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

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

[Out]

1/3*2^(5/2+n)*AppellF1(3/4,1/2+n,1,7/4,(-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)))^(3/2+n)*(a+a*sec(d*x+c))^n*tan(d*x+c)^(3/2)/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} \[ \frac {2^{n+\frac {5}{2}} \tan ^{\frac {3}{2}}(c+d x) \left (\frac {1}{\sec (c+d x)+1}\right )^{n+\frac {3}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac {3}{4};n+\frac {1}{2},1;\frac {7}{4};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [B]  time = 2.23, size = 238, normalized size = 2.09 \[ \frac {56 \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\tan (c+d x)} (a (\sec (c+d x)+1))^n F_1\left (\frac {3}{4};n+\frac {1}{2},1;\frac {7}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{d \left (6 (\cos (c+d x)-1) \left (2 F_1\left (\frac {7}{4};n+\frac {1}{2},2;\frac {11}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-(2 n+1) F_1\left (\frac {7}{4};n+\frac {3}{2},1;\frac {11}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )+21 (\cos (c+d x)+1) F_1\left (\frac {3}{4};n+\frac {1}{2},1;\frac {7}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(56*AppellF1[3/4, 1/2 + n, 1, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^3*(a*(1 + Sec[c +
d*x]))^n*Sin[(c + d*x)/2]*Sqrt[Tan[c + d*x]])/(d*(6*(2*AppellF1[7/4, 1/2 + n, 2, 11/4, Tan[(c + d*x)/2]^2, -T
an[(c + d*x)/2]^2] - (1 + 2*n)*AppellF1[7/4, 3/2 + n, 1, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 +
Cos[c + d*x]) + 21*AppellF1[3/4, 1/2 + n, 1, 7/4, 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 \[ {\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sqrt {\tan \left (d x + c\right )}, x\right ) \]

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

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 = 1.73, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{n} \left (\sqrt {\tan }\left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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