3.108 \(\int \frac{\sin ^7(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=144 \[ \frac{3 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{40 b d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}+\frac{\sin ^7(a+b x)}{7 b d (d \tan (a+b x))^{3/2}}-\frac{3 \sin ^5(a+b x)}{70 b d (d \tan (a+b x))^{3/2}}-\frac{\sin ^3(a+b x)}{20 b d (d \tan (a+b x))^{3/2}} \]

[Out]

-Sin[a + b*x]^3/(20*b*d*(d*Tan[a + b*x])^(3/2)) - (3*Sin[a + b*x]^5)/(70*b*d*(d*Tan[a + b*x])^(3/2)) + Sin[a +
b*x]^7/(7*b*d*(d*Tan[a + b*x])^(3/2)) + (3*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(40*b*d^2*Sqrt[Sin[2*a
+ 2*b*x]]*Sqrt[d*Tan[a + b*x]])

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Rubi [A]  time = 0.186241, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2596, 2598, 2601, 2572, 2639} \[ \frac{3 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{40 b d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}+\frac{\sin ^7(a+b x)}{7 b d (d \tan (a+b x))^{3/2}}-\frac{3 \sin ^5(a+b x)}{70 b d (d \tan (a+b x))^{3/2}}-\frac{\sin ^3(a+b x)}{20 b d (d \tan (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^7/(d*Tan[a + b*x])^(5/2),x]

[Out]

-Sin[a + b*x]^3/(20*b*d*(d*Tan[a + b*x])^(3/2)) - (3*Sin[a + b*x]^5)/(70*b*d*(d*Tan[a + b*x])^(3/2)) + Sin[a +
b*x]^7/(7*b*d*(d*Tan[a + b*x])^(3/2)) + (3*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(40*b*d^2*Sqrt[Sin[2*a
+ 2*b*x]]*Sqrt[d*Tan[a + b*x]])

Rule 2596

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

Rule 2598

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

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sin ^7(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx &=\frac{\sin ^7(a+b x)}{7 b d (d \tan (a+b x))^{3/2}}+\frac{3 \int \frac{\sin ^5(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx}{14 d^2}\\ &=-\frac{3 \sin ^5(a+b x)}{70 b d (d \tan (a+b x))^{3/2}}+\frac{\sin ^7(a+b x)}{7 b d (d \tan (a+b x))^{3/2}}+\frac{3 \int \frac{\sin ^3(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx}{20 d^2}\\ &=-\frac{\sin ^3(a+b x)}{20 b d (d \tan (a+b x))^{3/2}}-\frac{3 \sin ^5(a+b x)}{70 b d (d \tan (a+b x))^{3/2}}+\frac{\sin ^7(a+b x)}{7 b d (d \tan (a+b x))^{3/2}}+\frac{3 \int \frac{\sin (a+b x)}{\sqrt{d \tan (a+b x)}} \, dx}{40 d^2}\\ &=-\frac{\sin ^3(a+b x)}{20 b d (d \tan (a+b x))^{3/2}}-\frac{3 \sin ^5(a+b x)}{70 b d (d \tan (a+b x))^{3/2}}+\frac{\sin ^7(a+b x)}{7 b d (d \tan (a+b x))^{3/2}}+\frac{\left (3 \sqrt{\sin (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)} \, dx}{40 d^2 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{\sin ^3(a+b x)}{20 b d (d \tan (a+b x))^{3/2}}-\frac{3 \sin ^5(a+b x)}{70 b d (d \tan (a+b x))^{3/2}}+\frac{\sin ^7(a+b x)}{7 b d (d \tan (a+b x))^{3/2}}+\frac{(3 \sin (a+b x)) \int \sqrt{\sin (2 a+2 b x)} \, dx}{40 d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{\sin ^3(a+b x)}{20 b d (d \tan (a+b x))^{3/2}}-\frac{3 \sin ^5(a+b x)}{70 b d (d \tan (a+b x))^{3/2}}+\frac{\sin ^7(a+b x)}{7 b d (d \tan (a+b x))^{3/2}}+\frac{3 E\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sin (a+b x)}{40 b d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 1.54922, size = 122, normalized size = 0.85 \[ \frac{\sqrt{d \tan (a+b x)} \left (112 \tan (a+b x) \sec (a+b x) \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )-(15 \sin (a+b x)+29 \sin (3 (a+b x))+9 \sin (5 (a+b x))-5 \sin (7 (a+b x))) \sqrt{\sec ^2(a+b x)}\right )}{2240 b d^3 \sqrt{\sec ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^7/(d*Tan[a + b*x])^(5/2),x]

[Out]

(Sqrt[d*Tan[a + b*x]]*(-(Sqrt[Sec[a + b*x]^2]*(15*Sin[a + b*x] + 29*Sin[3*(a + b*x)] + 9*Sin[5*(a + b*x)] - 5*
Sin[7*(a + b*x)])) + 112*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sec[a + b*x]*Tan[a + b*x]))/(2240*b
*d^3*Sqrt[Sec[a + b*x]^2])

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Maple [B]  time = 0.187, size = 571, normalized size = 4. \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/(d*tan(b*x+a))^(5/2),x)

[Out]

-1/560/b*2^(1/2)*(cos(b*x+a)-1)^2*(40*2^(1/2)*cos(b*x+a)^8-108*cos(b*x+a)^6*2^(1/2)+82*cos(b*x+a)^4*2^(1/2)-21
*cos(b*x+a)*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(
1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)+42*cos(b*x+a)*
EllipticE((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b
*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)-21*EllipticF((-(cos(b*x+a)
-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin
(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)+42*EllipticE((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+
a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b
*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)+7*cos(b*x+a)^2*2^(1/2)-21*cos(b*x+a)*2^(1/2))*(cos(b*x+a)+1)^2/cos(b*x+a
)^3/sin(b*x+a)^2/(d*sin(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 )^{7}}{\left (d \tan \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^7/(d*tan(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)^7/(d*tan(b*x + a))^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (\cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4} + 3 \, \cos \left (b x + a\right )^{2} - 1\right )} \sqrt{d \tan \left (b x + a\right )} \sin \left (b x + a\right )}{d^{3} \tan \left (b x + a\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^7/(d*tan(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

integral(-(cos(b*x + a)^6 - 3*cos(b*x + a)^4 + 3*cos(b*x + a)^2 - 1)*sqrt(d*tan(b*x + a))*sin(b*x + a)/(d^3*ta
n(b*x + a)^3), 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(sin(b*x+a)**7/(d*tan(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )^{7}}{\left (d \tan \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^7/(d*tan(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^7/(d*tan(b*x + a))^(5/2), x)