∫ sin4(c + dx)(a + b tan(c + dx))2 dx

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

Optimal. Leaf size=113 \[ \frac {3}{8} x \left (a^2-5 b^2\right )+\frac {\cos ^2(c+d x) (7 b-5 a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {\sin (c+d x) \cos ^3(c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac {b^2 \tan (c+d x)}{d} \]

[Out]

3/8*(a^2-5*b^2)*x-2*a*b*ln(cos(d*x+c))/d+b^2*tan(d*x+c)/d+1/8*cos(d*x+c)^2*(7*b-5*a*tan(d*x+c))*(a+b*tan(d*x+c

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

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Rubi [A] time = 0.19, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3516, 1645, 1810, 635, 203, 260} \[ \frac {3}{8} x \left (a^2-5 b^2\right )+\frac {\cos ^2(c+d x) (7 b-5 a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {\sin (c+d x) \cos ^3(c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac {b^2 \tan (c+d x)}{d} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*(a^2 - 5*b^2)*x)/8 - (2*a*b*Log[Cos[c + d*x]])/d + (b^2*Tan[c + d*x])/d + (Cos[c + d*x]^2*(7*b - 5*a*Tan[c

+ d*x])*(a + b*Tan[c + d*x]))/(8*d) + (Cos[c + d*x]^3*Sin[c + d*x]*(a + b*Tan[c + d*x])^2)/(4*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 260

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 635

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 1645

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,

a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c

*x^2, x], x, 1]}, Simp[((d + e*x)^m*(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] + Dist[1/(2*a*c*(p

+ 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*c*(p + 1)*(d + e*x)*Q - a*e*g*m + c*d*f*(2*p

+ 3) + c*e*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &

& LtQ[p, -1] && GtQ[m, 0] && !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,

b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int

[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/

Rubi steps

\begin {align*} \int \sin ^4(c+d x) (a+b \tan (c+d x))^2 \, dx &=\frac {b \operatorname {Subst}\left (\int \frac {x^4 (a+x)^2}{\left (b^2+x^2\right )^3} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^2}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {(a+x) \left (a b^4+3 b^4 x-4 a b^2 x^2-4 b^2 x^3\right )}{\left (b^2+x^2\right )^2} \, dx,x,b \tan (c+d x)\right )}{4 b d}\\ &=\frac {\cos ^2(c+d x) (7 b-5 a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {b^4 \left (3 a^2-7 b^2\right )+16 a b^4 x+8 b^4 x^2}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{8 b^3 d}\\ &=\frac {\cos ^2(c+d x) (7 b-5 a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac {\operatorname {Subst}\left (\int \left (8 b^4+\frac {3 b^4 \left (a^2-5 b^2\right )+16 a b^4 x}{b^2+x^2}\right ) \, dx,x,b \tan (c+d x)\right )}{8 b^3 d}\\ &=\frac {b^2 \tan (c+d x)}{d}+\frac {\cos ^2(c+d x) (7 b-5 a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {3 b^4 \left (a^2-5 b^2\right )+16 a b^4 x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{8 b^3 d}\\ &=\frac {b^2 \tan (c+d x)}{d}+\frac {\cos ^2(c+d x) (7 b-5 a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac {(2 a b) \operatorname {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{d}+\frac {\left (3 b \left (a^2-5 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{8 d}\\ &=\frac {3}{8} \left (a^2-5 b^2\right ) x-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d}+\frac {\cos ^2(c+d x) (7 b-5 a \tan (c+d x)) (a+b \tan (c+d x))}{8 d}+\frac {\cos ^3(c+d x) \sin (c+d x) (a+b \tan (c+d x))^2}{4 d}\\ \end {align*}

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Mathematica [B] time = 3.51, size = 240, normalized size = 2.12 \[ \frac {b \left (\frac {2 \left (3 b^2-2 a^2\right ) \sin (2 (c+d x))}{b}+\frac {4 \left (3 b^2-2 a^2\right ) \tan ^{-1}(\tan (c+d x))}{b}+4 \left (\frac {a^2-3 b^2}{\sqrt {-b^2}}+2 a\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+4 \left (\frac {3 b^2-a^2}{\sqrt {-b^2}}+2 a\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+\frac {3 \left (a^2-b^2\right ) \left (\sin (2 (c+d x))+2 \tan ^{-1}(\tan (c+d x))\right )}{2 b}+\frac {2 \left (a^2-b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b}-4 a \cos ^4(c+d x)+16 a \cos ^2(c+d x)+8 b \tan (c+d x)\right )}{8 d} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(b*((4*(-2*a^2 + 3*b^2)*ArcTan[Tan[c + d*x]])/b + 16*a*Cos[c + d*x]^2 - 4*a*Cos[c + d*x]^4 + 4*(2*a + (a^2 - 3

*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] + 4*(2*a + (-a^2 + 3*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Ta

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

2 - b^2)*(2*ArcTan[Tan[c + d*x]] + Sin[2*(c + d*x)]))/(2*b) + 8*b*Tan[c + d*x]))/(8*d)

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fricas [A] time = 0.55, size = 137, normalized size = 1.21 \[ -\frac {8 \, a b \cos \left (d x + c\right )^{5} - 32 \, a b \cos \left (d x + c\right )^{3} + 32 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - {\left (6 \, {\left (a^{2} - 5 \, b^{2}\right )} d x - 13 \, a b\right )} \cos \left (d x + c\right ) - 2 \, {\left (2 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} - {\left (5 \, a^{2} - 9 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(8*a*b*cos(d*x + c)^5 - 32*a*b*cos(d*x + c)^3 + 32*a*b*cos(d*x + c)*log(-cos(d*x + c)) - (6*(a^2 - 5*b^2

)*d*x - 13*a*b)*cos(d*x + c) - 2*(2*(a^2 - b^2)*cos(d*x + c)^4 - (5*a^2 - 9*b^2)*cos(d*x + c)^2 + 8*b^2)*sin(d

*x + c))/(d*cos(d*x + c))

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/