3.137 \(\int \csc ^7(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=197 \[ -\frac{(5 a-b) (a+b)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{16 a^{3/2} f}-\frac{\cot (e+f x) \csc ^5(e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{5/2}}{6 a f}-\frac{(5 a-b) \cot (e+f x) \csc ^3(e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{24 a f}-\frac{(5 a-b) (a+b) \cot (e+f x) \csc (e+f x) \sqrt{a-b \cos ^2(e+f x)+b}}{16 a f} \]

[Out]

-((5*a - b)*(a + b)^2*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(16*a^(3/2)*f) - ((5*a -
b)*(a + b)*Sqrt[a + b - b*Cos[e + f*x]^2]*Cot[e + f*x]*Csc[e + f*x])/(16*a*f) - ((5*a - b)*(a + b - b*Cos[e +
f*x]^2)^(3/2)*Cot[e + f*x]*Csc[e + f*x]^3)/(24*a*f) - ((a + b - b*Cos[e + f*x]^2)^(5/2)*Cot[e + f*x]*Csc[e +
f*x]^5)/(6*a*f)

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Rubi [A]  time = 0.174983, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3186, 382, 378, 377, 206} \[ -\frac{(5 a-b) (a+b)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{16 a^{3/2} f}-\frac{\cot (e+f x) \csc ^5(e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{5/2}}{6 a f}-\frac{(5 a-b) \cot (e+f x) \csc ^3(e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{24 a f}-\frac{(5 a-b) (a+b) \cot (e+f x) \csc (e+f x) \sqrt{a-b \cos ^2(e+f x)+b}}{16 a f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^7*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

-((5*a - b)*(a + b)^2*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(16*a^(3/2)*f) - ((5*a -
b)*(a + b)*Sqrt[a + b - b*Cos[e + f*x]^2]*Cot[e + f*x]*Csc[e + f*x])/(16*a*f) - ((5*a - b)*(a + b - b*Cos[e +
f*x]^2)^(3/2)*Cot[e + f*x]*Csc[e + f*x]^3)/(24*a*f) - ((a + b - b*Cos[e + f*x]^2)^(5/2)*Cot[e + f*x]*Csc[e +
f*x]^5)/(6*a*f)

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 378

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] - Dist[(c*q)/(a*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 206

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

Rubi steps

\begin{align*} \int \csc ^7(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (a+b-b x^2\right )^{3/2}}{\left (1-x^2\right )^4} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\left (a+b-b \cos ^2(e+f x)\right )^{5/2} \cot (e+f x) \csc ^5(e+f x)}{6 a f}-\frac{(5 a-b) \operatorname{Subst}\left (\int \frac{\left (a+b-b x^2\right )^{3/2}}{\left (1-x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{6 a f}\\ &=-\frac{(5 a-b) \left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{24 a f}-\frac{\left (a+b-b \cos ^2(e+f x)\right )^{5/2} \cot (e+f x) \csc ^5(e+f x)}{6 a f}-\frac{((5 a-b) (a+b)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b-b x^2}}{\left (1-x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{8 a f}\\ &=-\frac{(5 a-b) (a+b) \sqrt{a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{16 a f}-\frac{(5 a-b) \left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{24 a f}-\frac{\left (a+b-b \cos ^2(e+f x)\right )^{5/2} \cot (e+f x) \csc ^5(e+f x)}{6 a f}-\frac{\left ((5 a-b) (a+b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{a+b-b x^2}} \, dx,x,\cos (e+f x)\right )}{16 a f}\\ &=-\frac{(5 a-b) (a+b) \sqrt{a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{16 a f}-\frac{(5 a-b) \left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{24 a f}-\frac{\left (a+b-b \cos ^2(e+f x)\right )^{5/2} \cot (e+f x) \csc ^5(e+f x)}{6 a f}-\frac{\left ((5 a-b) (a+b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{\cos (e+f x)}{\sqrt{a+b-b \cos ^2(e+f x)}}\right )}{16 a f}\\ &=-\frac{(5 a-b) (a+b)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a+b-b \cos ^2(e+f x)}}\right )}{16 a^{3/2} f}-\frac{(5 a-b) (a+b) \sqrt{a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{16 a f}-\frac{(5 a-b) \left (a+b-b \cos ^2(e+f x)\right )^{3/2} \cot (e+f x) \csc ^3(e+f x)}{24 a f}-\frac{\left (a+b-b \cos ^2(e+f x)\right )^{5/2} \cot (e+f x) \csc ^5(e+f x)}{6 a f}\\ \end{align*}

