3.353 \(\int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=83 \[ \frac{2^{m-\frac{3}{2}} \sec ^3(c+d x) (\sin (c+d x)+1)^{\frac{1}{2}-m} (a \sin (c+d x)+a)^{m+1} \, _2F_1\left (-\frac{3}{2},\frac{5}{2}-m;-\frac{1}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{3 a d} \]

[Out]

(2^(-3/2 + m)*Hypergeometric2F1[-3/2, 5/2 - m, -1/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^3*(1 + Sin[c + d*x])^(
1/2 - m)*(a + a*Sin[c + d*x])^(1 + m))/(3*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.0841911, antiderivative size = 83, 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 = {2689, 70, 69} \[ \frac{2^{m-\frac{3}{2}} \sec ^3(c+d x) (\sin (c+d x)+1)^{\frac{1}{2}-m} (a \sin (c+d x)+a)^{m+1} \, _2F_1\left (-\frac{3}{2},\frac{5}{2}-m;-\frac{1}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{3 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^m,x]

[Out]

(2^(-3/2 + m)*Hypergeometric2F1[-3/2, 5/2 - m, -1/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^3*(1 + Sin[c + d*x])^(
1/2 - m)*(a + a*Sin[c + d*x])^(1 + m))/(3*a*d)

Rule 2689

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 70

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

Rule 69

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

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx &=\frac{\left (a^2 \sec ^3(c+d x) (a-a \sin (c+d x))^{3/2} (a+a \sin (c+d x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{5}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{\left (2^{-\frac{5}{2}+m} \sec ^3(c+d x) (a-a \sin (c+d x))^{3/2} (a+a \sin (c+d x))^{1+m} \left (\frac{a+a \sin (c+d x)}{a}\right )^{\frac{1}{2}-m}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{x}{2}\right )^{-\frac{5}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{2^{-\frac{3}{2}+m} \, _2F_1\left (-\frac{3}{2},\frac{5}{2}-m;-\frac{1}{2};\frac{1}{2} (1-\sin (c+d x))\right ) \sec ^3(c+d x) (1+\sin (c+d x))^{\frac{1}{2}-m} (a+a \sin (c+d x))^{1+m}}{3 a d}\\ \end{align*}

Mathematica [C]  time = 13.1039, size = 307, normalized size = 3.7 \[ -\frac{4 \cos ^2\left (\frac{1}{8} (2 c+2 d x-\pi )\right ) \cot \left (\frac{1}{8} (2 c+2 d x-\pi )\right ) \sec ^4(c+d x) (a (\sin (c+d x)+1))^m F_1\left (-\frac{3}{2};4-2 m,2 m-7;-\frac{1}{2};\tan ^2\left (\frac{1}{8} (-2 c-2 d x+\pi )\right ),-\tan ^2\left (\frac{1}{8} (2 c+2 d x-\pi )\right )\right )}{3 d \left (2 (2 m-7) F_1\left (-\frac{1}{2};4-2 m,2 (m-3);\frac{1}{2};\tan ^2\left (\frac{1}{8} (-2 c-2 d x+\pi )\right ),-\tan ^2\left (\frac{1}{8} (2 c+2 d x-\pi )\right )\right )+4 (m-2) F_1\left (-\frac{1}{2};5-2 m,2 m-7;\frac{1}{2};\tan ^2\left (\frac{1}{8} (-2 c-2 d x+\pi )\right ),-\tan ^2\left (\frac{1}{8} (2 c+2 d x-\pi )\right )\right )+\cot ^2\left (\frac{1}{8} (2 c+2 d x-\pi )\right ) F_1\left (-\frac{3}{2};4-2 m,2 m-7;-\frac{1}{2};\tan ^2\left (\frac{1}{8} (-2 c-2 d x+\pi )\right ),-\tan ^2\left (\frac{1}{8} (2 c+2 d x-\pi )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^m,x]

[Out]

(-4*AppellF1[-3/2, 4 - 2*m, -7 + 2*m, -1/2, Tan[(-2*c + Pi - 2*d*x)/8]^2, -Tan[(2*c - Pi + 2*d*x)/8]^2]*Cos[(2
*c - Pi + 2*d*x)/8]^2*Cot[(2*c - Pi + 2*d*x)/8]*Sec[c + d*x]^4*(a*(1 + Sin[c + d*x]))^m)/(3*d*(2*(-7 + 2*m)*Ap
pellF1[-1/2, 4 - 2*m, 2*(-3 + m), 1/2, Tan[(-2*c + Pi - 2*d*x)/8]^2, -Tan[(2*c - Pi + 2*d*x)/8]^2] + 4*(-2 + m
)*AppellF1[-1/2, 5 - 2*m, -7 + 2*m, 1/2, Tan[(-2*c + Pi - 2*d*x)/8]^2, -Tan[(2*c - Pi + 2*d*x)/8]^2] + AppellF
1[-3/2, 4 - 2*m, -7 + 2*m, -1/2, Tan[(-2*c + Pi - 2*d*x)/8]^2, -Tan[(2*c - Pi + 2*d*x)/8]^2]*Cot[(2*c - Pi + 2
*d*x)/8]^2))

________________________________________________________________________________________

Maple [F]  time = 0.103, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{4} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x)

[Out]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^m*sec(d*x + c)^4, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((a*sin(d*x + c) + a)^m*sec(d*x + c)^4, x)

________________________________________________________________________________________

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(d*x+c)**4*(a+a*sin(d*x+c))**m,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^m*sec(d*x + c)^4, x)