3.155 \(\int (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=149 \[ \frac{a^2 \tan ^3(c+d x)}{3 d}-\frac{a^2 \tan (c+d x)}{d}+a^2 x-\frac{2 a b \cos (c+d x)}{d}+\frac{2 a b \sec ^3(c+d x)}{3 d}-\frac{4 a b \sec (c+d x)}{d}+\frac{5 b^2 \tan ^3(c+d x)}{6 d}-\frac{5 b^2 \tan (c+d x)}{2 d}-\frac{b^2 \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac{5 b^2 x}{2} \]

[Out]

a^2*x + (5*b^2*x)/2 - (2*a*b*Cos[c + d*x])/d - (4*a*b*Sec[c + d*x])/d + (2*a*b*Sec[c + d*x]^3)/(3*d) - (a^2*Ta
n[c + d*x])/d - (5*b^2*Tan[c + d*x])/(2*d) + (a^2*Tan[c + d*x]^3)/(3*d) + (5*b^2*Tan[c + d*x]^3)/(6*d) - (b^2*
Sin[c + d*x]^2*Tan[c + d*x]^3)/(2*d)

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Rubi [A]  time = 0.161982, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2722, 3473, 8, 2590, 270, 2591, 288, 302, 203} \[ \frac{a^2 \tan ^3(c+d x)}{3 d}-\frac{a^2 \tan (c+d x)}{d}+a^2 x-\frac{2 a b \cos (c+d x)}{d}+\frac{2 a b \sec ^3(c+d x)}{3 d}-\frac{4 a b \sec (c+d x)}{d}+\frac{5 b^2 \tan ^3(c+d x)}{6 d}-\frac{5 b^2 \tan (c+d x)}{2 d}-\frac{b^2 \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac{5 b^2 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*x + (5*b^2*x)/2 - (2*a*b*Cos[c + d*x])/d - (4*a*b*Sec[c + d*x])/d + (2*a*b*Sec[c + d*x]^3)/(3*d) - (a^2*Ta
n[c + d*x])/d - (5*b^2*Tan[c + d*x])/(2*d) + (a^2*Tan[c + d*x]^3)/(3*d) + (5*b^2*Tan[c + d*x]^3)/(6*d) - (b^2*
Sin[c + d*x]^2*Tan[c + d*x]^3)/(2*d)

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]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 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 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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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])

Rubi steps

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

Mathematica [A]  time = 0.686576, size = 176, normalized size = 1.18 \[ -\frac{\sec ^3(c+d x) \left (-36 \left (2 a^2+5 b^2\right ) (c+d x) \cos (c+d x)+32 a^2 \sin (3 (c+d x))-24 a^2 c \cos (3 (c+d x))-24 a^2 d x \cos (3 (c+d x))+288 a b \cos (2 (c+d x))+24 a b \cos (4 (c+d x))+200 a b+30 b^2 \sin (c+d x)+65 b^2 \sin (3 (c+d x))+3 b^2 \sin (5 (c+d x))-60 b^2 c \cos (3 (c+d x))-60 b^2 d x \cos (3 (c+d x))\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^3*(200*a*b - 36*(2*a^2 + 5*b^2)*(c + d*x)*Cos[c + d*x] + 288*a*b*Cos[2*(c + d*x)] - 24*a^2*c*Co
s[3*(c + d*x)] - 60*b^2*c*Cos[3*(c + d*x)] - 24*a^2*d*x*Cos[3*(c + d*x)] - 60*b^2*d*x*Cos[3*(c + d*x)] + 24*a*
b*Cos[4*(c + d*x)] + 30*b^2*Sin[c + d*x] + 32*a^2*Sin[3*(c + d*x)] + 65*b^2*Sin[3*(c + d*x)] + 3*b^2*Sin[5*(c
+ d*x)]))/(96*d)

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Maple [A]  time = 0.044, size = 185, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}-\tan \left ( dx+c \right ) +dx+c \right ) +2\,ab \left ( 1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{\cos \left ( dx+c \right ) }}- \left ( 8/3+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +{b}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{3\,\cos \left ( dx+c \right ) }}-{\frac{4\,\cos \left ( dx+c \right ) }{3} \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{2}}+{\frac{5\,c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.50033, size = 161, normalized size = 1.08 \begin{align*} \frac{2 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} +{\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 )} b^{2} - 4 \, a b{\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^2*tan(d*x+c)^4,x, algorithm="maxima")

[Out]

1/6*(2*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^2 + (2*tan(d*x + c)^3 + 15*d*x + 15*c - 3*tan(d*x + c
)/(tan(d*x + c)^2 + 1) - 12*tan(d*x + c))*b^2 - 4*a*b*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c))
)/d

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Fricas [A]  time = 1.46423, size = 281, normalized size = 1.89 \begin{align*} \frac{3 \,{\left (2 \, a^{2} + 5 \, b^{2}\right )} d x \cos \left (d x + c\right )^{3} - 12 \, a b \cos \left (d x + c\right )^{4} - 24 \, a b \cos \left (d x + c\right )^{2} + 4 \, a b -{\left (3 \, b^{2} \cos \left (d x + c\right )^{4} + 2 \,{\left (4 \, a^{2} + 7 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^2*tan(d*x+c)^4,x, algorithm="fricas")

[Out]

1/6*(3*(2*a^2 + 5*b^2)*d*x*cos(d*x + c)^3 - 12*a*b*cos(d*x + c)^4 - 24*a*b*cos(d*x + c)^2 + 4*a*b - (3*b^2*cos
(d*x + c)^4 + 2*(4*a^2 + 7*b^2)*cos(d*x + c)^2 - 2*a^2 - 2*b^2)*sin(d*x + c))/(d*cos(d*x + c)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))**2*tan(d*x+c)**4,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out