3.159 \(\int \frac{(a+a \sec (c+d x))^n}{\sin ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=105 \[ -\frac{(1-\cos (c+d x))^{5/4} \cos (c+d x) (\cos (c+d x)+1)^{\frac{5}{4}-n} (a \sec (c+d x)+a)^n F_1\left (1-n;\frac{5}{4},\frac{5}{4}-n;2-n;\cos (c+d x),-\cos (c+d x)\right )}{d (1-n) \sin ^{\frac{5}{2}}(c+d x)} \]

[Out]

-((AppellF1[1 - n, 5/4, 5/4 - n, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*(1 - Cos[c + d*x])^(5/4)*Cos[c + d*x]*(1
+ Cos[c + d*x])^(5/4 - n)*(a + a*Sec[c + d*x])^n)/(d*(1 - n)*Sin[c + d*x]^(5/2)))

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Rubi [A]  time = 0.266506, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3876, 2886, 135, 133} \[ -\frac{(1-\cos (c+d x))^{5/4} \cos (c+d x) (\cos (c+d x)+1)^{\frac{5}{4}-n} (a \sec (c+d x)+a)^n F_1\left (1-n;\frac{5}{4},\frac{5}{4}-n;2-n;\cos (c+d x),-\cos (c+d x)\right )}{d (1-n) \sin ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^n/Sin[c + d*x]^(3/2),x]

[Out]

-((AppellF1[1 - n, 5/4, 5/4 - n, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*(1 - Cos[c + d*x])^(5/4)*Cos[c + d*x]*(1
+ Cos[c + d*x])^(5/4 - n)*(a + a*Sec[c + d*x])^n)/(d*(1 - n)*Sin[c + d*x]^(5/2)))

Rule 3876

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

Rule 2886

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

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^n}{\sin ^{\frac{3}{2}}(c+d x)} \, dx &=\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int \frac{(-\cos (c+d x))^{-n} (-a-a \cos (c+d x))^n}{\sin ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{\frac{5}{4}-n} (-a+a \cos (c+d x))^{5/4} (a+a \sec (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{(-x)^{-n} (-a-a x)^{-\frac{5}{4}+n}}{(-a+a x)^{5/4}} \, dx,x,\cos (c+d x)\right )}{d \sin ^{\frac{5}{2}}(c+d x)}\\ &=-\left (-\frac{\left ((-\cos (c+d x))^n (1+\cos (c+d x))^{\frac{1}{4}-n} (-a-a \cos (c+d x)) (-a+a \cos (c+d x))^{5/4} (a+a \sec (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{(-x)^{-n} (1+x)^{-\frac{5}{4}+n}}{(-a+a x)^{5/4}} \, dx,x,\cos (c+d x)\right )}{a d \sin ^{\frac{5}{2}}(c+d x)}\right )\\ &=-\frac{\left (\sqrt [4]{1-\cos (c+d x)} (-\cos (c+d x))^n (1+\cos (c+d x))^{\frac{1}{4}-n} (-a-a \cos (c+d x)) (-a+a \cos (c+d x)) (a+a \sec (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{(-x)^{-n} (1+x)^{-\frac{5}{4}+n}}{(1-x)^{5/4}} \, dx,x,\cos (c+d x)\right )}{a^2 d \sin ^{\frac{5}{2}}(c+d x)}\\ &=-\frac{F_1\left (1-n;\frac{5}{4},\frac{5}{4}-n;2-n;\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{5/4} \cos (c+d x) (1+\cos (c+d x))^{\frac{5}{4}-n} (a+a \sec (c+d x))^n}{d (1-n) \sin ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [B]  time = 1.17057, size = 212, normalized size = 2.02 \[ -\frac{6 (\cos (c+d x)+1) (a (\sec (c+d x)+1))^n F_1\left (-\frac{1}{4};n,-\frac{1}{2};\frac{3}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}{d \sqrt{\sin (c+d x)} \left (3 (\cos (c+d x)+1) F_1\left (-\frac{1}{4};n,-\frac{1}{2};\frac{3}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-2 (\cos (c+d x)-1) \left (F_1\left (\frac{3}{4};n,\frac{1}{2};\frac{7}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+2 n F_1\left (\frac{3}{4};n+1,-\frac{1}{2};\frac{7}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[c + d*x])^n/Sin[c + d*x]^(3/2),x]

[Out]

(-6*AppellF1[-1/4, n, -1/2, 3/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])*(a*(1 + Sec[c + d
*x]))^n)/(d*(-2*(AppellF1[3/4, n, 1/2, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*n*AppellF1[3/4, 1 + n
, -1/2, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + 3*AppellF1[-1/4, n, -1/2, 3/4, Ta
n[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x]))*Sqrt[Sin[c + d*x]])

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Maple [F]  time = 0.16, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n} \left ( \sin \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^n/sin(d*x+c)^(3/2),x)

[Out]

int((a+a*sec(d*x+c))^n/sin(d*x+c)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sin \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n/sin(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n/sin(d*x + c)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sqrt{\sin \left (d x + c\right )}}{\cos \left (d x + c\right )^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n/sin(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sin \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n/sin(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n/sin(d*x + c)^(3/2), x)