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

Optimal. Leaf size=114 \[ \frac {2^{n+\frac {7}{2}} \tan ^{\frac {5}{2}}(c+d x) \left (\frac {1}{\sec (c+d x)+1}\right )^{n+\frac {5}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac {5}{4};n+\frac {3}{2},1;\frac {9}{4};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{5 d} \]

[Out]

1/5*2^(7/2+n)*AppellF1(5/4,3/2+n,1,9/4,(-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)))^(5/2+n)*(a+a*sec(d*x+c))^n*tan(d*x+c)^(5/2)/d

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Rubi [A]  time = 0.06, antiderivative size = 114, 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} \[ \frac {2^{n+\frac {7}{2}} \tan ^{\frac {5}{2}}(c+d x) \left (\frac {1}{\sec (c+d x)+1}\right )^{n+\frac {5}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac {5}{4};n+\frac {3}{2},1;\frac {9}{4};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(7/2 + n)*AppellF1[5/4, 3/2 + n, 1, 9/4, -((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))^(5/2 + n)*(a + a*Sec[c + d*x])^n*Tan[c + d*x]^(5/2))/(5*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 (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x) \, dx &=\frac {2^{\frac {7}{2}+n} F_1\left (\frac {5}{4};\frac {3}{2}+n,1;\frac {9}{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 {5}{2}+n} (a+a \sec (c+d x))^n \tan ^{\frac {5}{2}}(c+d x)}{5 d}\\ \end {align*}

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Mathematica [B]  time = 19.02, size = 2072, normalized size = 18.18 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(1 + n)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(a*(1 + Sec[c + d*x]))^n*(-1 + Tan[(c + d*x)/2])^(-1/2 - n)*(-2
*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]*Sec[(c + d*x)/2]^2)^(1/
2 + n)*(-1 + Tan[(c + d*x)/2])^(1/2 + n) + (AppellF1[1/2, 1/2 + n, 3/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c + d
*x)/2]] + AppellF1[1/2, 3/2 + n, 1/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]])*(1 - Tan[(c + d*x)/2])^(1
/2 + n)*(-1 + Tan[(c + d*x)/2]^2)^(1/2 + n))*Tan[c + d*x]^2)/(d*((2^n*Sec[c + d*x]^2*(Cos[(c + d*x)/2]^2*Sec[c
+ d*x])^n*(-1 + Tan[(c + d*x)/2])^(-1/2 - n)*(-2*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]*Sec[(c + d*x)/2]^2)^(1/2 + n)*(-1 + Tan[(c + d*x)/2])^(1/2 + n) + (AppellF1[1/2, 1/2
+ n, 3/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]] + AppellF1[1/2, 3/2 + n, 1/2 + n, 3/2, Tan[(c + d*x)/
2], -Tan[(c + d*x)/2]])*(1 - Tan[(c + d*x)/2])^(1/2 + n)*(-1 + Tan[(c + d*x)/2]^2)^(1/2 + n)))/Sqrt[Tan[c + d*
x]] + 2^n*(-1/2 - n)*Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(-1 + Tan[(c + d*x)/2])^(-3/2 - n)
*(-2*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]*Sec[(c + d*x)/2]^2)
^(1/2 + n)*(-1 + Tan[(c + d*x)/2])^(1/2 + n) + (AppellF1[1/2, 1/2 + n, 3/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c
+ d*x)/2]] + AppellF1[1/2, 3/2 + n, 1/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]])*(1 - Tan[(c + d*x)/2]
)^(1/2 + n)*(-1 + Tan[(c + d*x)/2]^2)^(1/2 + n))*Sqrt[Tan[c + d*x]] + 2^(1 + n)*(Cos[(c + d*x)/2]^2*Sec[c + d*
x])^n*(-1 + Tan[(c + d*x)/2])^(-1/2 - n)*(-((1/2 + n)*AppellF1[1/4, 1/2 + n, 1, 5/4, Tan[(c + d*x)/2]^2, -Tan[
(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(1/2 + n)*(-1 + Tan[(c + d*x)/2])^(-1/2 +
n)) - 2*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(1/2 + n)*(-1 + Tan[(c + d*x)/2])^(1/2 + n)*(-1/5*(AppellF1[5/4, 1/
2 + n, 2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]) + ((1/2 + n)*Appe
llF1[5/4, 3/2 + n, 1, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5) -
2*(1/2 + n)*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]*Sec[(c + d*x
)/2]^2)^(-1/2 + n)*(-1 + Tan[(c + d*x)/2])^(1/2 + 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]) + (1/2 + n)*(AppellF1[1/2, 1/2 + n, 3/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c + d
*x)/2]] + AppellF1[1/2, 3/2 + n, 1/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]])*Sec[(c + d*x)/2]^2*(1 - T
an[(c + d*x)/2])^(1/2 + n)*Tan[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]^2)^(-1/2 + n) - ((1/2 + n)*(AppellF1[1/2, 1
/2 + n, 3/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]] + AppellF1[1/2, 3/2 + n, 1/2 + n, 3/2, Tan[(c + d*x
)/2], -Tan[(c + d*x)/2]])*Sec[(c + d*x)/2]^2*(1 - Tan[(c + d*x)/2])^(-1/2 + n)*(-1 + Tan[(c + d*x)/2]^2)^(1/2
+ n))/2 + (-1/6*((3/2 + n)*AppellF1[3/2, 1/2 + n, 5/2 + n, 5/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]]*Sec[(c +
d*x)/2]^2) + ((3/2 + n)*AppellF1[3/2, 5/2 + n, 1/2 + n, 5/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]]*Sec[(c + d*x
)/2]^2)/6)*(1 - Tan[(c + d*x)/2])^(1/2 + n)*(-1 + Tan[(c + d*x)/2]^2)^(1/2 + n))*Sqrt[Tan[c + d*x]] + 2^(1 + n
)*n*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(-1 + n)*(-1 + Tan[(c + d*x)/2])^(-1/2 - n)*(-2*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]*Sec[(c + d*x)/2]^2)^(1/2 + n)*(-1 + Tan[(c + d*x
)/2])^(1/2 + n) + (AppellF1[1/2, 1/2 + n, 3/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]] + AppellF1[1/2, 3
/2 + n, 1/2 + n, 3/2, Tan[(c + d*x)/2], -Tan[(c + d*x)/2]])*(1 - Tan[(c + d*x)/2])^(1/2 + n)*(-1 + Tan[(c + d*
x)/2]^2)^(1/2 + n))*Sqrt[Tan[c + d*x]]*(-(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])))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac {3}{2}}, x\right ) \]

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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

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 = 1.76, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{n} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^(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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

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 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(3/2)*(a + a/cos(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^(3/2)*(a + a/cos(c + d*x))^n, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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