3.521 \(\int \frac{\sec ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=284 \[ -\frac{3 b \left (-7 a^2 b^2+2 a^4-15 b^4\right )}{16 d \left (a^2-b^2\right )^3 \sqrt{a+b \sin (c+d x)}}-\frac{3 \left (4 a^2-14 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 d (a-b)^{7/2}}+\frac{3 \left (4 a^2+14 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 d (a+b)^{7/2}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^2(c+d x) \left (2 a \left (3 a^2-8 b^2\right ) \sin (c+d x)+b \left (a^2+9 b^2\right )\right )}{16 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}} \]

[Out]

(-3*(4*a^2 - 14*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(32*(a - b)^(7/2)*d) + (3*(4*a^2
+ 14*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(32*(a + b)^(7/2)*d) - (3*b*(2*a^4 - 7*a^2*b
^2 - 15*b^4))/(16*(a^2 - b^2)^3*d*Sqrt[a + b*Sin[c + d*x]]) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x]))/(4*(a^2 -
b^2)*d*Sqrt[a + b*Sin[c + d*x]]) + (Sec[c + d*x]^2*(b*(a^2 + 9*b^2) + 2*a*(3*a^2 - 8*b^2)*Sin[c + d*x]))/(16*(
a^2 - b^2)^2*d*Sqrt[a + b*Sin[c + d*x]])

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Rubi [A]  time = 0.520548, antiderivative size = 284, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2668, 741, 823, 829, 827, 1166, 206} \[ -\frac{3 b \left (-7 a^2 b^2+2 a^4-15 b^4\right )}{16 d \left (a^2-b^2\right )^3 \sqrt{a+b \sin (c+d x)}}-\frac{3 \left (4 a^2-14 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 d (a-b)^{7/2}}+\frac{3 \left (4 a^2+14 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 d (a+b)^{7/2}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^2(c+d x) \left (2 a \left (3 a^2-8 b^2\right ) \sin (c+d x)+b \left (a^2+9 b^2\right )\right )}{16 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-3*(4*a^2 - 14*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(32*(a - b)^(7/2)*d) + (3*(4*a^2
+ 14*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(32*(a + b)^(7/2)*d) - (3*b*(2*a^4 - 7*a^2*b
^2 - 15*b^4))/(16*(a^2 - b^2)^3*d*Sqrt[a + b*Sin[c + d*x]]) - (Sec[c + d*x]^4*(b - a*Sin[c + d*x]))/(4*(a^2 -
b^2)*d*Sqrt[a + b*Sin[c + d*x]]) + (Sec[c + d*x]^2*(b*(a^2 + 9*b^2) + 2*a*(3*a^2 - 8*b^2)*Sin[c + d*x]))/(16*(
a^2 - b^2)^2*d*Sqrt[a + b*Sin[c + d*x]])

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 829

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[((e*f - d*g)*(d
+ e*x)^(m + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*Simp[c*d*f + a*
e*g - c*(e*f - d*g)*x, x])/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&
FractionQ[m] && LtQ[m, -1]

