ArcTan[Tan[c + d*x]])/d + (b*Tan[c + d*x]*(-3465*(3*a^2 + 3*a*b + b^2) + 1155*(3*a^2 + 3*a*b + b^2)
*Tan[c + d*x]^2 - 693*b*(3*a + b)*Tan[c + d*x]^4 + 495*b*(3*a + b)*Tan[c + d*x]^6 - 385*b^2*Tan[c + d*x]^8 + 3
15*b^2*Tan[c + d*x]^10))/(3465*d)

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Maple [A]
time = 0.12, size = 187, normalized size = 1.30

method	result	size
norman	\(\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) x -\frac {b^{3} \left (\tan ^{9}\left (d x +c \right )\right )}{9 d}+\frac {b^{3} \left (\tan ^{11}\left (d x +c \right )\right )}{11 d}-\frac {b \left (3 a^{2}+3 a b +b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {b \left (3 a^{2}+3 a b +b^{2}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {b^{2} \left (3 a +b \right ) \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}+\frac {b^{2} \left (3 a +b \right ) \left (\tan ^{7}\left (d x +c \right )\right )}{7 d}\)	\(149\)
derivativedivides	\(\frac {\frac {b^{3} \left (\tan ^{11}\left (d x +c \right )\right )}{11}-\frac {b^{3} \left (\tan ^{9}\left (d x +c \right )\right )}{9}+\frac {3 a \,b^{2} \left (\tan ^{7}\left (d x +c \right )\right )}{7}+\frac {b^{3} \left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {3 a \,b^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {b^{3} \left (\tan ^{5}\left (d x +c \right )\right )}{5}+a^{2} b \left (\tan ^{3}\left (d x +c \right )\right )+a \,b^{2} \left (\tan ^{3}\left (d x +c \right )\right )+\frac {b^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{3}-3 a^{2} b \tan \left (d x +c \right )-3 a \,b^{2} \tan \left (d x +c \right )-b^{3} \tan \left (d x +c \right )+\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(187\)
default	\(\frac {\frac {b^{3} \left (\tan ^{11}\left (d x +c \right )\right )}{11}-\frac {b^{3} \left (\tan ^{9}\left (d x +c \right )\right )}{9}+\frac {3 a \,b^{2} \left (\tan ^{7}\left (d x +c \right )\right )}{7}+\frac {b^{3} \left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {3 a \,b^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {b^{3} \left (\tan ^{5}\left (d x +c \right )\right )}{5}+a^{2} b \left (\tan ^{3}\left (d x +c \right )\right )+a \,b^{2} \left (\tan ^{3}\left (d x +c \right )\right )+\frac {b^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{3}-3 a^{2} b \tan \left (d x +c \right )-3 a \,b^{2} \tan \left (d x +c \right )-b^{3} \tan \left (d x +c \right )+\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(187\)
risch	\(a^{3} x +3 a^{2} b x +3 a \,b^{2} x +b^{3} x -\frac {4 i b \left (751674 b^{2} {\mathrm e}^{12 i \left (d x +c \right )}+751674 b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+1358280 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+6930 a^{2}+3254 b^{2}+10395 b^{2} {\mathrm e}^{20 i \left (d x +c \right )}+93555 a^{2} {\mathrm e}^{18 i \left (d x +c \right )}+438900 b^{2} {\mathrm e}^{14 i \left (d x +c \right )}+1503810 a^{2} {\mathrm e}^{12 i \left (d x +c \right )}+51975 b^{2} {\mathrm e}^{18 i \left (d x +c \right )}+381150 a^{2} {\mathrm e}^{16 i \left (d x +c \right )}+219450 b^{2} {\mathrm e}^{16 i \left (d x +c \right )}+928620 a^{2} {\mathrm e}^{14 i \left (d x +c \right )}+1697850 a^{2} {\mathrm e}^{10 i \left (d x +c \right )}+317460 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+634920 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+526680 a b \,{\mathrm e}^{16 i \left (d x +c \right )}+145530 a b \,{\mathrm e}^{18 i \left (d x +c \right )}+2109492 a b \,{\mathrm e}^{10 i \left (d x +c \right )}+20790 a b \,{\mathrm e}^{20 i \left (d x +c \right )}+1219680 a b \,{\mathrm e}^{14 i \left (d x +c \right )}+1915452 a b \,{\mathrm e}^{12 i \left (d x +c \right )}+910800 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+333630 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+75042 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+1655280 a b \,{\mathrm e}^{8 i \left (d x +c \right )}+8712 a b +10395 a^{2} {\mathrm e}^{20 i \left (d x +c \right )}+762300 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+287595 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+126995 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+65835 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+25399 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3465 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{11}}\)	\(471\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/11*b^3*tan(d*x+c)^11-1/9*b^3*tan(d*x+c)^9+3/7*a*b^2*tan(d*x+c)^7+1/7*b^3*tan(d*x+c)^7-3/5*a*b^2*tan(d*x
+c)^5-1/5*b^3*tan(d*x+c)^5+a^2*b*tan(d*x+c)^3+a*b^2*tan(d*x+c)^3+1/3*b^3*tan(d*x+c)^3-3*a^2*b*tan(d*x+c)-3*a*b
^2*tan(d*x+c)-b^3*tan(d*x+c)+(a^3+3*a^2*b+3*a*b^2+b^3)*arctan(tan(d*x+c)))

