3.2.55 \(\int (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx\) [155]

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

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 149, normalized size of antiderivative = 1.00, 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 = {2801, 3554, 8, 2670, 276, 2671, 294, 308, 209} \begin {gather*} \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} \end {gather*}

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 8

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

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 276

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 294

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 308

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 2670

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 2671

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]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 2801

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 3554

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))^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) \text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^4} \, dx,x,\cos (c+d x)\right )}{d}+\frac {b^2 \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.47, size = 176, normalized size = 1.18 \begin {gather*} -\frac {\sec ^3(c+d x) \left (200 a b-36 \left (2 a^2+5 b^2\right ) (c+d x) \cos (c+d x)+288 a b \cos (2 (c+d x))-24 a^2 c \cos (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))\right )}{96 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*(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*Cos[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)]))/d

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Maple [A]
time = 0.22, size = 185, normalized size = 1.24

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+2 a b \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+b^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\)	\(185\)
default	\(\frac {a^{2} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+2 a b \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+b^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\)	\(185\)
risch	\(a^{2} x +\frac {5 b^{2} x}{2}+\frac {i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {i b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 \left (6 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+6 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+16 a b \,{\mathrm e}^{3 i \left (d x +c \right )}+4 i a^{2}+7 i b^{2}+12 a \,{\mathrm e}^{i \left (d x +c \right )} b \right )}{3 d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{3}}\)	\(211\)

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,method=_RETURNVERBOSE)

[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 = 0.57, size = 119, normalized size = 0.80 \begin {gather*} \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 {gather*}

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 = 0.35, size = 118, normalized size = 0.79 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \tan ^{4}{\left (c + d x \right )}\, dx \end {gather*}

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]

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

[Out]

Timed out

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Mupad [B]
time = 10.04, size = 235, normalized size = 1.58 \begin {gather*} \frac {x\,\left (2\,a^2+5\,b^2\right )}{2}-\frac {\left (-2\,a^2-5\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,a^2}{3}+\frac {20\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {28\,a^2}{3}+\frac {22\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {64\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\left (\frac {8\,a^2}{3}+\frac {20\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {32\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\left (-2\,a^2-5\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {32\,a\,b}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-{\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 )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(2*a^2 + 5*b^2))/2 - (tan(c/2 + (d*x)/2)^3*((8*a^2)/3 + (20*b^2)/3) - tan(c/2 + (d*x)/2)^9*(2*a^2 + 5*b^2)
- (32*a*b)/3 + tan(c/2 + (d*x)/2)^7*((8*a^2)/3 + (20*b^2)/3) + tan(c/2 + (d*x)/2)^5*((28*a^2)/3 + (22*b^2)/3)
- tan(c/2 + (d*x)/2)*(2*a^2 + 5*b^2) + (32*a*b*tan(c/2 + (d*x)/2)^2)/3 + (64*a*b*tan(c/2 + (d*x)/2)^4)/3)/(d*(
tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 - tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x
)/2)^10 - 1))

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