3.344 \(\int \sec ^3(a+b x) (c \sin (a+b x))^m \, dx\)

Optimal. Leaf size=48 \[ \frac{(c \sin (a+b x))^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};\sin ^2(a+b x)\right )}{b c (m+1)} \]

[Out]

(Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, Sin[a + b*x]^2]*(c*Sin[a + b*x])^(1 + m))/(b*c*(1 + m))

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Rubi [A]  time = 0.046838, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2564, 364} \[ \frac{(c \sin (a+b x))^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};\sin ^2(a+b x)\right )}{b c (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, Sin[a + b*x]^2]*(c*Sin[a + b*x])^(1 + m))/(b*c*(1 + m))

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

Mathematica [A]  time = 0.0240761, size = 51, normalized size = 1.06 \[ \frac{\sin (a+b x) (c \sin (a+b x))^m \, _2F_1\left (2,\frac{m+1}{2};\frac{m+1}{2}+1;\sin ^2(a+b x)\right )}{b (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[2, (1 + m)/2, 1 + (1 + m)/2, Sin[a + b*x]^2]*Sin[a + b*x]*(c*Sin[a + b*x])^m)/(b*(1 + m))

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Maple [F]  time = 0.259, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( bx+a \right ) \right ) ^{3} \left ( c\sin \left ( bx+a \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (c \sin \left (b x + a\right )\right )^{m} \sec \left (b x + a\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*sin(b*x + a))^m*sec(b*x + a)^3, 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(sec(b*x+a)**3*(c*sin(b*x+a))**m,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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