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

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

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 \frac{\sec ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{1}{(a+x)^{3/2} \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}+\frac{b^3 \operatorname{Subst}\left (\int \frac{\frac{3}{2} \left (2 a^2-3 b^2\right )+\frac{7 a x}{2}}{(a+x)^{3/2} \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^2(c+d x) \left (b \left (a^2+9 b^2\right )+2 a \left (3 a^2-8 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}-\frac{b \operatorname{Subst}\left (\int \frac{-\frac{3}{4} \left (4 a^4-9 a^2 b^2+15 b^4\right )-\frac{3}{2} a \left (3 a^2-8 b^2\right ) x}{(a+x)^{3/2} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac{3 b \left (2 a^4-7 a^2 b^2-15 b^4\right )}{16 \left (a^2-b^2\right )^3 d \sqrt{a+b \sin (c+d x)}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^2(c+d x) \left (b \left (a^2+9 b^2\right )+2 a \left (3 a^2-8 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}+\frac{b \operatorname{Subst}\left (\int \frac{\frac{3}{4} a \left (4 a^4-15 a^2 b^2+31 b^4\right )+\frac{3}{4} \left (2 a^4-7 a^2 b^2-15 b^4\right ) x}{\sqrt{a+x} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^3 d}\\ &=-\frac{3 b \left (2 a^4-7 a^2 b^2-15 b^4\right )}{16 \left (a^2-b^2\right )^3 d \sqrt{a+b \sin (c+d x)}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^2(c+d x) \left (b \left (a^2+9 b^2\right )+2 a \left (3 a^2-8 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}+\frac{b \operatorname{Subst}\left (\int \frac{-\frac{3}{4} a \left (2 a^4-7 a^2 b^2-15 b^4\right )+\frac{3}{4} a \left (4 a^4-15 a^2 b^2+31 b^4\right )+\frac{3}{4} \left (2 a^4-7 a^2 b^2-15 b^4\right ) x^2}{-a^2+b^2+2 a x^2-x^4} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{4 \left (a^2-b^2\right )^3 d}\\ &=-\frac{3 b \left (2 a^4-7 a^2 b^2-15 b^4\right )}{16 \left (a^2-b^2\right )^3 d \sqrt{a+b \sin (c+d x)}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^2(c+d x) \left (b \left (a^2+9 b^2\right )+2 a \left (3 a^2-8 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}-\frac{\left (3 \left (4 a^2-14 a b+15 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a-b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{32 (a-b)^3 d}+\frac{\left (3 \left (4 a^2+14 a b+15 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{32 (a+b)^3 d}\\ &=-\frac{3 \left (4 a^2-14 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 (a-b)^{7/2} d}+\frac{3 \left (4 a^2+14 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 (a+b)^{7/2} d}-\frac{3 b \left (2 a^4-7 a^2 b^2-15 b^4\right )}{16 \left (a^2-b^2\right )^3 d \sqrt{a+b \sin (c+d x)}}-\frac{\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^2(c+d x) \left (b \left (a^2+9 b^2\right )+2 a \left (3 a^2-8 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.19539, size = 324, normalized size = 1.14 \[ \frac{\frac{3}{2} \left (-7 a^2 b^2+2 a^4-15 b^4\right ) \left ((a+b) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b}\right )+(b-a) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{a+b \sin (c+d x)}{a+b}\right )\right )+3 a \sqrt{a-b} \sqrt{a+b} \left (3 a^2-8 b^2\right ) \sqrt{a+b \sin (c+d x)} \left (\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )-\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )\right )-(a-b) (a+b) \sec ^2(c+d x) \left (2 a \left (3 a^2-8 b^2\right ) \sin (c+d x)+b \left (a^2+9 b^2\right )\right )-4 (a-b)^2 (a+b)^2 \sec ^4(c+d x) (a \sin (c+d x)-b)}{16 d \left (a^2-b^2\right )^2 \left (b^2-a^2\right ) \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

((3*(2*a^4 - 7*a^2*b^2 - 15*b^4)*((a + b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a - b)] + (-a
+ b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a + b)]))/2 - 4*(a - b)^2*(a + b)^2*Sec[c + d*x]^4*
(-b + a*Sin[c + d*x]) + 3*a*Sqrt[a - b]*Sqrt[a + b]*(3*a^2 - 8*b^2)*(Sqrt[a + b]*ArcTanh[Sqrt[a + b*Sin[c + d*
x]]/Sqrt[a - b]] - Sqrt[a - b]*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])*Sqrt[a + b*Sin[c + d*x]] - (a -
b)*(a + b)*Sec[c + d*x]^2*(b*(a^2 + 9*b^2) + 2*a*(3*a^2 - 8*b^2)*Sin[c + d*x]))/(16*(a^2 - b^2)^2*(-a^2 + b^2)
*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B]  time = 0.749, size = 649, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+b*sin(d*x+c))^(3/2),x)

[Out]

