3.251 \(\int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx\)

Optimal. Leaf size=137 \[ \frac {9 \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}-\frac {9 \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}+\frac {9}{10 b d (d \cos (a+b x))^{5/2}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}} \]

[Out]

9/4*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(7/2)-9/4*arctanh((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(7/2)+9/10/b/
d/(d*cos(b*x+a))^(5/2)-1/2*csc(b*x+a)^2/b/d/(d*cos(b*x+a))^(5/2)+9/2/b/d^3/(d*cos(b*x+a))^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2565, 290, 325, 329, 298, 203, 206} \[ \frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}+\frac {9 \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}-\frac {9 \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}+\frac {9}{10 b d (d \cos (a+b x))^{5/2}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3/(d*Cos[a + b*x])^(7/2),x]

[Out]

(9*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(7/2)) - (9*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(7/2
)) + 9/(10*b*d*(d*Cos[a + b*x])^(5/2)) + 9/(2*b*d^3*Sqrt[d*Cos[a + b*x]]) - Csc[a + b*x]^2/(2*b*d*(d*Cos[a + b
*x])^(5/2))

Rule 203

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

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])

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rubi steps

\begin {align*} \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {1}{x^{7/2} \left (1-\frac {x^2}{d^2}\right )^2} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}}-\frac {9 \operatorname {Subst}\left (\int \frac {1}{x^{7/2} \left (1-\frac {x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{4 b d}\\ &=\frac {9}{10 b d (d \cos (a+b x))^{5/2}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}}-\frac {9 \operatorname {Subst}\left (\int \frac {1}{x^{3/2} \left (1-\frac {x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{4 b d^3}\\ &=\frac {9}{10 b d (d \cos (a+b x))^{5/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}}-\frac {9 \operatorname {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{4 b d^5}\\ &=\frac {9}{10 b d (d \cos (a+b x))^{5/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}}-\frac {9 \operatorname {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{d^2}} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{2 b d^5}\\ &=\frac {9}{10 b d (d \cos (a+b x))^{5/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}}-\frac {9 \operatorname {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{4 b d^3}+\frac {9 \operatorname {Subst}\left (\int \frac {1}{d+x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{4 b d^3}\\ &=\frac {9 \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}-\frac {9 \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}+\frac {9}{10 b d (d \cos (a+b x))^{5/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.47, size = 102, normalized size = 0.74 \[ \frac {45 \cot ^2(a+b x) \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\csc ^2(a+b x)\right )+\left (-\cot ^2(a+b x)\right )^{3/4} \left (-5 \cot ^2(a+b x)+4 \sec ^2(a+b x)+40\right )}{10 b d^3 \left (-\cot ^2(a+b x)\right )^{3/4} \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3/(d*Cos[a + b*x])^(7/2),x]

[Out]

(45*Cot[a + b*x]^2*Hypergeometric2F1[1/4, 1/4, 5/4, Csc[a + b*x]^2] + (-Cot[a + b*x]^2)^(3/4)*(40 - 5*Cot[a +
b*x]^2 + 4*Sec[a + b*x]^2))/(10*b*d^3*Sqrt[d*Cos[a + b*x]]*(-Cot[a + b*x]^2)^(3/4))

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fricas [A]  time = 0.54, size = 438, normalized size = 3.20 \[ \left [\frac {90 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) - 45 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \sqrt {-d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, {\left (45 \, \cos \left (b x + a\right )^{4} - 36 \, \cos \left (b x + a\right )^{2} - 4\right )} \sqrt {d \cos \left (b x + a\right )}}{80 \, {\left (b d^{4} \cos \left (b x + a\right )^{5} - b d^{4} \cos \left (b x + a\right )^{3}\right )}}, \frac {90 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) + 45 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \sqrt {d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, {\left (45 \, \cos \left (b x + a\right )^{4} - 36 \, \cos \left (b x + a\right )^{2} - 4\right )} \sqrt {d \cos \left (b x + a\right )}}{80 \, {\left (b d^{4} \cos \left (b x + a\right )^{5} - b d^{4} \cos \left (b x + a\right )^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(7/2),x, algorithm="fricas")

[Out]

[1/80*(90*(cos(b*x + a)^5 - cos(b*x + a)^3)*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) +
1)/(d*cos(b*x + a))) - 45*(cos(b*x + a)^5 - cos(b*x + a)^3)*sqrt(-d)*log((d*cos(b*x + a)^2 - 4*sqrt(d*cos(b*x
+ a))*sqrt(-d)*(cos(b*x + a) - 1) - 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 + 2*cos(b*x + a) + 1)) + 8*(45*cos(b
*x + a)^4 - 36*cos(b*x + a)^2 - 4)*sqrt(d*cos(b*x + a)))/(b*d^4*cos(b*x + a)^5 - b*d^4*cos(b*x + a)^3), 1/80*(
90*(cos(b*x + a)^5 - cos(b*x + a)^3)*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*cos(b
*x + a))) + 45*(cos(b*x + a)^5 - cos(b*x + a)^3)*sqrt(d)*log((d*cos(b*x + a)^2 - 4*sqrt(d*cos(b*x + a))*sqrt(d
)*(cos(b*x + a) + 1) + 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 - 2*cos(b*x + a) + 1)) + 8*(45*cos(b*x + a)^4 - 3
6*cos(b*x + a)^2 - 4)*sqrt(d*cos(b*x + a)))/(b*d^4*cos(b*x + a)^5 - b*d^4*cos(b*x + a)^3)]

