3.287 \(\int \frac{\sin ^{\frac{7}{2}}(a+b x)}{\cos ^{\frac{7}{2}}(a+b x)} \, dx\)

Optimal. Leaf size=226 \[ \frac{2 \sin ^{\frac{5}{2}}(a+b x)}{5 b \cos ^{\frac{5}{2}}(a+b x)}-\frac{2 \sqrt{\sin (a+b x)}}{b \sqrt{\cos (a+b x)}}+\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{\sqrt{2} b}-\frac{\log \left (\cot (a+b x)-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{2 \sqrt{2} b}+\frac{\log \left (\cot (a+b x)+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{2 \sqrt{2} b} \]

[Out]

ArcTan[1 - (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/(Sqrt[2]*b) - ArcTan[1 + (Sqrt[2]*Sqrt[Cos[a + b*x
]])/Sqrt[Sin[a + b*x]]]/(Sqrt[2]*b) - Log[1 + Cot[a + b*x] - (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/
(2*Sqrt[2]*b) + Log[1 + Cot[a + b*x] + (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/(2*Sqrt[2]*b) - (2*Sqr
t[Sin[a + b*x]])/(b*Sqrt[Cos[a + b*x]]) + (2*Sin[a + b*x]^(5/2))/(5*b*Cos[a + b*x]^(5/2))

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Rubi [A]  time = 0.149362, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {2566, 2575, 297, 1162, 617, 204, 1165, 628} \[ \frac{2 \sin ^{\frac{5}{2}}(a+b x)}{5 b \cos ^{\frac{5}{2}}(a+b x)}-\frac{2 \sqrt{\sin (a+b x)}}{b \sqrt{\cos (a+b x)}}+\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{\sqrt{2} b}-\frac{\log \left (\cot (a+b x)-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{2 \sqrt{2} b}+\frac{\log \left (\cot (a+b x)+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}+1\right )}{2 \sqrt{2} b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^(7/2)/Cos[a + b*x]^(7/2),x]

[Out]

ArcTan[1 - (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/(Sqrt[2]*b) - ArcTan[1 + (Sqrt[2]*Sqrt[Cos[a + b*x
]])/Sqrt[Sin[a + b*x]]]/(Sqrt[2]*b) - Log[1 + Cot[a + b*x] - (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/
(2*Sqrt[2]*b) + Log[1 + Cot[a + b*x] + (Sqrt[2]*Sqrt[Cos[a + b*x]])/Sqrt[Sin[a + b*x]]]/(2*Sqrt[2]*b) - (2*Sqr
t[Sin[a + b*x]])/(b*Sqrt[Cos[a + b*x]]) + (2*Sin[a + b*x]^(5/2))/(5*b*Cos[a + b*x]^(5/2))

Rule 2566

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(a*Sin[e
+ f*x])^(m - 1)*(b*Cos[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Sin[e +
f*x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Integ
ersQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2575