Mathematica [A]  time = 1.25992, size = 161, normalized size = 0.82 \[ \frac{-\sqrt{2} \sqrt{a} \csc ^2(e+f x) \sqrt{2 a-b \cos (2 (e+f x))+b} \left (\left (15 a^2+22 a b+3 b^2\right ) \cos (e+f x)+2 a \cot (e+f x) \csc (e+f x) \left (4 a \csc ^2(e+f x)+5 a+7 b\right )\right )-6 (5 a-b) (a+b)^2 \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \cos (e+f x)}{\sqrt{2 a-b \cos (2 (e+f x))+b}}\right )}{96 a^{3/2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^7*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(-6*(5*a - b)*(a + b)^2*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]] - Sqrt[2]*S
qrt[a]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]*Csc[e + f*x]^2*((15*a^2 + 22*a*b + 3*b^2)*Cos[e + f*x] + 2*a*Cot[e +
f*x]*Csc[e + f*x]*(5*a + 7*b + 4*a*Csc[e + f*x]^2)))/(96*a^(3/2)*f)

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Maple [B]  time = 2.063, size = 565, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

-1/96*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(15*a^4*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*
cos(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^6+27*a^3*b*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4
+(a+b)*cos(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^6+9*b^2*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+
e)^4+(a+b)*cos(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^6*a^2-3*b^3*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*
cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)*sin(f*x+e)^6*a+30*a^(7/2)*(cos(f*x+e)^2*(a+b*sin(f*x
+e)^2))^(1/2)*sin(f*x+e)^4+44*b*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*sin(f*x+e)^4*a^(5/2)+6*b^2*(cos(f*x+e)
^2*(a+b*sin(f*x+e)^2))^(1/2)*sin(f*x+e)^4*a^(3/2)+20*a^(7/2)*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*sin(f*x+e
)^2+28*b*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*sin(f*x+e)^2*a^(5/2)+16*a^(7/2)*(cos(f*x+e)^2*(a+b*sin(f*x+e)
^2))^(1/2))/sin(f*x+e)^6/a^(5/2)/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \csc \left (f x + e\right )^{7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^7, x)

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Fricas [A]  time = 10.5865, size = 1773, normalized size = 9. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/192*(3*((5*a^3 + 9*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^6 - 3*(5*a^3 + 9*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e
)^4 - 5*a^3 - 9*a^2*b - 3*a*b^2 + b^3 + 3*(5*a^3 + 9*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^2)*sqrt(a)*log(2*((a^
2 - 6*a*b + b^2)*cos(f*x + e)^4 + 2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + (a + b)
*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) + a^2 + 2*a*b + b^2)/(cos(f*x + e)^4 - 2*cos(f*x + e)^2
+ 1)) - 4*((15*a^3 + 22*a^2*b + 3*a*b^2)*cos(f*x + e)^5 - 2*(20*a^3 + 29*a^2*b + 3*a*b^2)*cos(f*x + e)^3 + 3*
(11*a^3 + 12*a^2*b + a*b^2)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^2*f*cos(f*x + e)^6 - 3*a^2*f*cos
(f*x + e)^4 + 3*a^2*f*cos(f*x + e)^2 - a^2*f), 1/96*(3*((5*a^3 + 9*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^6 - 3*(
5*a^3 + 9*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^4 - 5*a^3 - 9*a^2*b - 3*a*b^2 + b^3 + 3*(5*a^3 + 9*a^2*b + 3*a*b
^2 - b^3)*cos(f*x + e)^2)*sqrt(-a)*arctan(-1/2*((a - b)*cos(f*x + e)^2 + a + b)*sqrt(-b*cos(f*x + e)^2 + a + b
)*sqrt(-a)/(a*b*cos(f*x + e)^3 - (a^2 + a*b)*cos(f*x + e))) + 2*((15*a^3 + 22*a^2*b + 3*a*b^2)*cos(f*x + e)^5
- 2*(20*a^3 + 29*a^2*b + 3*a*b^2)*cos(f*x + e)^3 + 3*(11*a^3 + 12*a^2*b + a*b^2)*cos(f*x + e))*sqrt(-b*cos(f*x
+ e)^2 + a + b))/(a^2*f*cos(f*x + e)^6 - 3*a^2*f*cos(f*x + e)^4 + 3*a^2*f*cos(f*x + e)^2 - a^2*f)]

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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(csc(f*x+e)**7*(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \csc \left (f x + e\right )^{7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^7*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^7, x)