-3/16/d*b/(a-b)^3/(b*sin(d*x+c)+b)^2*(a+b*sin(d*x+c))^(3/2)*a+13/32/d*b^2/(a-b)^3/(b*sin(d*x+c)+b)^2*(a+b*sin(
d*x+c))^(3/2)+3/16/d*b/(a-b)^3/(b*sin(d*x+c)+b)^2*(a+b*sin(d*x+c))^(1/2)*a^2-21/32/d*b^2/(a-b)^3/(b*sin(d*x+c)
+b)^2*(a+b*sin(d*x+c))^(1/2)*a+15/32/d*b^3/(a-b)^3/(b*sin(d*x+c)+b)^2*(a+b*sin(d*x+c))^(1/2)+3/8/d/(a-b)^3/(-a
+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a^2-21/16/d*b/(a-b)^3/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c
))^(1/2)/(-a+b)^(1/2))*a+45/32/d*b^2/(a-b)^3/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))+2/d*b^5/
(a-b)^3/(a+b)^3/(a+b*sin(d*x+c))^(1/2)-3/16/d*b/(a+b)^3/(b*sin(d*x+c)-b)^2*(a+b*sin(d*x+c))^(3/2)*a-13/32/d*b^
2/(a+b)^3/(b*sin(d*x+c)-b)^2*(a+b*sin(d*x+c))^(3/2)+3/16/d*b/(a+b)^3/(b*sin(d*x+c)-b)^2*(a+b*sin(d*x+c))^(1/2)
*a^2+21/32/d*b^2/(a+b)^3/(b*sin(d*x+c)-b)^2*(a+b*sin(d*x+c))^(1/2)*a+15/32/d*b^3/(a+b)^3/(b*sin(d*x+c)-b)^2*(a
+b*sin(d*x+c))^(1/2)+3/8/d/(a+b)^(7/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a^2+21/16/d*b/(a+b)^(7/2)*a
rctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a+45/32/d*b^2/(a+b)^(7/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2
))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{5}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*sec(d*x + c)^5/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^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(sec(d*x+c)**5/(a+b*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 1.19322, size = 733, normalized size = 2.58 \begin{align*} \frac{1}{32} \, b^{5}{\left (\frac{3 \,{\left (4 \, a^{2} - 14 \, a b + 15 \, b^{2}\right )} \arctan \left (\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{\sqrt{-a + b}}\right )}{{\left (a^{3} b^{5} d - 3 \, a^{2} b^{6} d + 3 \, a b^{7} d - b^{8} d\right )} \sqrt{-a + b}} - \frac{3 \,{\left (4 \, a^{2} + 14 \, a b + 15 \, b^{2}\right )} \arctan \left (\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{\sqrt{-a - b}}\right )}{{\left (a^{3} b^{5} d + 3 \, a^{2} b^{6} d + 3 \, a b^{7} d + b^{8} d\right )} \sqrt{-a - b}} + \frac{64}{{\left (a^{6} d - 3 \, a^{4} b^{2} d + 3 \, a^{2} b^{4} d - b^{6} d\right )} \sqrt{b \sin \left (d x + c\right ) + a}} - \frac{2 \,{\left (6 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a^{4} - 18 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{5} + 18 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{6} - 6 \, \sqrt{b \sin \left (d x + c\right ) + a} a^{7} - 21 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a^{2} b^{2} + 62 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{3} b^{2} - 71 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{4} b^{2} + 30 \, \sqrt{b \sin \left (d x + c\right ) + a} a^{5} b^{2} - 13 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} b^{4} + 68 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a b^{4} - 76 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{2} b^{4} + 30 \, \sqrt{b \sin \left (d x + c\right ) + a} a^{3} b^{4} + 17 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} b^{6} - 54 \, \sqrt{b \sin \left (d x + c\right ) + a} a b^{6}\right )}}{{\left (a^{6} b^{4} d - 3 \, a^{4} b^{6} d + 3 \, a^{2} b^{8} d - b^{10} d\right )}{\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{2} - 2 \,{\left (b \sin \left (d x + c\right ) + a\right )} a + a^{2} - b^{2}\right )}^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/32*b^5*(3*(4*a^2 - 14*a*b + 15*b^2)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a + b))/((a^3*b^5*d - 3*a^2*b^6*d
+ 3*a*b^7*d - b^8*d)*sqrt(-a + b)) - 3*(4*a^2 + 14*a*b + 15*b^2)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a - b))
/((a^3*b^5*d + 3*a^2*b^6*d + 3*a*b^7*d + b^8*d)*sqrt(-a - b)) + 64/((a^6*d - 3*a^4*b^2*d + 3*a^2*b^4*d - b^6*d
)*sqrt(b*sin(d*x + c) + a)) - 2*(6*(b*sin(d*x + c) + a)^(7/2)*a^4 - 18*(b*sin(d*x + c) + a)^(5/2)*a^5 + 18*(b*
sin(d*x + c) + a)^(3/2)*a^6 - 6*sqrt(b*sin(d*x + c) + a)*a^7 - 21*(b*sin(d*x + c) + a)^(7/2)*a^2*b^2 + 62*(b*s
in(d*x + c) + a)^(5/2)*a^3*b^2 - 71*(b*sin(d*x + c) + a)^(3/2)*a^4*b^2 + 30*sqrt(b*sin(d*x + c) + a)*a^5*b^2 -
13*(b*sin(d*x + c) + a)^(7/2)*b^4 + 68*(b*sin(d*x + c) + a)^(5/2)*a*b^4 - 76*(b*sin(d*x + c) + a)^(3/2)*a^2*b
^4 + 30*sqrt(b*sin(d*x + c) + a)*a^3*b^4 + 17*(b*sin(d*x + c) + a)^(3/2)*b^6 - 54*sqrt(b*sin(d*x + c) + a)*a*b
^6)/((a^6*b^4*d - 3*a^4*b^6*d + 3*a^2*b^8*d - b^10*d)*((b*sin(d*x + c) + a)^2 - 2*(b*sin(d*x + c) + a)*a + a^2
- b^2)^2))