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{k = Denomina
tor[m]}, -Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Cos[e + f*x])^(1/k)/(b*Si
n[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\sin ^{\frac{7}{2}}(a+b x)}{\cos ^{\frac{7}{2}}(a+b x)} \, dx &=\frac{2 \sin ^{\frac{5}{2}}(a+b x)}{5 b \cos ^{\frac{5}{2}}(a+b x)}-\int \frac{\sin ^{\frac{3}{2}}(a+b x)}{\cos ^{\frac{3}{2}}(a+b x)} \, dx\\ &=-\frac{2 \sqrt{\sin (a+b x)}}{b \sqrt{\cos (a+b x)}}+\frac{2 \sin ^{\frac{5}{2}}(a+b x)}{5 b \cos ^{\frac{5}{2}}(a+b x)}+\int \frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}} \, dx\\ &=-\frac{2 \sqrt{\sin (a+b x)}}{b \sqrt{\cos (a+b x)}}+\frac{2 \sin ^{\frac{5}{2}}(a+b x)}{5 b \cos ^{\frac{5}{2}}(a+b x)}-\frac{2 \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{b}\\ &=-\frac{2 \sqrt{\sin (a+b x)}}{b \sqrt{\cos (a+b x)}}+\frac{2 \sin ^{\frac{5}{2}}(a+b x)}{5 b \cos ^{\frac{5}{2}}(a+b x)}+\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{b}-\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{b}\\ &=-\frac{2 \sqrt{\sin (a+b x)}}{b \sqrt{\cos (a+b x)}}+\frac{2 \sin ^{\frac{5}{2}}(a+b x)}{5 b \cos ^{\frac{5}{2}}(a+b x)}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 b}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 b}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}\\ &=-\frac{\log \left (1+\cot (a+b x)-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}+\frac{\log \left (1+\cot (a+b x)+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}-\frac{2 \sqrt{\sin (a+b x)}}{b \sqrt{\cos (a+b x)}}+\frac{2 \sin ^{\frac{5}{2}}(a+b x)}{5 b \cos ^{\frac{5}{2}}(a+b x)}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}\\ &=\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}-\frac{\tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{\sqrt{2} b}-\frac{\log \left (1+\cot (a+b x)-\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}+\frac{\log \left (1+\cot (a+b x)+\frac{\sqrt{2} \sqrt{\cos (a+b x)}}{\sqrt{\sin (a+b x)}}\right )}{2 \sqrt{2} b}-\frac{2 \sqrt{\sin (a+b x)}}{b \sqrt{\cos (a+b x)}}+\frac{2 \sin ^{\frac{5}{2}}(a+b x)}{5 b \cos ^{\frac{5}{2}}(a+b x)}\\ \end{align*}

Mathematica [C]  time = 0.0563914, size = 57, normalized size = 0.25 \[ \frac{2 \sin ^{\frac{9}{2}}(a+b x) \sqrt [4]{\cos ^2(a+b x)} \, _2F_1\left (\frac{9}{4},\frac{9}{4};\frac{13}{4};\sin ^2(a+b x)\right )}{9 b \sqrt{\cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^(7/2)/Cos[a + b*x]^(7/2),x]

[Out]

(2*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[9/4, 9/4, 13/4, Sin[a + b*x]^2]*Sin[a + b*x]^(9/2))/(9*b*Sqrt[Cos[
a + b*x]])

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Maple [C]  time = 0.178, size = 702, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^(7/2)/cos(b*x+a)^(7/2),x)

[Out]

-1/10/b*2^(1/2)*(5*I*cos(b*x+a)^2*sin(b*x+a)*EllipticPi((-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2
*I,1/2*2^(1/2))*(-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*(
(-1+cos(b*x+a))/sin(b*x+a))^(1/2)-5*I*cos(b*x+a)^2*sin(b*x+a)*(-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*(
(-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi((-(-1+cos(b*x+a)-si
n(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))+5*cos(b*x+a)^2*sin(b*x+a)*EllipticPi((-(-1+cos(b*x+a)-sin(b
*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)
+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)+5*cos(b*x+a)^2*sin(b*x+a)*(-(-1+cos(b*x+a)-s
in(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*
EllipticPi((-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-10*cos(b*x+a)^2*sin(b*x+a)*(-
(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/si
n(b*x+a))^(1/2)*EllipticF((-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+12*cos(b*x+a)^3*2^(1/2)-
12*cos(b*x+a)^2*2^(1/2)-2*cos(b*x+a)*2^(1/2)+2*2^(1/2))*sin(b*x+a)^(1/2)/(-1+cos(b*x+a))/cos(b*x+a)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )^{\frac{7}{2}}}{\cos \left (b x + a\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^(7/2)/cos(b*x+a)^(7/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)^(7/2)/cos(b*x + a)^(7/2), x)

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Fricas [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(sin(b*x+a)^(7/2)/cos(b*x+a)^(7/2),x, algorithm="fricas")

[Out]

Timed out

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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(sin(b*x+a)**(7/2)/cos(b*x+a)**(7/2),x)

[Out]

Timed out

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Giac [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(sin(b*x+a)^(7/2)/cos(b*x+a)^(7/2),x, algorithm="giac")

[Out]

Timed out