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

Optimal. Leaf size=76 \[ \frac{1}{2} x \left (a^2-3 b^2\right )-\frac{2 a b \log (\cos (c+d x))}{d}-\frac{\sin (c+d x) \cos (c+d x) (a+b \tan (c+d x))^2}{2 d}+\frac{3 b^2 \tan (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.113079, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3516, 1645, 774, 635, 203, 260} \[ \frac{1}{2} x \left (a^2-3 b^2\right )-\frac{2 a b \log (\cos (c+d x))}{d}-\frac{\sin (c+d x) \cos (c+d x) (a+b \tan (c+d x))^2}{2 d}+\frac{3 b^2 \tan (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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/
2]

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 774

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

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 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]

Rubi steps

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

Mathematica [B]  time = 2.36223, size = 162, normalized size = 2.13 \[ \frac{b \left (\frac{\left (b^2-a^2\right ) \sin (2 (c+d x))}{2 b}+\frac{\left (b^2-a^2\right ) \tan ^{-1}(\tan (c+d x))}{b}+\left (\frac{a^2-2 b^2}{\sqrt{-b^2}}+2 a\right ) \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+\left (\frac{2 b^2-a^2}{\sqrt{-b^2}}+2 a\right ) \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )+2 a \cos ^2(c+d x)+2 b \tan (c+d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(((-a^2 + b^2)*ArcTan[Tan[c + d*x]])/b + 2*a*Cos[c + d*x]^2 + (2*a + (a^2 - 2*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^
2] - b*Tan[c + d*x]] + (2*a + (-a^2 + 2*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + ((-a^2 + b^2)*Sin[
2*(c + d*x)])/(2*b) + 2*b*Tan[c + d*x]))/(2*d)

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Maple [B]  time = 0.039, size = 145, normalized size = 1.9 \begin{align*}{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{d\cos \left ( dx+c \right ) }}+{\frac{{b}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{3\,{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,{b}^{2}x}{2}}-{\frac{3\,{b}^{2}c}{2\,d}}-{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}-2\,{\frac{ab\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}x}{2}}+{\frac{{a}^{2}c}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.47601, size = 111, normalized size = 1.46 \begin{align*} \frac{2 \, a b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, b^{2} \tan \left (d x + c\right ) +{\left (a^{2} - 3 \, b^{2}\right )}{\left (d x + c\right )} + \frac{2 \, a b -{\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.01632, size = 240, normalized size = 3.16 \begin{align*} \frac{2 \, a b \cos \left (d x + c\right )^{3} - 4 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) +{\left ({\left (a^{2} - 3 \, b^{2}\right )} d x - a b\right )} \cos \left (d x + c\right ) -{\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{2} \sin ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.87871, size = 1432, normalized size = 18.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*d*x*tan(d*x)^3*tan(c)^3 - 3*b^2*d*x*tan(d*x)^3*tan(c)^3 - 2*a*b*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(
c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^3*tan(c)^3 +
a^2*d*x*tan(d*x)^3*tan(c) - 3*b^2*d*x*tan(d*x)^3*tan(c) - a^2*d*x*tan(d*x)^2*tan(c)^2 + 3*b^2*d*x*tan(d*x)^2*t
an(c)^2 + a^2*d*x*tan(d*x)*tan(c)^3 - 3*b^2*d*x*tan(d*x)*tan(c)^3 + a*b*tan(d*x)^3*tan(c)^3 - 2*a*b*log(4*(tan
(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) +
1))*tan(d*x)^3*tan(c) + 2*a*b*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*ta
n(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^2*tan(c)^2 + a^2*tan(d*x)^3*tan(c)^2 - 3*b^2*tan(d*x)^3
*tan(c)^2 - 2*a*b*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(
d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)*tan(c)^3 + a^2*tan(d*x)^2*tan(c)^3 - 3*b^2*tan(d*x)^2*tan(c)^3 - a^2
*d*x*tan(d*x)^2 + 3*b^2*d*x*tan(d*x)^2 + a^2*d*x*tan(d*x)*tan(c) - 3*b^2*d*x*tan(d*x)*tan(c) - a*b*tan(d*x)^3*
tan(c) - a^2*d*x*tan(c)^2 + 3*b^2*d*x*tan(c)^2 - 5*a*b*tan(d*x)^2*tan(c)^2 - a*b*tan(d*x)*tan(c)^3 + 2*a*b*log
(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*t
an(c) + 1))*tan(d*x)^2 - 2*b^2*tan(d*x)^3 - 2*a*b*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan
(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)*tan(c) - 2*a^2*tan(d*x)^2*tan(c) + 2
*a*b*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*ta
n(d*x)*tan(c) + 1))*tan(c)^2 - 2*a^2*tan(d*x)*tan(c)^2 - 2*b^2*tan(c)^3 - a^2*d*x + 3*b^2*d*x + a*b*tan(d*x)^2
+ 5*a*b*tan(d*x)*tan(c) + a*b*tan(c)^2 + 2*a*b*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c
) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)) + a^2*tan(d*x) - 3*b^2*tan(d*x) + a^2*tan(c) -
3*b^2*tan(c) - a*b)/(d*tan(d*x)^3*tan(c)^3 + d*tan(d*x)^3*tan(c) - d*tan(d*x)^2*tan(c)^2 + d*tan(d*x)*tan(c)^3
- d*tan(d*x)^2 + d*tan(d*x)*tan(c) - d*tan(c)^2 - d)