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Maxima [A]
time = 0.51, size = 167, normalized size = 1.16 \begin {gather*} a^{3} x + \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} b}{d} + \frac {{\left (15 \, \tan \left (d x + c\right )^{7} - 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 105 \, d x + 105 \, c - 105 \, \tan \left (d x + c\right )\right )} a b^{2}}{35 \, d} + \frac {{\left (315 \, \tan \left (d x + c\right )^{11} - 385 \, \tan \left (d x + c\right )^{9} + 495 \, \tan \left (d x + c\right )^{7} - 693 \, \tan \left (d x + c\right )^{5} + 1155 \, \tan \left (d x + c\right )^{3} + 3465 \, d x + 3465 \, c - 3465 \, \tan \left (d x + c\right )\right )} b^{3}}{3465 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x + (tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^2*b/d + 1/35*(15*tan(d*x + c)^7 - 21*tan(d*x + c)^5
+ 35*tan(d*x + c)^3 + 105*d*x + 105*c - 105*tan(d*x + c))*a*b^2/d + 1/3465*(315*tan(d*x + c)^11 - 385*tan(d*x
+ c)^9 + 495*tan(d*x + c)^7 - 693*tan(d*x + c)^5 + 1155*tan(d*x + c)^3 + 3465*d*x + 3465*c - 3465*tan(d*x + c)
)*b^3/d

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Fricas [A]
time = 3.67, size = 145, normalized size = 1.01 \begin {gather*} \frac {315 \, b^{3} \tan \left (d x + c\right )^{11} - 385 \, b^{3} \tan \left (d x + c\right )^{9} + 495 \, {\left (3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )^{7} - 693 \, {\left (3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )^{5} + 1155 \, {\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )^{3} + 3465 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x - 3465 \, {\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )}{3465 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(315*b^3*tan(d*x + c)^11 - 385*b^3*tan(d*x + c)^9 + 495*(3*a*b^2 + b^3)*tan(d*x + c)^7 - 693*(3*a*b^2 +
b^3)*tan(d*x + c)^5 + 1155*(3*a^2*b + 3*a*b^2 + b^3)*tan(d*x + c)^3 + 3465*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*
x - 3465*(3*a^2*b + 3*a*b^2 + b^3)*tan(d*x + c))/d

