3.1458 \(\int (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=146 \[ \frac {a^3 \tan (c+d x)}{d}+a^3 (-x)+\frac {3 a^2 b \cos (c+d x)}{d}+\frac {3 a^2 b \sec (c+d x)}{d}+\frac {9 a b^2 \tan (c+d x)}{2 d}-\frac {3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {9}{2} a b^2 x-\frac {b^3 \cos ^3(c+d x)}{3 d}+\frac {2 b^3 \cos (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d} \]

[Out]

-a^3*x-9/2*a*b^2*x+3*a^2*b*cos(d*x+c)/d+2*b^3*cos(d*x+c)/d-1/3*b^3*cos(d*x+c)^3/d+3*a^2*b*sec(d*x+c)/d+b^3*sec
(d*x+c)/d+a^3*tan(d*x+c)/d+9/2*a*b^2*tan(d*x+c)/d-3/2*a*b^2*sin(d*x+c)^2*tan(d*x+c)/d

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Rubi [A]  time = 0.21, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2722, 3473, 8, 2590, 14, 2591, 288, 321, 203, 270} \[ \frac {3 a^2 b \cos (c+d x)}{d}+\frac {3 a^2 b \sec (c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d}+a^3 (-x)+\frac {9 a b^2 \tan (c+d x)}{2 d}-\frac {3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {9}{2} a b^2 x-\frac {b^3 \cos ^3(c+d x)}{3 d}+\frac {2 b^3 \cos (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^3*Tan[c + d*x]^2,x]

[Out]

-(a^3*x) - (9*a*b^2*x)/2 + (3*a^2*b*Cos[c + d*x])/d + (2*b^3*Cos[c + d*x])/d - (b^3*Cos[c + d*x]^3)/(3*d) + (3
*a^2*b*Sec[c + d*x])/d + (b^3*Sec[c + d*x])/d + (a^3*Tan[c + d*x])/d + (9*a*b^2*Tan[c + d*x])/(2*d) - (3*a*b^2
*Sin[c + d*x]^2*Tan[c + d*x])/(2*d)

Rule 8

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 203

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 2722

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

\begin {align*} \int (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx &=\int \left (a^3 \tan ^2(c+d x)+3 a^2 b \sin (c+d x) \tan ^2(c+d x)+3 a b^2 \sin ^2(c+d x) \tan ^2(c+d x)+b^3 \sin ^3(c+d x) \tan ^2(c+d x)\right ) \, dx\\ &=a^3 \int \tan ^2(c+d x) \, dx+\left (3 a^2 b\right ) \int \sin (c+d x) \tan ^2(c+d x) \, dx+\left (3 a b^2\right ) \int \sin ^2(c+d x) \tan ^2(c+d x) \, dx+b^3 \int \sin ^3(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {a^3 \tan (c+d x)}{d}-a^3 \int 1 \, dx-\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (3 a b^2\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}-\frac {b^3 \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-a^3 x+\frac {a^3 \tan (c+d x)}{d}-\frac {3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \left (-1+\frac {1}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (9 a b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {b^3 \operatorname {Subst}\left (\int \left (-2+\frac {1}{x^2}+x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-a^3 x+\frac {3 a^2 b \cos (c+d x)}{d}+\frac {2 b^3 \cos (c+d x)}{d}-\frac {b^3 \cos ^3(c+d x)}{3 d}+\frac {3 a^2 b \sec (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d}+\frac {9 a b^2 \tan (c+d x)}{2 d}-\frac {3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {\left (9 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-a^3 x-\frac {9}{2} a b^2 x+\frac {3 a^2 b \cos (c+d x)}{d}+\frac {2 b^3 \cos (c+d x)}{d}-\frac {b^3 \cos ^3(c+d x)}{3 d}+\frac {3 a^2 b \sec (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d}+\frac {9 a b^2 \tan (c+d x)}{2 d}-\frac {3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.74, size = 113, normalized size = 0.77 \[ \frac {3 a \left (\left (8 a^2+27 b^2\right ) \tan (c+d x)-4 \left (2 a^2+9 b^2\right ) (c+d x)\right )+b \sec (c+d x) \left (4 \left (9 a^2+5 b^2\right ) \cos (2 (c+d x))+108 a^2+9 a b \sin (3 (c+d x))-b^2 \cos (4 (c+d x))+45 b^2\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^3*Tan[c + d*x]^2,x]

