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

Optimal. Leaf size=112 \[ -\frac{2^{n+\frac{1}{2}} \left (\frac{1}{\sec (c+d x)+1}\right )^{n-\frac{1}{2}} (a \sec (c+d x)+a)^n F_1\left (-\frac{1}{4};n-\frac{3}{2},1;\frac{3}{4};-\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d \sqrt{\tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0673982, antiderivative size = 112, normalized size of antiderivative = 1., 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} \[ -\frac{2^{n+\frac{1}{2}} \left (\frac{1}{\sec (c+d x)+1}\right )^{n-\frac{1}{2}} (a \sec (c+d x)+a)^n F_1\left (-\frac{1}{4};n-\frac{3}{2},1;\frac{3}{4};-\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 15.55, size = 2164, normalized size = 19.32 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(2^(1/2 + n)*Cot[c + d*x]^2*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^n*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(a*(1 + S
ec[c + d*x]))^n*(21*Hypergeometric2F1[-1/4, -1/2 + n, 3/4, Tan[(c + d*x)/2]^2] + 7*AppellF1[3/4, -1/2 + n, 1,
7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2 + 7*Hypergeometric2F1[3/4, -1/2 + n, 7/4, Tan
[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2 - 3*AppellF1[7/4, -1/2 + n, 1, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2]*Tan[(c + d*x)/2]^4))/(21*d*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*((2^(-1/2 + n)*(Cos[c + d*x]*Sec[(c + d*x
)/2]^2)^n*Sec[c + d*x]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(21*Hypergeometric2F1[-1/4, -1/2 + n, 3/4, Tan[(c
+ d*x)/2]^2] + 7*AppellF1[3/4, -1/2 + n, 1, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2
+ 7*Hypergeometric2F1[3/4, -1/2 + n, 7/4, Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2 - 3*AppellF1[7/4, -1/2 + n, 1
, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^4))/(21*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])
]*Tan[c + d*x]^(3/2)) + (2^(-1/2 + n)*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^n*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*
((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x]))*(21*Hypergeometric2F1[-1/
4, -1/2 + n, 3/4, Tan[(c + d*x)/2]^2] + 7*AppellF1[3/4, -1/2 + n, 1, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2
]^2]*Tan[(c + d*x)/2]^2 + 7*Hypergeometric2F1[3/4, -1/2 + n, 7/4, Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2 - 3*A
ppellF1[7/4, -1/2 + n, 1, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^4))/(21*(Cos[c + d*x
]/(1 + Cos[c + d*x]))^(3/2)*Sqrt[Tan[c + d*x]]) - (2^(1/2 + n)*n*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^n*(Cos[(c +
d*x)/2]^2*Sec[c + d*x])^(1 + n)*(-(Sec[(c + d*x)/2]^2*Sin[c + d*x]) + Cos[c + d*x]*Sec[(c + d*x)/2]^2*Tan[(c
+ d*x)/2])*(21*Hypergeometric2F1[-1/4, -1/2 + n, 3/4, Tan[(c + d*x)/2]^2] + 7*AppellF1[3/4, -1/2 + n, 1, 7/4,
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2 + 7*Hypergeometric2F1[3/4, -1/2 + n, 7/4, Tan[(c +
d*x)/2]^2]*Tan[(c + d*x)/2]^2 - 3*AppellF1[7/4, -1/2 + n, 1, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*T
an[(c + d*x)/2]^4))/(21*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[Tan[c + d*x]]) - (2^(1/2 + n)*(Cos[c + d*x]
*Sec[(c + d*x)/2]^2)^n*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(7*AppellF1[3/4, -1/2 + n, 1, 7/4, Tan[(c + d*x)/2]
^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2] + 7*Hypergeometric2F1[3/4, -1/2 + n, 7/4, Tan[(c
+ d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2] - 6*AppellF1[7/4, -1/2 + n, 1, 11/4, Tan[(c + d*x)/2]^2, -Tan
[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^3 + 7*Tan[(c + d*x)/2]^2*((-3*AppellF1[7/4, -1/2 + n, 2,
11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/7 + (3*(-1/2 + n)*AppellF1
[7/4, 1/2 + n, 1, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/7) - 3*T
an[(c + d*x)/2]^4*((-7*AppellF1[11/4, -1/2 + n, 2, 15/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x
)/2]^2*Tan[(c + d*x)/2])/11 + (7*(-1/2 + n)*AppellF1[11/4, 1/2 + n, 1, 15/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x
)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/11) + (21*Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(Hypergeometric2F1[-1
/4, -1/2 + n, 3/4, Tan[(c + d*x)/2]^2] - (1 - Tan[(c + d*x)/2]^2)^(1/2 - n)))/4 + (21*Sec[(c + d*x)/2]^2*Tan[(
c + d*x)/2]*(-Hypergeometric2F1[3/4, -1/2 + n, 7/4, Tan[(c + d*x)/2]^2] + (1 - Tan[(c + d*x)/2]^2)^(1/2 - n)))
/4))/(21*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[Tan[c + d*x]]) - (2^(1/2 + n)*n*(Cos[c + d*x]*Sec[(c + d*x
)/2]^2)^n*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(-1 + n)*(21*Hypergeometric2F1[-1/4, -1/2 + n, 3/4, Tan[(c + d*x)/
2]^2] + 7*AppellF1[3/4, -1/2 + n, 1, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2 + 7*Hype
rgeometric2F1[3/4, -1/2 + n, 7/4, Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2 - 3*AppellF1[7/4, -1/2 + n, 1, 11/4,
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^4)*(-(Cos[(c + d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]
) + Cos[(c + d*x)/2]^2*Sec[c + d*x]*Tan[c + d*x]))/(21*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[Tan[c + d*x]
])))

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Maple [F]  time = 0.256, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n} \left ( \tan \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/tan(d*x+c)^(3/2),x)

[Out]

int((a+a*sec(d*x+c))^n/tan(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}}{\tan \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/tan(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n/tan(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}}{\tan \left (d x + c\right )^{\frac{3}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*sec(d*x + c) + a)^n/tan(d*x + c)^(3/2), 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/tan(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}}{\tan \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/tan(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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