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Sympy [A]
time = 0.50, size = 224, normalized size = 1.56 \begin {gather*} \begin {cases} a^{3} x + 3 a^{2} b x + \frac {a^{2} b \tan ^{3}{\left (c + d x \right )}}{d} - \frac {3 a^{2} b \tan {\left (c + d x \right )}}{d} + 3 a b^{2} x + \frac {3 a b^{2} \tan ^{7}{\left (c + d x \right )}}{7 d} - \frac {3 a b^{2} \tan ^{5}{\left (c + d x \right )}}{5 d} + \frac {a b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac {3 a b^{2} \tan {\left (c + d x \right )}}{d} + b^{3} x + \frac {b^{3} \tan ^{11}{\left (c + d x \right )}}{11 d} - \frac {b^{3} \tan ^{9}{\left (c + d x \right )}}{9 d} + \frac {b^{3} \tan ^{7}{\left (c + d x \right )}}{7 d} - \frac {b^{3} \tan ^{5}{\left (c + d x \right )}}{5 d} + \frac {b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{3} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan ^{4}{\left (c \right )}\right )^{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x + 3*a**2*b*x + a**2*b*tan(c + d*x)**3/d - 3*a**2*b*tan(c + d*x)/d + 3*a*b**2*x + 3*a*b**2*ta
n(c + d*x)**7/(7*d) - 3*a*b**2*tan(c + d*x)**5/(5*d) + a*b**2*tan(c + d*x)**3/d - 3*a*b**2*tan(c + d*x)/d + b*
*3*x + b**3*tan(c + d*x)**11/(11*d) - b**3*tan(c + d*x)**9/(9*d) + b**3*tan(c + d*x)**7/(7*d) - b**3*tan(c + d
*x)**5/(5*d) + b**3*tan(c + d*x)**3/(3*d) - b**3*tan(c + d*x)/d, Ne(d, 0)), (x*(a + b*tan(c)**4)**3, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 3499 vs. \(2 (134) = 268\).
time = 15.36, size = 3499, normalized size = 24.30 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(3465*a^3*d*x*tan(d*x)^11*tan(c)^11 + 10395*a^2*b*d*x*tan(d*x)^11*tan(c)^11 + 10395*a*b^2*d*x*tan(d*x)^
11*tan(c)^11 + 3465*b^3*d*x*tan(d*x)^11*tan(c)^11 - 38115*a^3*d*x*tan(d*x)^10*tan(c)^10 - 114345*a^2*b*d*x*tan
(d*x)^10*tan(c)^10 - 114345*a*b^2*d*x*tan(d*x)^10*tan(c)^10 - 38115*b^3*d*x*tan(d*x)^10*tan(c)^10 + 10395*a^2*
b*tan(d*x)^11*tan(c)^10 + 10395*a*b^2*tan(d*x)^11*tan(c)^10 + 3465*b^3*tan(d*x)^11*tan(c)^10 + 10395*a^2*b*tan
(d*x)^10*tan(c)^11 + 10395*a*b^2*tan(d*x)^10*tan(c)^11 + 3465*b^3*tan(d*x)^10*tan(c)^11 + 190575*a^3*d*x*tan(d
*x)^9*tan(c)^9 + 571725*a^2*b*d*x*tan(d*x)^9*tan(c)^9 + 571725*a*b^2*d*x*tan(d*x)^9*tan(c)^9 + 190575*b^3*d*x*
tan(d*x)^9*tan(c)^9 - 3465*a^2*b*tan(d*x)^11*tan(c)^8 - 3465*a*b^2*tan(d*x)^11*tan(c)^8 - 1155*b^3*tan(d*x)^11
*tan(c)^8 - 114345*a^2*b*tan(d*x)^10*tan(c)^9 - 114345*a*b^2*tan(d*x)^10*tan(c)^9 - 38115*b^3*tan(d*x)^10*tan(
c)^9 - 114345*a^2*b*tan(d*x)^9*tan(c)^10 - 114345*a*b^2*tan(d*x)^9*tan(c)^10 - 38115*b^3*tan(d*x)^9*tan(c)^10
