Optimal. Leaf size=391 \[ -\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{7/2} f}+\frac {i (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{7/2} f}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt {a+b \tan (e+f x)}} \]
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Rubi [A]
time = 1.42, antiderivative size = 391, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3648, 3730,
3697, 3696, 95, 214} \begin {gather*} -\frac {4 \left (-2 a^2 d+5 a b c+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))^{3/2}}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 f \left (a^2+b^2\right ) (a+b \tan (e+f x))^{5/2}}+\frac {2 \left (-8 a^4 d^2+50 a^3 b c d-a^2 b^2 \left (45 c^2-49 d^2\right )-70 a b^3 c d+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 f \left (a^2+b^2\right )^3 (b c-a d) \sqrt {a+b \tan (e+f x)}}-\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a-i b)^{7/2}}+\frac {i (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f (a+i b)^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 95
Rule 214
Rule 3648
Rule 3696
Rule 3697
Rule 3730
Rubi steps
\begin {align*} \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{7/2}} \, dx &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {2 \int \frac {\frac {1}{2} \left (-6 b c d-a \left (5 c^2-d^2\right )\right )-\frac {5}{2} \left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)+2 d (b c-a d) \tan ^2(e+f x)}{(a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx}{5 \left (a^2+b^2\right )}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {4 \int \frac {\frac {1}{4} (b c-a d) \left (15 a^2 c^2-15 b^2 c^2+40 a b c d-7 a^2 d^2+3 b^2 d^2\right )-\frac {15}{2} (b c-a d)^2 (a c+b d) \tan (e+f x)-d (b c-a d) \left (5 a b c-2 a^2 d+3 b^2 d\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx}{15 \left (a^2+b^2\right )^2 (b c-a d)}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt {a+b \tan (e+f x)}}-\frac {8 \int \frac {-\frac {15}{8} (b c-a d)^2 \left (6 a^2 b c d-2 b^3 c d+a^3 \left (c^2-d^2\right )-3 a b^2 \left (c^2-d^2\right )\right )+\frac {15}{8} (b c-a d)^2 \left (3 a^2 b c^2-b^3 c^2-2 a^3 c d+6 a b^2 c d-3 a^2 b d^2+b^3 d^2\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{15 \left (a^2+b^2\right )^3 (b c-a d)^2}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^2 \int \frac {1+i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b)^3}+\frac {(c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b)^3}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^2 \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a-i b)^3 f}+\frac {(c+i d)^2 \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 (a+i b)^3 f}\\ &=-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt {a+b \tan (e+f x)}}+\frac {(c-i d)^2 \text {Subst}\left (\int \frac {1}{i a+b-(i c+d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^3 f}+\frac {(c+i d)^2 \text {Subst}\left (\int \frac {1}{-i a+b-(-i c+d) x^2} \, dx,x,\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^3 f}\\ &=-\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{7/2} f}+\frac {i (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{7/2} f}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{5 \left (a^2+b^2\right ) f (a+b \tan (e+f x))^{5/2}}-\frac {4 \left (5 a b c-2 a^2 d+3 b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^{3/2}}+\frac {2 \left (50 a^3 b c d-70 a b^3 c d-8 a^4 d^2-a^2 b^2 \left (45 c^2-49 d^2\right )+3 b^4 \left (5 c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 \left (a^2+b^2\right )^3 (b c-a d) f \sqrt {a+b \tan (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 6.27, size = 498, normalized size = 1.27 \begin {gather*} \frac {b (c+d \tan (e+f x))^{5/2}}{5 (i a-b) (b c-a d) f (a+b \tan (e+f x))^{5/2}}-\frac {b (c+d \tan (e+f x))^{5/2}}{5 (i a+b) (b c-a d) f (a+b \tan (e+f x))^{5/2}}-\frac {\frac {(c+d \tan (e+f x))^{3/2}}{(a-i b) (a+b \tan (e+f x))^{3/2}}+\frac {3 (c-i d) \left (\frac {\sqrt {-c+i d} \tanh ^{-1}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-a+i b)^{3/2}}+\frac {\sqrt {c+d \tan (e+f x)}}{(a-i b) \sqrt {a+b \tan (e+f x)}}\right )}{a-i b}}{3 (i a+b) f}+\frac {\frac {(c+d \tan (e+f x))^{3/2}}{(a+i b) (a+b \tan (e+f x))^{3/2}}-\frac {3 (c+i d) \left (\frac {\sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2}}-\frac {\sqrt {c+d \tan (e+f x)}}{(a+i b) \sqrt {a+b \tan (e+f x)}}\right )}{a+i b}}{3 (i a-b) f} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {7}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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