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(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/d/tan(d*x+c)^(1/2)
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Rubi [A] time = 0.07, antiderivative size = 112, 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} \[ -\frac {2^{n+\frac {1}{2}} \left (\frac {1}{\sec (c+d x)+1}\right )^{n-\frac {1}{2}} (a \sec (c+d x)+a)^n F_1\left (-\frac {1}{4};n-\frac {3}{2},1;\frac {3}{4};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d \sqrt {\tan (c+d x)}} \]
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*}
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Mathematica [B] time = 15.86, 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]
-1/21*(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 + 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))/(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
*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*(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]
*Tan[(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, -T
an[(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)*Appell
F1[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
*Tan[(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*Hy
pergeometric2F1[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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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\tan \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]
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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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\tan \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^n/tan(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate((a*sec(d*x + c) + a)^n/tan(d*x + c)^(3/2), x)
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maple [F] time = 1.44, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sec \left (d x +c \right )\right )^{n}}{\tan \left (d x +c \right )^{\frac {3}{2}}}\, dx \]
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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\tan \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^n/tan(d*x+c)^(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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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a/cos(c + d*x))^n/tan(c + d*x)^(3/2),x)
[Out]
int((a + a/cos(c + d*x))^n/tan(c + d*x)^(3/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))**n/tan(d*x+c)**(3/2),x)
[Out]
Integral((a*(sec(c + d*x) + 1))**n/tan(c + d*x)**(3/2), x)
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