- 3465*a^2*b*tan(d*x)^8*tan(c)^11 - 3465*a*b^2*tan(d*x)^8*tan(c)^11 - 1155*b^3*tan(d*x)^8*tan(c)^11 - 571725*a
^3*d*x*tan(d*x)^8*tan(c)^8 - 1715175*a^2*b*d*x*tan(d*x)^8*tan(c)^8 - 1715175*a*b^2*d*x*tan(d*x)^8*tan(c)^8 - 5
71725*b^3*d*x*tan(d*x)^8*tan(c)^8 + 2079*a*b^2*tan(d*x)^11*tan(c)^6 + 693*b^3*tan(d*x)^11*tan(c)^6 + 27720*a^2
*b*tan(d*x)^10*tan(c)^7 + 38115*a*b^2*tan(d*x)^10*tan(c)^7 + 12705*b^3*tan(d*x)^10*tan(c)^7 + 550935*a^2*b*tan
(d*x)^9*tan(c)^8 + 571725*a*b^2*tan(d*x)^9*tan(c)^8 + 190575*b^3*tan(d*x)^9*tan(c)^8 + 550935*a^2*b*tan(d*x)^8
*tan(c)^9 + 571725*a*b^2*tan(d*x)^8*tan(c)^9 + 190575*b^3*tan(d*x)^8*tan(c)^9 + 27720*a^2*b*tan(d*x)^7*tan(c)^
10 + 38115*a*b^2*tan(d*x)^7*tan(c)^10 + 12705*b^3*tan(d*x)^7*tan(c)^10 + 2079*a*b^2*tan(d*x)^6*tan(c)^11 + 693
*b^3*tan(d*x)^6*tan(c)^11 + 1143450*a^3*d*x*tan(d*x)^7*tan(c)^7 + 3430350*a^2*b*d*x*tan(d*x)^7*tan(c)^7 + 3430
350*a*b^2*d*x*tan(d*x)^7*tan(c)^7 + 1143450*b^3*d*x*tan(d*x)^7*tan(c)^7 - 1485*a*b^2*tan(d*x)^11*tan(c)^4 - 49
5*b^3*tan(d*x)^11*tan(c)^4 - 22869*a*b^2*tan(d*x)^10*tan(c)^5 - 7623*b^3*tan(d*x)^10*tan(c)^5 - 97020*a^2*b*ta
n(d*x)^9*tan(c)^6 - 190575*a*b^2*tan(d*x)^9*tan(c)^6 - 63525*b^3*tan(d*x)^9*tan(c)^6 - 1538460*a^2*b*tan(d*x)^
8*tan(c)^7 - 1715175*a*b^2*tan(d*x)^8*tan(c)^7 - 571725*b^3*tan(d*x)^8*tan(c)^7 - 1538460*a^2*b*tan(d*x)^7*tan
(c)^8 - 1715175*a*b^2*tan(d*x)^7*tan(c)^8 - 571725*b^3*tan(d*x)^7*tan(c)^8 - 97020*a^2*b*tan(d*x)^6*tan(c)^9 -
190575*a*b^2*tan(d*x)^6*tan(c)^9 - 63525*b^3*tan(d*x)^6*tan(c)^9 - 22869*a*b^2*tan(d*x)^5*tan(c)^10 - 7623*b^
3*tan(d*x)^5*tan(c)^10 - 1485*a*b^2*tan(d*x)^4*tan(c)^11 - 495*b^3*tan(d*x)^4*tan(c)^11 - 1600830*a^3*d*x*tan(
d*x)^6*tan(c)^6 - 4802490*a^2*b*d*x*tan(d*x)^6*tan(c)^6 - 4802490*a*b^2*d*x*tan(d*x)^6*tan(c)^6 - 1600830*b^3*
d*x*tan(d*x)^6*tan(c)^6 + 385*b^3*tan(d*x)^11*tan(c)^2 + 5940*a*b^2*tan(d*x)^10*tan(c)^3 + 5445*b^3*tan(d*x)^1
0*tan(c)^3 + 72765*a*b^2*tan(d*x)^9*tan(c)^4 + 38115*b^3*tan(d*x)^9*tan(c)^4 + 194040*a^2*b*tan(d*x)^8*tan(c)^
5 + 474705*a*b^2*tan(d*x)^8*tan(c)^5 + 190575*b^3*tan(d*x)^8*tan(c)^5 + 2765070*a^2*b*tan(d*x)^7*tan(c)^6 + 32
84820*a*b^2*tan(d*x)^7*tan(c)^6 + 1143450*b^3*tan(d*x)^7*tan(c)^6 + 2765070*a^2*b*tan(d*x)^6*tan(c)^7 + 328482
0*a*b^2*tan(d*x)^6*tan(c)^7 + 1143450*b^3*tan(d*x)^6*tan(c)^7 + 194040*a^2*b*tan(d*x)^5*tan(c)^8 + 474705*a*b^
2*tan(d*x)^5*tan(c)^8 + 190575*b^3*tan(d*x)^5*tan(c)^8 + 72765*a*b^2*tan(d*x)^4*tan(c)^9 + 38115*b^3*tan(d*x)^
4*tan(c)^9 + 5940*a*b^2*tan(d*x)^3*tan(c)^10 + 5445*b^3*tan(d*x)^3*tan(c)^10 + 385*b^3*tan(d*x)^2*tan(c)^11 +
