3.269 \(\int (a+b \sec (c+d x))^n \sin ^3(c+d x) \, dx\)

Optimal. Leaf size=121 \[ \frac{b \left (6 a^2-b^2 \left (n^2-3 n+2\right )\right ) (a+b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (2,n+1,n+2,\frac{b \sec (c+d x)}{a}+1\right )}{6 a^4 d (n+1)}+\frac{\cos ^3(c+d x) (2 a-b (2-n) \sec (c+d x)) (a+b \sec (c+d x))^{n+1}}{6 a^2 d} \]

[Out]

(b*(6*a^2 - b^2*(2 - 3*n + n^2))*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x
])^(1 + n))/(6*a^4*d*(1 + n)) + (Cos[c + d*x]^3*(a + b*Sec[c + d*x])^(1 + n)*(2*a - b*(2 - n)*Sec[c + d*x]))/(
6*a^2*d)

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Rubi [A]  time = 0.10479, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3874, 145, 65} \[ \frac{b \left (6 a^2-b^2 \left (n^2-3 n+2\right )\right ) (a+b \sec (c+d x))^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{b \sec (c+d x)}{a}+1\right )}{6 a^4 d (n+1)}+\frac{\cos ^3(c+d x) (2 a-b (2-n) \sec (c+d x)) (a+b \sec (c+d x))^{n+1}}{6 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^n*Sin[c + d*x]^3,x]

[Out]

(b*(6*a^2 - b^2*(2 - 3*n + n^2))*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x
])^(1 + n))/(6*a^4*d*(1 + n)) + (Cos[c + d*x]^3*(a + b*Sec[c + d*x])^(1 + n)*(2*a - b*(2 - n)*Sec[c + d*x]))/(
6*a^2*d)

Rule 3874

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Dist[f^(-1), Subs
t[Int[((-1 + x)^((p - 1)/2)*(1 + x)^((p - 1)/2)*(a + b*x)^m)/x^(p + 1), x], x, Csc[e + f*x]], x] /; FreeQ[{a,
b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 145

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), x] + Dist[(f*h)/b^2 - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int (a+b \sec (c+d x))^n \sin ^3(c+d x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(-1+x) (1+x) (a-b x)^n}{x^4} \, dx,x,-\sec (c+d x)\right )}{d}\\ &=\frac{\cos ^3(c+d x) (a+b \sec (c+d x))^{1+n} (2 a-b (2-n) \sec (c+d x))}{6 a^2 d}-\frac{\left (6-\frac{b^2 (1-n) (2-n)}{a^2}\right ) \operatorname{Subst}\left (\int \frac{(a-b x)^n}{x^2} \, dx,x,-\sec (c+d x)\right )}{6 d}\\ &=\frac{b \left (6 a^2-b^2 \left (2-3 n+n^2\right )\right ) \, _2F_1\left (2,1+n;2+n;1+\frac{b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{6 a^4 d (1+n)}+\frac{\cos ^3(c+d x) (a+b \sec (c+d x))^{1+n} (2 a-b (2-n) \sec (c+d x))}{6 a^2 d}\\ \end{align*}

Mathematica [A]  time = 1.70088, size = 155, normalized size = 1.28 \[ \frac{\cos (c+d x) (a+b \sec (c+d x))^n \left (-\frac{2 b \left (b^2 \left (n^2-3 n+2\right )-6 a^2\right ) \text{Hypergeometric2F1}\left (2,1-n,2-n,\frac{a \cos (c+d x)}{a \cos (c+d x)+b}\right )}{a (n-1)}-\frac{2 (2 a-b (n-2)) (a \cos (c+d x)+b)^2}{a}+8 \cos ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2\right )}{12 a d (a \cos (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^n*Sin[c + d*x]^3,x]

[Out]

(Cos[c + d*x]*((-2*(2*a - b*(-2 + n))*(b + a*Cos[c + d*x])^2)/a + 8*Cos[(c + d*x)/2]^2*(b + a*Cos[c + d*x])^2
- (2*b*(-6*a^2 + b^2*(2 - 3*n + n^2))*Hypergeometric2F1[2, 1 - n, 2 - n, (a*Cos[c + d*x])/(b + a*Cos[c + d*x])
])/(a*(-1 + n)))*(a + b*Sec[c + d*x])^n)/(12*a*d*(b + a*Cos[c + d*x]))

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Maple [F]  time = 0.635, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sec \left ( dx+c \right ) \right ) ^{n} \left ( \sin \left ( dx+c \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^n*sin(d*x+c)^3,x)

[Out]

int((a+b*sec(d*x+c))^n*sin(d*x+c)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^n*sin(d*x+c)^3,x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^n*sin(d*x + c)^3, 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 \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^n*sin(d*x+c)^3,x, algorithm="fricas")

[Out]

integral(-(cos(d*x + c)^2 - 1)*(b*sec(d*x + c) + a)^n*sin(d*x + c), 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((a+b*sec(d*x+c))**n*sin(d*x+c)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^n*sin(d*x+c)^3,x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^n*sin(d*x + c)^3, x)