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

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

[Out]

-3/32*(4*a^2-14*a*b+15*b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(7/2)/d+3/32*(4*a^2+14*a*b+15*b^
2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(7/2)/d-3/16*b*(2*a^4-7*a^2*b^2-15*b^4)/(a^2-b^2)^3/d/(a+
b*sin(d*x+c))^(1/2)-1/4*sec(d*x+c)^4*(b-a*sin(d*x+c))/(a^2-b^2)/d/(a+b*sin(d*x+c))^(1/2)+1/16*sec(d*x+c)^2*(b*
(a^2+9*b^2)+2*a*(3*a^2-8*b^2)*sin(d*x+c))/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.52, antiderivative size = 284, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 2.24, size = 324, normalized size = 1.14 \[ \frac {3 a \sqrt {a-b} \sqrt {a+b} \left (3 a^2-8 b^2\right ) \sqrt {a+b \sin (c+d x)} \left (\sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )-\sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )\right )-(a-b) (a+b) \sec ^2(c+d x) \left (2 a \left (3 a^2-8 b^2\right ) \sin (c+d x)+b \left (a^2+9 b^2\right )\right )+\frac {3}{2} \left (2 a^4-7 a^2 b^2-15 b^4\right ) \left ((a+b) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \sin (c+d x)}{a-b}\right )+(b-a) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \sin (c+d x)}{a+b}\right )\right )-4 (a-b)^2 (a+b)^2 \sec ^4(c+d x) (a \sin (c+d x)-b)}{16 d \left (a^2-b^2\right )^2 \left (b^2-a^2\right ) \sqrt {a+b \sin (c+d x)}} \]

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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fricas [F]  time = 1.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{5}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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]

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

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maple [B]  time = 1.10, size = 649, normalized size = 2.29 \[ \frac {2 b^{5}}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {a +b \sin \left (d x +c \right )}}-\frac {3 b \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a}{16 d \left (a -b \right )^{3} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {13 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{32 d \left (a -b \right )^{3} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {3 b \sqrt {a +b \sin \left (d x +c \right )}\, a^{2}}{16 d \left (a -b \right )^{3} \left (b \sin \left (d x +c \right )+b \right )^{2}}-\frac {21 b^{2} \sqrt {a +b \sin \left (d x +c \right )}\, a}{32 d \left (a -b \right )^{3} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {15 b^{3} \sqrt {a +b \sin \left (d x +c \right )}}{32 d \left (a -b \right )^{3} \left (b \sin \left (d x +c \right )+b \right )^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a^{2}}{8 d \left (a -b \right )^{3} \sqrt {-a +b}}-\frac {21 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{16 d \left (a -b \right )^{3} \sqrt {-a +b}}+\frac {45 b^{2} \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{32 d \left (a -b \right )^{3} \sqrt {-a +b}}-\frac {3 b \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a}{16 d \left (a +b \right )^{3} \left (b \sin \left (d x +c \right )-b \right )^{2}}-\frac {13 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{32 d \left (a +b \right )^{3} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {3 b \sqrt {a +b \sin \left (d x +c \right )}\, a^{2}}{16 d \left (a +b \right )^{3} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {21 b^{2} \sqrt {a +b \sin \left (d x +c \right )}\, a}{32 d \left (a +b \right )^{3} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {15 b^{3} \sqrt {a +b \sin \left (d x +c \right )}}{32 d \left (a +b \right )^{3} \left (b \sin \left (d x +c \right )-b \right )^{2}}+\frac {3 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a^{2}}{8 d \left (a +b \right )^{\frac {7}{2}}}+\frac {21 b \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{16 d \left (a +b \right )^{\frac {7}{2}}}+\frac {45 b^{2} \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{32 d \left (a +b \right )^{\frac {7}{2}}} \]

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]

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)^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))-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)*ar
ctanh((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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for
more details)Is 4*a-4*b positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^5\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + b*sin(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^5*(a + b*sin(c + d*x))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

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]

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

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