1600830*a^3*d*x*tan(d*x)^5*tan(c)^5 + 4802490*a^2*b*d*x*tan(d*x)^5*tan(c)^5 + 4802490*a*b^2*d*x*tan(d*x)^5*tan
(c)^5 + 1600830*b^3*d*x*tan(d*x)^5*tan(c)^5 - 315*b^3*tan(d*x)^11 - 4235*b^3*tan(d*x)^10*tan(c) - 8910*a*b^2*t
an(d*x)^9*tan(c)^2 - 27225*b^3*tan(d*x)^9*tan(c)^2 - 103950*a*b^2*tan(d*x)^8*tan(c)^3 - 114345*b^3*tan(d*x)^8*
tan(c)^3 - 242550*a^2*b*tan(d*x)^7*tan(c)^4 - 637560*a*b^2*tan(d*x)^7*tan(c)^4 - 381150*b^3*tan(d*x)^7*tan(c)^
4 - 3347190*a^2*b*tan(d*x)^6*tan(c)^5 - 4074840*a*b^2*tan(d*x)^6*tan(c)^5 - 1600830*b^3*tan(d*x)^6*tan(c)^5 -
3347190*a^2*b*tan(d*x)^5*tan(c)^6 - 4074840*a*b^2*tan(d*x)^5*tan(c)^6 - 1600830*b^3*tan(d*x)^5*tan(c)^6 - 2425
50*a^2*b*tan(d*x)^4*tan(c)^7 - 637560*a*b^2*tan(d*x)^4*tan(c)^7 - 381150*b^3*tan(d*x)^4*tan(c)^7 - 103950*a*b^
2*tan(d*x)^3*tan(c)^8 - 114345*b^3*tan(d*x)^3*tan(c)^8 - 8910*a*b^2*tan(d*x)^2*tan(c)^9 - 27225*b^3*tan(d*x)^2
*tan(c)^9 - 4235*b^3*tan(d*x)*tan(c)^10 - 315*b^3*tan(c)^11 - 1143450*a^3*d*x*tan(d*x)^4*tan(c)^4 - 3430350*a^
2*b*d*x*tan(d*x)^4*tan(c)^4 - 3430350*a*b^2*d*x*tan(d*x)^4*tan(c)^4 - 1143450*b^3*d*x*tan(d*x)^4*tan(c)^4 + 38
5*b^3*tan(d*x)^9 + 5940*a*b^2*tan(d*x)^8*tan(c) + 5445*b^3*tan(d*x)^8*tan(c) + 72765*a*b^2*tan(d*x)^7*tan(c)^2
+ 38115*b^3*tan(d*x)^7*tan(c)^2 + 194040*a^2*b*tan(d*x)^6*tan(c)^3 + 474705*a*b^2*tan(d*x)^6*tan(c)^3 + 19057
5*b^3*tan(d*x)^6*tan(c)^3 + 2765070*a^2*b*tan(d*x)^5*tan(c)^4 + 3284820*a*b^2*tan(d*x)^5*tan(c)^4 + 1143450*b^
3*tan(d*x)^5*tan(c)^4 + 2765070*a^2*b*tan(d*x)^...

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Mupad [B]
time = 11.66, size = 180, normalized size = 1.25 \begin {gather*} \frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a^2\,b+a\,b^2+\frac {b^3}{3}\right )}{d}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,{\left (a+b\right )}^3}{a^3+3\,a^2\,b+3\,a\,b^2+b^3}\right )\,{\left (a+b\right )}^3}{d}-\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^9}{9\,d}+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^{11}}{11\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {b^3}{5}+\frac {3\,a\,b^2}{5}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^7\,\left (\frac {b^3}{7}+\frac {3\,a\,b^2}{7}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x)^4)^3,x)

[Out]

(tan(c + d*x)^3*(a*b^2 + a^2*b + b^3/3))/d + (atan((tan(c + d*x)*(a + b)^3)/(3*a*b^2 + 3*a^2*b + a^3 + b^3))*(
a + b)^3)/d - (b^3*tan(c + d*x)^9)/(9*d) + (b^3*tan(c + d*x)^11)/(11*d) - (tan(c + d*x)*(3*a*b^2 + 3*a^2*b + b
^3))/d - (tan(c + d*x)^5*((3*a*b^2)/5 + b^3/5))/d + (tan(c + d*x)^7*((3*a*b^2)/7 + b^3/7))/d

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