[Out]

(b*Sec[c + d*x]*(108*a^2 + 45*b^2 + 4*(9*a^2 + 5*b^2)*Cos[2*(c + d*x)] - b^2*Cos[4*(c + d*x)] + 9*a*b*Sin[3*(c
+ d*x)]) + 3*a*(-4*(2*a^2 + 9*b^2)*(c + d*x) + (8*a^2 + 27*b^2)*Tan[c + d*x]))/(24*d)

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fricas [A]  time = 0.43, size = 116, normalized size = 0.79 \[ -\frac {2 \, b^{3} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} d x \cos \left (d x + c\right ) - 18 \, a^{2} b - 6 \, b^{3} - 6 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, a b^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{3} + 6 \, a b^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/6*(2*b^3*cos(d*x + c)^4 + 3*(2*a^3 + 9*a*b^2)*d*x*cos(d*x + c) - 18*a^2*b - 6*b^3 - 6*(3*a^2*b + 2*b^3)*cos
(d*x + c)^2 - 3*(3*a*b^2*cos(d*x + c)^2 + 2*a^3 + 6*a*b^2)*sin(d*x + c))/(d*cos(d*x + c))

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giac [A]  time = 0.20, size = 207, normalized size = 1.42 \[ -\frac {3 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} {\left (d x + c\right )} + \frac {12 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b + b^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, a^{2} b - 10 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/6*(3*(2*a^3 + 9*a*b^2)*(d*x + c) + 12*(a^3*tan(1/2*d*x + 1/2*c) + 3*a*b^2*tan(1/2*d*x + 1/2*c) + 3*a^2*b +
b^3)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(9*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 18*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 6*b^
3*tan(1/2*d*x + 1/2*c)^4 - 36*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 24*b^3*tan(1/2*d*x + 1/2*c)^2 - 9*a*b^2*tan(1/2*d
*x + 1/2*c) - 18*a^2*b - 10*b^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 0.55, size = 169, normalized size = 1.16 \[ \frac {a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+3 a^{2} b \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a \,b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+b^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)^2*(a+b*sin(d*x+c))^3,x)

[Out]

1/d*(a^3*(tan(d*x+c)-d*x-c)+3*a^2*b*(sin(d*x+c)^4/cos(d*x+c)+(2+sin(d*x+c)^2)*cos(d*x+c))+3*a*b^2*(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)+b^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)))

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maxima [A]  time = 0.47, size = 119, normalized size = 0.82 \[ -\frac {6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} + 9 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a b^{2} + 2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} b^{3} - 18 \, a^{2} b {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/6*(6*(d*x + c - tan(d*x + c))*a^3 + 9*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c))*a*
b^2 + 2*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*b^3 - 18*a^2*b*(1/cos(d*x + c) + cos(d*x + c)))/d

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mupad [B]  time = 16.66, size = 249, normalized size = 1.71 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^3+9\,a\,b^2\right )+12\,a^2\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (2\,a^3+9\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (6\,a^3+15\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (6\,a^3+15\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (24\,a^2\,b+\frac {32\,b^3}{3}\right )+\frac {16\,b^3}{3}+12\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+9\,b^2\right )}{2\,a^3+9\,a\,b^2}\right )\,\left (2\,a^2+9\,b^2\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^2*(a + b*sin(c + d*x))^3)/cos(c + d*x)^2,x)

[Out]

(tan(c/2 + (d*x)/2)*(9*a*b^2 + 2*a^3) + 12*a^2*b + tan(c/2 + (d*x)/2)^7*(9*a*b^2 + 2*a^3) + tan(c/2 + (d*x)/2)
^3*(15*a*b^2 + 6*a^3) + tan(c/2 + (d*x)/2)^5*(15*a*b^2 + 6*a^3) + tan(c/2 + (d*x)/2)^2*(24*a^2*b + (32*b^3)/3)
+ (16*b^3)/3 + 12*a^2*b*tan(c/2 + (d*x)/2)^4)/(d*(2*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2)^6 - tan(c/2 +
(d*x)/2)^8 + 1)) - (a*atan((a*tan(c/2 + (d*x)/2)*(2*a^2 + 9*b^2))/(9*a*b^2 + 2*a^3))*(2*a^2 + 9*b^2))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)**2*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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