\(\int \sin ^4(c+d x) (a+b \tan (c+d x)) \, dx\) [12]

Optimal result

Integrand size = 19, antiderivative size = 83 \[ \int \sin ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 a x}{8}-\frac {b \log (\cos (c+d x))}{d}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \tan (c+d x))}{4 d}-\frac {\cos (c+d x) \sin (c+d x) (3 a+4 b \tan (c+d x))}{8 d} \]

[Out]

3/8*a*x-b*ln(cos(d*x+c))/d-1/4*cos(d*x+c)*sin(d*x+c)^3*(a+b*tan(d*x+c))/d-1/8*cos(d*x+c)*sin(d*x+c)*(3*a+4*b*t
an(d*x+c))/d

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {833, 649, 209, 266} \[ \int \sin ^4(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \tan (c+d x))}{4 d}-\frac {\sin (c+d x) \cos (c+d x) (3 a+4 b \tan (c+d x))}{8 d}+\frac {3 a x}{8}-\frac {b \log (\cos (c+d x))}{d} \]

[In]

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

[Out]

(3*a*x)/8 - (b*Log[Cos[c + d*x]])/d - (Cos[c + d*x]*Sin[c + d*x]^3*(a + b*Tan[c + d*x]))/(4*d) - (Cos[c + d*x]
*Sin[c + d*x]*(3*a + 4*b*Tan[c + d*x]))/(8*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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
- 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4 (a+b x)}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \tan (c+d x))}{4 d}+\frac {\text {Subst}\left (\int \frac {x^2 (3 a+4 b x)}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 d} \\ & = -\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \tan (c+d x))}{4 d}-\frac {\cos (c+d x) \sin (c+d x) (3 a+4 b \tan (c+d x))}{8 d}+\frac {\text {Subst}\left (\int \frac {3 a+8 b x}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d} \\ & = -\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \tan (c+d x))}{4 d}-\frac {\cos (c+d x) \sin (c+d x) (3 a+4 b \tan (c+d x))}{8 d}+\frac {(3 a) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}+\frac {b \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {3 a x}{8}-\frac {b \log (\cos (c+d x))}{d}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \tan (c+d x))}{4 d}-\frac {\cos (c+d x) \sin (c+d x) (3 a+4 b \tan (c+d x))}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \sin ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 a (c+d x)}{8 d}-\frac {b \left (-\cos ^2(c+d x)+\frac {1}{4} \cos ^4(c+d x)+\log (\cos (c+d x))\right )}{d}-\frac {a \sin (2 (c+d x))}{4 d}+\frac {a \sin (4 (c+d x))}{32 d} \]

[In]

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

[Out]

(3*a*(c + d*x))/(8*d) - (b*(-Cos[c + d*x]^2 + Cos[c + d*x]^4/4 + Log[Cos[c + d*x]]))/d - (a*Sin[2*(c + d*x)])/
(4*d) + (a*Sin[4*(c + d*x)])/(32*d)

Maple [A] (verified)

Time = 1.73 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(73\)
default	\(\frac {a \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(73\)
risch	\(i b x +\frac {3 a x}{8}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} b}{16 d}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a}{8 d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} b}{16 d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a}{8 d}+\frac {2 i b c}{d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {b \cos \left (4 d x +4 c \right )}{32 d}+\frac {a \sin \left (4 d x +4 c \right )}{32 d}\)	\(129\)

[In]

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

[Out]

1/d*(a*(-1/4*(sin(d*x+c)^3+3/2*sin(d*x+c))*cos(d*x+c)+3/8*d*x+3/8*c)+b*(-1/4*sin(d*x+c)^4-1/2*sin(d*x+c)^2-ln(
cos(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.89 \[ \int \sin ^4(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {2 \, b \cos \left (d x + c\right )^{4} - 3 \, a d x - 8 \, b \cos \left (d x + c\right )^{2} + 8 \, b \log \left (-\cos \left (d x + c\right )\right ) - {\left (2 \, a \cos \left (d x + c\right )^{3} - 5 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

[In]

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

[Out]

-1/8*(2*b*cos(d*x + c)^4 - 3*a*d*x - 8*b*cos(d*x + c)^2 + 8*b*log(-cos(d*x + c)) - (2*a*cos(d*x + c)^3 - 5*a*c
os(d*x + c))*sin(d*x + c))/d

Sympy [F]

\[ \int \sin ^4(c+d x) (a+b \tan (c+d x)) \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right ) \sin ^{4}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05 \[ \int \sin ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 \, {\left (d x + c\right )} a + 4 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - \frac {5 \, a \tan \left (d x + c\right )^{3} - 8 \, b \tan \left (d x + c\right )^{2} + 3 \, a \tan \left (d x + c\right ) - 6 \, b}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \]

[In]

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

[Out]

1/8*(3*(d*x + c)*a + 4*b*log(tan(d*x + c)^2 + 1) - (5*a*tan(d*x + c)^3 - 8*b*tan(d*x + c)^2 + 3*a*tan(d*x + c)
- 6*b)/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 976 vs. \(2 (77) = 154\).

