3.158 \(\int \frac {(a+a \sec (c+d x))^n}{\sqrt {\sin (c+d x)}} \, dx\)

Optimal. Leaf size=105 \[ -\frac {(1-\cos (c+d x))^{3/4} \cos (c+d x) (\cos (c+d x)+1)^{\frac {3}{4}-n} (a \sec (c+d x)+a)^n F_1\left (1-n;\frac {3}{4},\frac {3}{4}-n;2-n;\cos (c+d x),-\cos (c+d x)\right )}{d (1-n) \sin ^{\frac {3}{2}}(c+d x)} \]

[Out]

-AppellF1(1-n,3/4-n,3/4,2-n,-cos(d*x+c),cos(d*x+c))*(1-cos(d*x+c))^(3/4)*cos(d*x+c)*(1+cos(d*x+c))^(3/4-n)*(a+
a*sec(d*x+c))^n/d/(1-n)/sin(d*x+c)^(3/2)

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Rubi [A]  time = 0.25, antiderivative size = 105, normalized size of antiderivative = 1.00, 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} \[ -\frac {(1-\cos (c+d x))^{3/4} \cos (c+d x) (\cos (c+d x)+1)^{\frac {3}{4}-n} (a \sec (c+d x)+a)^n F_1\left (1-n;\frac {3}{4},\frac {3}{4}-n;2-n;\cos (c+d x),-\cos (c+d x)\right )}{d (1-n) \sin ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 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])

Rubi steps

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

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Mathematica [B]  time = 1.04, size = 212, normalized size = 2.02 \[ \frac {10 \sqrt {\sin (c+d x)} (\cos (c+d x)+1) (a (\sec (c+d x)+1))^n F_1\left (\frac {1}{4};n,\frac {1}{2};\frac {5}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{d \left (2 (\cos (c+d x)-1) \left (F_1\left (\frac {5}{4};n,\frac {3}{2};\frac {9}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 n F_1\left (\frac {5}{4};n+1,\frac {1}{2};\frac {9}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )+5 (\cos (c+d x)+1) F_1\left (\frac {1}{4};n,\frac {1}{2};\frac {5}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(10*AppellF1[1/4, n, 1/2, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])*(a*(1 + Sec[c + d*x
]))^n*Sqrt[Sin[c + d*x]])/(d*(2*(AppellF1[5/4, n, 3/2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*n*App
ellF1[5/4, 1 + n, 1/2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + 5*AppellF1[1/4, n,
1/2, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])))

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\sin \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n/sin(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n/sqrt(sin(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\sin \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n/sin(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.84, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sec \left (d x +c \right )\right )^{n}}{\sqrt {\sin \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^n/sin(d*x+c)^(1/2),x)

[Out]

int((a+a*sec(d*x+c))^n/sin(d*x+c)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\sin \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n/sin(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n}{\sqrt {\sin \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^n/sin(c + d*x)^(1/2),x)

[Out]

int((a + a/cos(c + d*x))^n/sin(c + d*x)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n}}{\sqrt {\sin {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**n/sin(d*x+c)**(1/2),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**n/sqrt(sin(c + d*x)), x)

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