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

Optimal. Leaf size=276 \[ -\frac{\left (\sqrt{-b^2} \left (a^2+b^2 (n+1)\right )+a b^2 n\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{4 b d (n+1) \left (a^2+b^2\right ) \left (a-\sqrt{-b^2}\right )}-\frac{\left (a b^2 n-\sqrt{-b^2} \left (a^2+b^2 (n+1)\right )\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{4 b d (n+1) \left (a^2+b^2\right ) \left (a+\sqrt{-b^2}\right )}-\frac{\cos ^2(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{2 d \left (a^2+b^2\right )} \]

[Out]

-((a*b^2*n + Sqrt[-b^2]*(a^2 + b^2*(1 + n)))*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt
[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(4*b*(a^2 + b^2)*(a - Sqrt[-b^2])*d*(1 + n)) - ((a*b^2*n - Sqrt[-b^2]*(
a^2 + b^2*(1 + n)))*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*(a + b*Tan[c + d
*x])^(1 + n))/(4*b*(a^2 + b^2)*(a + Sqrt[-b^2])*d*(1 + n)) - (Cos[c + d*x]^2*(b + a*Tan[c + d*x])*(a + b*Tan[c
+ d*x])^(1 + n))/(2*(a^2 + b^2)*d)

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Rubi [A]  time = 0.367371, antiderivative size = 276, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3516, 1649, 831, 68} \[ -\frac{\left (\sqrt{-b^2} \left (a^2+b^2 (n+1)\right )+a b^2 n\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{4 b d (n+1) \left (a^2+b^2\right ) \left (a-\sqrt{-b^2}\right )}-\frac{\left (a b^2 n-\sqrt{-b^2} \left (a^2+b^2 (n+1)\right )\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{4 b d (n+1) \left (a^2+b^2\right ) \left (a+\sqrt{-b^2}\right )}-\frac{\cos ^2(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*b^2*n + Sqrt[-b^2]*(a^2 + b^2*(1 + n)))*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt
[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(4*b*(a^2 + b^2)*(a - Sqrt[-b^2])*d*(1 + n)) - ((a*b^2*n - Sqrt[-b^2]*(
a^2 + b^2*(1 + n)))*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*(a + b*Tan[c + d
*x])^(1 + n))/(4*b*(a^2 + b^2)*(a + Sqrt[-b^2])*d*(1 + n)) - (Cos[c + d*x]^2*(b + a*Tan[c + d*x])*(a + b*Tan[c
+ d*x])^(1 + n))/(2*(a^2 + b^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 1649

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 + 1)*(a + c*x^2)^(p + 1)*(a*(e*f - d*g) + (c*d*f + a*e*g)*x))/(2*a*(p +
1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSu
m[2*a*(p + 1)*(c*d^2 + a*e^2)*Q + c*d^2*f*(2*p + 3) - a*e*(d*g*m - e*f*(m + 2*p + 3)) + e*(c*d*f + a*e*g)*(m +
2*p + 4)*x, x], x], x]] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] &
&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 831

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \sin ^2(c+d x) (a+b \tan (c+d x))^n \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{x^2 (a+x)^n}{\left (b^2+x^2\right )^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac{\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}-\frac{\operatorname{Subst}\left (\int \frac{(a+x)^n \left (-b^2 \left (a^2+b^2 (1+n)\right )-a b^2 n x\right )}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 b \left (a^2+b^2\right ) d}\\ &=-\frac{\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}-\frac{\operatorname{Subst}\left (\int \left (\frac{\left (a b^4 n-b^2 \sqrt{-b^2} \left (a^2+b^2 (1+n)\right )\right ) (a+x)^n}{2 b^2 \left (\sqrt{-b^2}-x\right )}+\frac{\left (-a b^4 n-b^2 \sqrt{-b^2} \left (a^2+b^2 (1+n)\right )\right ) (a+x)^n}{2 b^2 \left (\sqrt{-b^2}+x\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{2 b \left (a^2+b^2\right ) d}\\ &=-\frac{\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}-\frac{\left (a b^2 n-\sqrt{-b^2} \left (a^2+b^2 (1+n)\right )\right ) \operatorname{Subst}\left (\int \frac{(a+x)^n}{\sqrt{-b^2}-x} \, dx,x,b \tan (c+d x)\right )}{4 b \left (a^2+b^2\right ) d}+\frac{\left (a b^2 n+\sqrt{-b^2} \left (a^2+b^2 (1+n)\right )\right ) \operatorname{Subst}\left (\int \frac{(a+x)^n}{\sqrt{-b^2}+x} \, dx,x,b \tan (c+d x)\right )}{4 b \left (a^2+b^2\right ) d}\\ &=-\frac{\left (a b^2 n+\sqrt{-b^2} \left (a^2+b^2 (1+n)\right )\right ) \, _2F_1\left (1,1+n;2+n;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{4 b \left (a^2+b^2\right ) \left (a-\sqrt{-b^2}\right ) d (1+n)}-\frac{\left (a b^2 n-\sqrt{-b^2} \left (a^2+b^2 (1+n)\right )\right ) \, _2F_1\left (1,1+n;2+n;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{4 b \left (a^2+b^2\right ) \left (a+\sqrt{-b^2}\right ) d (1+n)}-\frac{\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 1.08754, size = 270, normalized size = 0.98 \[ \frac{(a+b \tan (c+d x))^{n+1} \left (\left (a^2 b^2 (n-1)+a^3 \sqrt{-b^2}-a \left (-b^2\right )^{3/2} (2 n+1)+b^4 (-(n+1))\right ) \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )-\left (-a^2 b^2 (n-1)+a^3 \sqrt{-b^2}-a \left (-b^2\right )^{3/2} (2 n+1)+b^4 (n+1)\right ) \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )+2 b (n+1) \left (a^2+b^2\right ) \cos (c+d x) (a \sin (c+d x)+b \cos (c+d x))\right )}{4 b d (n+1) \left (a^2+b^2\right ) \left (\sqrt{-b^2}-a\right ) \left (a+\sqrt{-b^2}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((a^3*Sqrt[-b^2] + a^2*b^2*(-1 + n) - b^4*(1 + n) - a*(-b^2)^(3/2)*(1 + 2*n))*Hypergeometric2F1[1, 1 + n, 2 +
n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2])] - (a^3*Sqrt[-b^2] - a^2*b^2*(-1 + n) + b^4*(1 + n) - a*(-b^2)^(3/2)
*(1 + 2*n))*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])] + 2*b*(a^2 + b^2)*(1 + n
)*Cos[c + d*x]*(b*Cos[c + d*x] + a*Sin[c + d*x]))*(a + b*Tan[c + d*x])^(1 + n))/(4*b*(a^2 + b^2)*(-a + Sqrt[-b
^2])*(a + Sqrt[-b^2])*d*(1 + n))

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Maple [F]  time = 0.473, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{n}\, dx \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))^n,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2}\,{d x} \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))^n,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (\cos \left (d x + c\right )^{2} - 1\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{n}, x\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))^n,x, algorithm="fricas")

[Out]

integral(-(cos(d*x + c)^2 - 1)*(b*tan(d*x + c) + a)^n, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**n,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2}\,{d x} \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))^n,x, algorithm="giac")

[Out]

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