Time = 0.54 (sec) , antiderivative size = 976, normalized size of antiderivative = 11.76 \[ \int \sin ^4(c+d x) (a+b \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/32*(12*a*d*x*tan(d*x)^4*tan(c)^4 - 16*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(
c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4*tan(c)^4 + 24*a*d*x*tan(d*x)^4*tan(c)^2 + 24*a*d*x*tan(d*x)^2*ta
n(c)^4 + 11*b*tan(d*x)^4*tan(c)^4 - 32*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c
)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4*tan(c)^2 + 12*a*tan(d*x)^4*tan(c)^3 - 32*b*log(4*(tan(d*x)^2*tan(
c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^4 + 12*a*ta
n(d*x)^3*tan(c)^4 + 12*a*d*x*tan(d*x)^4 + 48*a*d*x*tan(d*x)^2*tan(c)^2 + 6*b*tan(d*x)^4*tan(c)^2 - 32*b*tan(d*
x)^3*tan(c)^3 + 12*a*d*x*tan(c)^4 + 6*b*tan(d*x)^2*tan(c)^4 - 16*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan
(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4 + 20*a*tan(d*x)^4*tan(c) - 64*b*log(4*(
tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan
(c)^2 + 24*a*tan(d*x)^3*tan(c)^2 + 24*a*tan(d*x)^2*tan(c)^3 - 16*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan
(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(c)^4 + 20*a*tan(d*x)*tan(c)^4 + 24*a*d*x*tan(d
*x)^2 - 13*b*tan(d*x)^4 - 64*b*tan(d*x)^3*tan(c) + 24*a*d*x*tan(c)^2 - 36*b*tan(d*x)^2*tan(c)^2 - 64*b*tan(d*x
)*tan(c)^3 - 13*b*tan(c)^4 - 32*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + t
an(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2 - 20*a*tan(d*x)^3 - 24*a*tan(d*x)^2*tan(c) - 32*b*log(4*(tan(d*x)^2*tan(
c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(c)^2 - 24*a*tan(d*x)*tan(
c)^2 - 20*a*tan(c)^3 + 12*a*d*x + 6*b*tan(d*x)^2 - 32*b*tan(d*x)*tan(c) + 6*b*tan(c)^2 - 16*b*log(4*(tan(d*x)^
2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1)) - 12*a*tan(d*x) - 12*a*
tan(c) + 11*b)/(d*tan(d*x)^4*tan(c)^4 + 2*d*tan(d*x)^4*tan(c)^2 + 2*d*tan(d*x)^2*tan(c)^4 + d*tan(d*x)^4 + 4*d
*tan(d*x)^2*tan(c)^2 + d*tan(c)^4 + 2*d*tan(d*x)^2 + 2*d*tan(c)^2 + d)

Mupad [B] (verification not implemented)

Time = 4.45 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.87 \[ \int \sin ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3\,a\,x}{8}+\frac {b\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2\,d}+\frac {3\,b}{4\,d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}-\frac {5\,a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{8\,d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}-\frac {3\,a\,\mathrm {tan}\left (c+d\,x\right )}{8\,d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )} \]

[In]

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

[Out]

(3*a*x)/8 + (b*log(tan(c + d*x)^2 + 1))/(2*d) + (3*b)/(4*d*(2*tan(c + d*x)^2 + tan(c + d*x)^4 + 1)) - (5*a*tan
(c + d*x)^3)/(8*d*(2*tan(c + d*x)^2 + tan(c + d*x)^4 + 1)) + (b*tan(c + d*x)^2)/(d*(2*tan(c + d*x)^2 + tan(c +
d*x)^4 + 1)) - (3*a*tan(c + d*x))/(8*d*(2*tan(c + d*x)^2 + tan(c + d*x)^4 + 1))