3.796 \(\int \sin (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=101 \[ -\frac {a \cos (c+d x)}{d}+\frac {5 a \tan ^3(c+d x)}{6 d}-\frac {5 a \tan (c+d x)}{2 d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {2 a \sec (c+d x)}{d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac {5 a x}{2} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2838, 2590, 270, 2591, 288, 302, 203} \[ -\frac {a \cos (c+d x)}{d}+\frac {5 a \tan ^3(c+d x)}{6 d}-\frac {5 a \tan (c+d x)}{2 d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {2 a \sec (c+d x)}{d}-\frac {a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac {5 a x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 76, normalized size = 0.75 \[ \frac {a \left (-3 \sin (2 (c+d x))-12 \cos (c+d x)-28 \tan (c+d x)+4 \sec ^3(c+d x)-24 \sec (c+d x)+4 \tan (c+d x) \sec ^2(c+d x)+30 c+30 d x\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(30*c + 30*d*x - 12*Cos[c + d*x] - 24*Sec[c + d*x] + 4*Sec[c + d*x]^3 - 3*Sin[2*(c + d*x)] - 28*Tan[c + d*x
] + 4*Sec[c + d*x]^2*Tan[c + d*x]))/(12*d)

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fricas [A]  time = 0.46, size = 98, normalized size = 0.97 \[ \frac {3 \, a \cos \left (d x + c\right )^{4} - 15 \, a d x \cos \left (d x + c\right ) + 17 \, a \cos \left (d x + c\right )^{2} + {\left (15 \, a d x \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right ) - 4 \, a}{6 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(3*a*cos(d*x + c)^4 - 15*a*d*x*cos(d*x + c) + 17*a*cos(d*x + c)^2 + (15*a*d*x*cos(d*x + c) - 3*a*cos(d*x +
c)^2 + 2*a)*sin(d*x + c) - 4*a)/(d*cos(d*x + c)*sin(d*x + c) - d*cos(d*x + c))

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giac [A]  time = 0.20, size = 134, normalized size = 1.33 \[ \frac {15 \, {\left (d x + c\right )} a + \frac {3 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {6 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} + \frac {21 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 48 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 23 \, a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/6*(15*(d*x + c)*a + 3*a/(tan(1/2*d*x + 1/2*c) + 1) + 6*(a*tan(1/2*d*x + 1/2*c)^3 - 2*a*tan(1/2*d*x + 1/2*c)^
2 - a*tan(1/2*d*x + 1/2*c) - 2*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 + (21*a*tan(1/2*d*x + 1/2*c)^2 - 48*a*tan(1/2
*d*x + 1/2*c) + 23*a)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

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maple [A]  time = 0.46, size = 154, normalized size = 1.52 \[ \frac {a \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+a \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^5*(a+a*sin(d*x+c)),x)

[Out]

1/d*(a*(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)+a*(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)))

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maxima [A]  time = 0.42, size = 87, normalized size = 0.86 \[ \frac {{\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac {3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a - 2 \, a {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/6*((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))*a - 2*a*((6*co
s(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c)))/d

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mupad [B]  time = 14.62, size = 243, normalized size = 2.41 \[ \frac {5\,a\,x}{2}-\frac {\left (\frac {a\,\left (30\,d\,x-30\right )}{6}-5\,a\,d\,x\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (5\,a\,d\,x-\frac {a\,\left (30\,d\,x-60\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {a\,\left (30\,d\,x-50\right )}{6}-5\,a\,d\,x\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\left (5\,a\,d\,x-\frac {a\,\left (30\,d\,x-14\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {a\,\left (30\,d\,x-4\right )}{6}-5\,a\,d\,x\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (5\,a\,d\,x-\frac {a\,\left (30\,d\,x-34\right )}{6}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,\left (15\,d\,x-32\right )}{6}-\frac {5\,a\,d\,x}{2}}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^5*(a + a*sin(c + d*x)))/cos(c + d*x)^4,x)

[Out]

(5*a*x)/2 - ((a*(15*d*x - 32))/6 - tan(c/2 + (d*x)/2)*((a*(30*d*x - 34))/6 - 5*a*d*x) + tan(c/2 + (d*x)/2)^2*(
(a*(30*d*x - 4))/6 - 5*a*d*x) - tan(c/2 + (d*x)/2)^3*((a*(30*d*x - 14))/6 - 5*a*d*x) + tan(c/2 + (d*x)/2)^7*((
a*(30*d*x - 30))/6 - 5*a*d*x) + tan(c/2 + (d*x)/2)^5*((a*(30*d*x - 50))/6 - 5*a*d*x) - tan(c/2 + (d*x)/2)^6*((
a*(30*d*x - 60))/6 - 5*a*d*x) + (20*a*tan(c/2 + (d*x)/2)^4)/3 - (5*a*d*x)/2)/(d*(tan(c/2 + (d*x)/2) - 1)^3*(ta
n(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d*x)/2)^2 + 1)^2)

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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)**4*sin(d*x+c)**5*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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