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giac [B]  time = 1.41, size = 417, normalized size = 3.04 \[ \frac {\frac {90 \, \arctan \left (-\frac {\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} + \frac {45 \, \log \left ({\left | -\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d} \right |}\right )}{\sqrt {-d}} + \frac {10 \, \sqrt {-d}}{{\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{2} - d} + \frac {5 \, \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}}{d} - \frac {32 \, {\left (15 \, {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{4} - 40 \, {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{3} \sqrt {-d} - 70 \, {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )}^{2} d + 40 \, {\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}\right )} \sqrt {-d} d + 11 \, d^{2}\right )}}{{\left (\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d} - \sqrt {-d}\right )}^{5}}}{40 \, b d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(7/2),x, algorithm="giac")

[Out]

1/40*(90*arctan(-(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))/sqrt(-d))/sqrt(-d) +
45*log(abs(-sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 + sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d)))/sqrt(-d) + 10*sqrt(-d)/((s
qrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))^2 - d) + 5*sqrt(-d*tan(1/2*b*x + 1/2*a)^
4 + d)/d - 32*(15*(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))^4 - 40*(sqrt(-d)*tan
(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))^3*sqrt(-d) - 70*(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 -
sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))^2*d + 40*(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^
4 + d))*sqrt(-d)*d + 11*d^2)/(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d) - sqrt(-d)
)^5)/(b*d^3)

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maple [B]  time = 0.52, size = 1165, normalized size = 8.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^3/(d*cos(b*x+a))^(7/2),x)

[Out]

1/40/d^(15/2)/(-d)^(1/2)/sin(1/2*b*x+1/2*a)^2/(8*sin(1/2*b*x+1/2*a)^8-20*sin(1/2*b*x+1/2*a)^6+18*sin(1/2*b*x+1
/2*a)^4-7*sin(1/2*b*x+1/2*a)^2+1)*(-5*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(7/2)*(-d)^(1/2)-360*(2*ln(2/cos(1
/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(9/2)+ln(2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)
*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^4+ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^
(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^4)*sin(1/2*b*x+1/2*a)^10+180
*(10*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(9/2)-4*(-2*sin(1/2*b*x+1/2
*a)^2*d+d)^(1/2)*d^(7/2)*(-d)^(1/2)+5*ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)
-2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^4+5*ln(2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d
)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^4)*sin(1/2*b*x+1/2*a)^8-90*(18*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(
1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(9/2)-16*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(7/2)*(-d)^(1/2)
+9*ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1
/2)*d^4+9*ln(2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))*
(-d)^(1/2)*d^4)*sin(1/2*b*x+1/2*a)^6+9*(70*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(
1/2)-d))*d^(9/2)-104*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(7/2)*(-d)^(1/2)+35*ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^
(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^4+35*ln(2/(cos(1/2*b*x+1/2*a
)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^4)*sin(1/2*b*x+1/2*a
)^4-9*(10*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(9/2)-24*(-2*sin(1/2*b
*x+1/2*a)^2*d+d)^(1/2)*d^(7/2)*(-d)^(1/2)+5*ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)
^(1/2)-2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^4+5*ln(2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)
^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))*(-d)^(1/2)*d^4)*sin(1/2*b*x+1/2*a)^2)/b

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maxima [A]  time = 0.62, size = 134, normalized size = 0.98 \[ \frac {\frac {4 \, {\left (45 \, d^{4} \cos \left (b x + a\right )^{4} - 36 \, d^{4} \cos \left (b x + a\right )^{2} - 4 \, d^{4}\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} d^{2} - \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} d^{4}} + \frac {90 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {45 \, \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{d^{\frac {5}{2}}}}{40 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

1/40*(4*(45*d^4*cos(b*x + a)^4 - 36*d^4*cos(b*x + a)^2 - 4*d^4)/((d*cos(b*x + a))^(9/2)*d^2 - (d*cos(b*x + a))
^(5/2)*d^4) + 90*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(5/2) + 45*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(
d*cos(b*x + a)) + sqrt(d)))/d^(5/2))/(b*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\sin \left (a+b\,x\right )}^3\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)^3*(d*cos(a + b*x))^(7/2)),x)

[Out]

int(1/(sin(a + b*x)^3*(d*cos(a + b*x))^(7/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**3/(d*cos(b*x+a))**(7/2),x)

[Out]

Timed out

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