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

Optimal. Leaf size=105 \[ -\frac{\sqrt{\sin (c+d x)} \cos (c+d x) (\cos (c+d x)+1)^{-n-\frac{1}{4}} (a \sec (c+d x)+a)^n F_1\left (1-n;-\frac{1}{4},-n-\frac{1}{4};2-n;\cos (c+d x),-\cos (c+d x)\right )}{d (1-n) \sqrt [4]{1-\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.261743, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3876, 2886, 135, 133} \[ -\frac{\sqrt{\sin (c+d x)} \cos (c+d x) (\cos (c+d x)+1)^{-n-\frac{1}{4}} (a \sec (c+d x)+a)^n F_1\left (1-n;-\frac{1}{4},-n-\frac{1}{4};2-n;\cos (c+d x),-\cos (c+d x)\right )}{d (1-n) \sqrt [4]{1-\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3876

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(Sin[
e + f*x]^FracPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(b + a*Sin[e + f*x])^FracPart[m], Int[((g*Cos[e + f*x])
^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && (EqQ[a^2 - b^2, 0] ||
IntegersQ[2*m, p])

Rule 2886

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(g*(g*Cos[e + f*x])^(p - 1))/(f*(a + b*Sin[e + f*x])^((p - 1)/2)*(a - b*Sin[e +
f*x])^((p - 1)/2)), Subst[Int[(d*x)^n*(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]],
x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

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

Mathematica [B]  time = 3.18768, size = 382, normalized size = 3.64 \[ \frac{10 \sin ^{\frac{5}{2}}(c+d x) (\cos (c+d x)+1) (a (\sec (c+d x)+1))^n \left (F_1\left (\frac{1}{4};n,\frac{3}{2};\frac{5}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-F_1\left (\frac{1}{4};n,\frac{5}{2};\frac{5}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )}{d \left (2 (\cos (c+d x)-1) \left (3 F_1\left (\frac{5}{4};n,\frac{5}{2};\frac{9}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-5 F_1\left (\frac{5}{4};n,\frac{7}{2};\frac{9}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-2 n F_1\left (\frac{5}{4};n+1,\frac{3}{2};\frac{9}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+2 n F_1\left (\frac{5}{4};n+1,\frac{5}{2};\frac{9}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )+5 (\cos (c+d x)+1) F_1\left (\frac{1}{4};n,\frac{3}{2};\frac{5}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-5 (\cos (c+d x)+1) F_1\left (\frac{1}{4};n,\frac{5}{2};\frac{5}{4};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(10*(AppellF1[1/4, n, 3/2, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[1/4, n, 5/2, 5/4, Tan[(c +
d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(1 + Cos[c + d*x])*(a*(1 + Sec[c + d*x]))^n*Sin[c + d*x]^(5/2))/(d*(2*(3*App
ellF1[5/4, n, 5/2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 5*AppellF1[5/4, n, 7/2, 9/4, Tan[(c + d*x)/
2]^2, -Tan[(c + d*x)/2]^2] - 2*n*AppellF1[5/4, 1 + n, 3/2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*n
*AppellF1[5/4, 1 + n, 5/2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + 5*AppellF1[1/4
, n, 3/2, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x]) - 5*AppellF1[1/4, n, 5/2, 5/4, Tan[
(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])))

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Maple [F]  time = 0.179, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n} \left ( \sin \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*sin(d*x+c)^(3/2),x)

[Out]

int((a+a*sec(d*x+c))^n*sin(d*x+c)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \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*sin(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \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*sin(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

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