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

Optimal. Leaf size=111 \[ \frac {2^{n+\frac {3}{2}} \sqrt {\tan (c+d x)} \left (\frac {1}{\sec (c+d x)+1}\right )^{n+\frac {1}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac {1}{4};n-\frac {1}{2},1;\frac {5}{4};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d} \]

[Out]

2^(3/2+n)*AppellF1(1/4,-1/2+n,1,5/4,(-a+a*sec(d*x+c))/(a+a*sec(d*x+c)),(a-a*sec(d*x+c))/(a+a*sec(d*x+c)))*(1/(
1+sec(d*x+c)))^(1/2+n)*(a+a*sec(d*x+c))^n*tan(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.06, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {3889} \[ \frac {2^{n+\frac {3}{2}} \sqrt {\tan (c+d x)} \left (\frac {1}{\sec (c+d x)+1}\right )^{n+\frac {1}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac {1}{4};n-\frac {1}{2},1;\frac {5}{4};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3889

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(2^(m
+ n + 1)*(e*Cot[c + d*x])^(m + 1)*(a + b*Csc[c + d*x])^n*(a/(a + b*Csc[c + d*x]))^(m + n + 1)*AppellF1[(m + 1
)/2, m + n, 1, (m + 3)/2, -((a - b*Csc[c + d*x])/(a + b*Csc[c + d*x])), (a - b*Csc[c + d*x])/(a + b*Csc[c + d*
x])])/(d*e*(m + 1)), x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rubi steps

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

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Mathematica [B]  time = 1.61, size = 229, normalized size = 2.06 \[ \frac {10 \cos (c+d x) (\cos (c+d x)+1) \sqrt {\tan (c+d x)} (a (\sec (c+d x)+1))^n F_1\left (\frac {1}{4};n-\frac {1}{2},1;\frac {5}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{d \left (2 (\cos (c+d x)-1) \left (2 F_1\left (\frac {5}{4};n-\frac {1}{2},2;\frac {9}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+(1-2 n) F_1\left (\frac {5}{4};n+\frac {1}{2},1;\frac {9}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )+5 (\cos (c+d x)+1) F_1\left (\frac {1}{4};n-\frac {1}{2},1;\frac {5}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(10*AppellF1[1/4, -1/2 + n, 1, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*x]*(1 + Cos[c + d*x])*(
a*(1 + Sec[c + d*x]))^n*Sqrt[Tan[c + d*x]])/(d*(2*(2*AppellF1[5/4, -1/2 + n, 2, 9/4, Tan[(c + d*x)/2]^2, -Tan[
(c + d*x)/2]^2] + (1 - 2*n)*AppellF1[5/4, 1/2 + n, 1, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos
[c + d*x]) + 5*AppellF1[1/4, -1/2 + n, 1, 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 = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\tan \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.54, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sec \left (d x +c \right )\right )^{n}}{\sqrt {\tan \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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